1000 Basic Electrical Engineering MCQs - Best For Practice

1000 Basic Electrical Engineering MCQs - Best For Practice 1000 BASIC ELECTRICAL ENGINEERING MCQS Basic Electrical Engineering Questions and Answers – Oscillation of Energy at Resonance « Prev Next » This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Oscillation of Energy at Resonance”. 1. The energy stored in the capacitor is of _________ nature. a) Electrostatic b) Magnetic c) Neither electrostatic nor magnetic d) Either electrostatic or magnetic View Answer Answer: a Explanation: Since capacitor stores charge in between the plates and energy associated with static charge is of electrostatic nature, so we can say energy stored in the capacitor is of electrostatic nature. 2. The energy stored in the inductor is of _________ nature. a) Electrostatic b) Magnetic c) Neither electrostatic nor magnetic d) Either electrostatic or magnetic View Answer Answer: b Explanation: Since inductor stores current which involves moving charge and energy associated with moving charge is of magnetic nature so we can say energy stored in the inductor is of magnetic nature. 3. At resonance, the circuit appears __________ a) Inductive b) Capacitive c) Either inductive or capacitive d) Resistive View Answer Answer: d Explanation: At resonance, the circuit appears resistive because the capacitive and inductive energies are equal to each other. 4. At resonance, the capacitive energy is ___________ inductive energy. a) Greater than b) Less than c) Equal to d) Depends on the circuit View Answer Answer: c Explanation: At resonance, energy stored in the capacitor is equal to energy stored in the inductor because capacitive reactance and inductive reactance are equal at resonance. So, at resonance, capacitive energy is equal to inductive energy. 5. At resonance, electrostatic energy is ___________ the magnetic energy. a) Greater than b) Less than c) Equal to d) Depends on the circuit View Answer Answer: c Explanation: At resonance, energy stored in the capacitor is equal to energy stored in the inductor because capacitive reactance and inductive reactance are equal at resonance. The capacitor stores electrostatic energy and the inductor stores magnetic energy hence they are equal. 6. The maximum magnetic energy stored in an inductor at any instance is? a) E=LIm2/2 b) E=LIm/2 c) E=LIm2 d) E=LIm2*2 View Answer Answer: a Explanation: At any instant, the magnetic energy stored in an inductor is E=LIm2/2, where Im is the maximum current and L is the value of the inductor. 7. The maximum electrostatic energy stored in a capacitor at any instance is? a) CVm2 b) 1/2*CVm2 c) CVm d) CVm/2 View Answer Answer: b Explanation: The maximum electrostatic energy stored in a capacitor at any instance is 1/2*CVm2, where C is the capacitance value and Vm is the peak voltage. 8. Q is the ratio of? a) Active power to reactive power b) Reactive power to active power c) Reactive power to average power d) Reactive power to capacitive power View Answer Answer: c Explanation: Q is the ratio of the reactive power to the average power. The reactive power is due to the inductance or capacitance and the average power is due to the resistance. 9. Find the value of Q if the reactive power is 10W and the average power is 5W. a) 10 b) 5 c) 2 d) 1 View Answer Answer: c Explanation: Q is the ratio of the reactive power to the average power. Q = Reactive power / Average power = 10/5 = 2. 10. Find the reactive power when the average power is 5W and Q=2. a) 10W b) 5W c) 2W d) 1W View Answer Answer: a Explanation: Q is the ratio of the reactive power to the average power. Q = Reactive power / Average power 2 = Reactive power / 5 Reactive Power = 2*5 = 10W. Basic Electrical Engineering Questions and Answers – Bandwidth « Prev Next » This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Bandwidth”. 1. The SI unit for bandwidth is? a) Hz b) Watt c) kHz d) kW View Answer Answer: a Explanation: The SI unit for bandwidth is Hz. Hertz is the SI unit because bandwidth is basically frequency and the unit for frequency is Hz. 2. At bandwidth frequency range, the value of the current I is? a) I=Im/2 b) I=Im2 c) I=Im d) I=Im/√2 View Answer Answer: d Explanation: At the bandwidth frequency range, the value of the current is equal to the maximum value of current divided by √2. 3. At bandwidth frequency range, the value of the voltage V is? a) V=Vm/2 2 b) V=Vm2 c) V=Vm d) V=Vm/√2 View Answer Answer: d Explanation: At the bandwidth frequency range, the value of the voltage is equal to the maximum value of voltage divided by √2. 4. At resonance, bandwidth includes the frequency range that allows _____ percent of the maximum current to flow. a) 33.33 b) 66.67 c) 50 d) 70.7 View Answer Answer: d Explanation: At resonance, bandwidth includes the frequency range that allows 70.2 percent of the maximum current to flow. This is because in the bandwidth frequency range, the value of the current is equal to the maximum value of current divided by √2. 5. At resonance, bandwidth includes the frequency range that allows _____ percent of the maximum voltage to flow. a) 33.33 b) 66.67 c) 50 d) 70.7 View Answer Answer: d Explanation: At resonance, bandwidth includes the frequency range that allows 70.2 percent of the maximum voltage to flow. This is because in the bandwidth frequency range, the value of the voltage is equal to the maximum value of voltage divided by √2. 6. Find the value of current in the bandwidth range when the maximum value of current is 50A. a) 56.65A b) 35.36A c) 45.34A d) 78.76A View Answer Answer: b Explanation: At the bandwidth frequency range, the value of the current is equal to the maximum value of current divided by √2. Hence I =50/√2= 35.36A. 7. Find the value of voltage in the bandwidth range when the maximum value of voltage is 100 V. a) 56.65 V b) 35.36 V c) 45.34 V d) 70.72 V View Answer Answer: d Explanation: At the bandwidth frequency range, the value of the voltage is equal to the maximum value of voltage divided by √2. Hence V =100/√2= 70.72V. 8. Bandwidth is the difference of_____________________ frequencies. a) half power b) full power c) double power d) wattless View Answer Answer: a Explanation: Current for the end frequencies of bandwidth is 1/√2 times the maximum current. So, power at the end frequencies of bandwidth is half the maximum power. So, bandwidth is the difference of half power frequencies. 9. For a sharp resonance, bandwidth is ______________ a) low b) high c) zero d) infinity View Answer Answer: a Explanation: For sharp resonance quality factor is high and the quality factor is inversely proportional to bandwidth so bandwidth is low for sharp resonance. 10. Current is maximum at __________ frequency of bandwidth. a) left end b) middle c) right end d) all end View Answer Answer: b Explanation: Current will be maximum at a frequency which is at the middle of bandwidth. On both sides, it decreases and is 1/√2 times the maximum current at the ends of bandwidth. Basic Electrical Engineering Questions and Answers – Selectivity « Prev Next » This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Selectivity”. 1. Shape of the resonance curve depends upon the? a) Q-factor b) Voltage c) Current d) Either voltage or current View Answer Answer: a Explanation: The shape of the resonance curve depends on the Q factor because of the equation: Q=Resonance frequency / Bandwidth. Sharp resonance means high quality factor. 2. A circuit is said to be selective if it has a _____ peak and ____ bandwidth. a) Blunt, narrow b) Sharp, narrow c) Sharp, broad d) Blunt, broad View Answer Answer: b Explanation: For a circuit to be selective, it should have high quality factor. And we know that for high quality factor, resonance frequency should be high(peak should be sharp) and bandwidth should be narrow. 3. What is the Q factor of a selective circuit? a) Very low b) Very high c) Zero d) Infinity View Answer Answer: b Explanation: For a circuit to be selective, it should have high quality factor. It should have a sharp peak with narrow bandwidth. 4. In selective circuits, higher the Q factor _________ the peak. a) Sharper b) Blunter c) Neither sharper nor blunter d) Either sharper or blunter View Answer Answer: a Explanation: Q=Resonance frequency / Bandwidth. Higher the quality factor, sharper the peak of resonance curve. 5. Q is a measure of _________ a) Resonance b) Bandwidth c) Selectivity d) Either resonance or bandwidth View Answer Answer: c Explanation: For a circuit to be selective, it should have a high quality factor. It should have a sharp peak with narrow bandwidth. 6. In selective circuits, the resonant frequency lies in the ________ of the bandwidth frequency range. a) Beginning b) End c) Midpoint d) Cannot be determined View Answer Answer: c Explanation: In selective circuits, the resonant frequency lies in the midpoint of the bandwidth frequency range. 7. In order for high selectivity, the resistance must be? a) Small b) Large c) Negative d) Positive View Answer Answer: a Explanation: For high selectivity, the Q factor should be large and for Q factor to be large, the resistance would be small because Q is inversely proportional to the resistance. Basic Electrical Engineering Questions and Answers – Voltages in a Series RLC Circuit « Prev Next » This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Voltages in a Series RLC Circuit”. 1. In a series RLC circuit, the phase difference between the voltage across the capacitor and the voltage across the resistor is? a) 00 b) 900 c) 1800 d) 3600 View Answer Answer: b Explanation: In a series RLC circuit, voltage across capacitor lag the current by 900 and voltage across resistor is in phase with current so, the phase difference between the voltage across the capacitor and the voltage across the resistor is 900. 2. In a series RLC circuit, the phase difference between the voltage across the inductor and the voltage across the resistor is? a) 00 b) 900 c) 1800 d) 3600 View Answer Answer: b Explanation: In a series RLC circuit, voltage across inductor lead the current by 900 and voltage across resistor is in phase with current so, the phase difference between the voltage across the inductor and the voltage across the resistor is 900. 3. In a series RLC circuit, the phase difference between the voltage across the capacitor and the voltage across the inductor is? a) 00 b) 900 c) 1800 d) 3600 View Answer Answer: c Explanation: In a series RLC circuit, voltage across inductor lead the current by 900 and voltage across capacitor lag the current by 900 so, the phase difference between the voltage across the inductor and the voltage across the capacitor is 1800. 4. In a series RLC circuit, the phase difference between the voltage across the resistor and the current in the circuit is? a) 00 b) 900 c) 1800 d) 3600 View Answer Answer: a Explanation: In a series RLC circuit, the phase difference between the voltage across the resistor and the current in the circuit is 0 degrees because they are in phase. 5. In a series RLC circuit, the phase difference between the voltage across the capacitor and the current in the circuit is? a) 00 b) 900 c) 1800 0 d) 3600 View Answer Answer: b Explanation: In a series RLC circuit, voltage across capacitor lag the current by 900 so, the phase difference between the voltage across the capacitor and current is 900. 6. In a series RLC circuit, the phase difference between the voltage across the inductor and the current in the circuit is? a) 00 b) 900 c) 1800 d) 3600 View Answer Answer: b Explanation: In a series RLC circuit, voltage across inductor lead the current by 900 so, the phase difference between the voltage across the inductor and the current is 900. 7. The current in the inductor lags the voltage in a series RLC circuit ___________ resonant frequency. a) Above b) Below c) Equal to d) Depends on the circuit View Answer Answer: a Explanation: The current in the inductor lags the voltage in a series RLC circuit if circuit is inductive dominant i.e. if XL XC ωL 1/ωC = ω 1/√LC = ω ω0. So, the current in the inductor lags the voltage in a series RLC circuit above the resonant frequency. 8. The current in the capacitor leads the voltage in a series RLC circuit ___________ resonant frequency. a) Above b) Below c) Equal to d) Depends on the circuit View Answer Answer: b Explanation: The current in the capacitor leads the voltage in a series RLC circuit if circuit is capacitive dominant i.e. if XL XC ωL 1/ωC = ω 1/√LC = ω ω0. So, the current in the capacitor leads the voltage in a series RLC circuit below the resonant frequency. 9. The current in the inductor ___________ the voltage in a series RLC circuit above the resonant frequency. a) Leads b) Lags c) Equal to d) Depends on the circuit View Answer Answer: b Explanation: ω ω0 = ω 1/√LC = ωL 1/ωC = XL XC The circuit is inductive dominant so, the current in the inductor lags the voltage in a series RLC circuit above the resonant frequency. 10. The current in the capacitor ___________ the voltage in a series RLC circuit below the resonant frequency. a) Leads b) Lags c) Equal to d) Depends on the circuit View Answer Answer: a Explanation: ω ω0 = ω 1/√LC = ωL 1/ωC = XL XC The circuit is capacitive dominant so, the current in the capacitor leads the voltage in a series RLC circuit above the resonant frequency. Basic Electrical Engineering Questions and Answers – The Current in a Series RLC Circuit « Prev Next » This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “The Current in a Series RLC Circuit”. 1. In a series RLC circuit, the phase difference between the current in the capacitor and the current in the resistor is? a) 00 b) 900 c) 1800 d) 3600 View Answer Answer: a Explanation: In a series RLC circuit, the phase difference between the current in the capacitor and the current in the resistor is 00 because same current flows in the capacitor as well as the resistor. 2. In a series RLC circuit, the phase difference between the current in the inductor and the current in the resistor is? a) 00 b) 900 c) 1800 d) 3600 View Answer Answer: a Explanation: In a series RLC circuit, the phase difference between the current in the inductor and the current in the resistor is 00 because same current flows in the inductor as well as the resistor. 3. In a series RLC circuit, the phase difference between the current in the capacitor and the current in the inductor is? a) 00 b) 900 c) 1800 d) 3600 View Answer Answer: a Explanation: In a series RLC circuit, the phase difference between the current in the inductor and the current in the capacitor is 00 because same current flows in the inductor as well as the capacitor. 4. In a series RLC circuit, the phase difference between the current in the circuit and the voltage across the resistor is? a) 00 b) 900 c) 1800 d) 3600 View Answer Answer: a Explanation: In a series RLC circuit, the phase difference between the voltage across the resistor and the current in the circuit is 00 because they are in phase. 5. In a series RLC circuit, the phase difference between the current in the circuit and the voltage across the capacitor is? a) 00 b) 900 c) 1800 d) 3600 View Answer Answer: b Explanation: In a series RLC circuit, voltage across capacitor lags the current in the circuit by 900 so, 0 the phase difference between the voltage across the capacitor and the current in the circuit is 900. 6. _________ the resonant frequency, the current in the inductor lags the voltage in a series RLC circuit. a) Above b) Below c) Equal to d) Depends on the circuit View Answer Answer: a Explanation: The current in the inductor lags the voltage in a series RLC circuit if a circuit is inductive dominant i.e. if XL XC ωL 1/ωC = ω 1/√LC = ω ω0. So, the current in the inductor lags the voltage in a series RLC circuit above the resonant frequency. 7. _________ the resonant frequency, the current in the capacitor leads the voltage in a series RLC circuit. a) Above b) Below c) Equal to d) Depends on the circuit View Answer Answer: b Explanation: The current in the capacitor leads the voltage in a series RLC circuit if circuit is capacitive dominant i.e.i.e. if XL XC ωL 1/ωC = ω 1/√LC = ω ω0. So, the current in the capacitor leads the voltage in a series RLC circuit below the resonant frequency. Basic Electrical Engineering Questions and Answers – Basic AC Parallel Circuits « Prev Next » This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Basic AC Parallel Circuits”. 1. In a parallel circuit, we consider _____________ instead of impedance. a) Resistance b) Capacitance c) Inductance d) Admittance View Answer Answer: d Explanation: In a parallel circuit, we consider admittance instead of impedance, where admittance is the reciprocal of impedance. 2. In a parallel circuit, we consider admittance instead of _________ a) Resistance b) Capacitance c) Inductance d) Impedance View Answer Answer: d Explanation: In a parallel circuit, we consider admittance instead of impedance, where admittance is the reciprocal of impedance. 3. Which, among the following is the correct expression for impedance? a) Z=Y b) Z=1/Y c) Z=Y2 d) Z=1/Y2 View Answer Answer: b Explanation: We know that impedance is the reciprocal of admittance, hence the correct expression for impedance is: Z=1/Y. 4. Which, among the following is the correct expression for admittance? a) Y=Z b) Y=1/Z c) Y=Z2 d) Y=1/Z2 View Answer Answer: b Explanation: We know that admittance is the reciprocal of impedance, hence the correct expression for admittance is: Y=1/Z. 5. What is the unit of admittance? a) ohm b) henry c) farad d) ohm-1 View Answer Answer: d Explanation: The unit for admittance is ohm-1 because the unit of impedance is ohm and admittance is the reciprocal of impedance. 6. As the impedance increases, the admittance ____________ a) Increases b) Decreases c) Remains the same d) Becomes zero View Answer Answer: b Explanation: As the impedance increases, the admittance decreases because admittance is equal to 1/impedance. 7. if the impedance of a system is 4 ohm, calculate its admittance. a) 0.25 ohm-1 b) 4 ohm-1 c) 25 ohm-1 d) 0.4 ohm-1 View Answer Answer: a Explanation: We know that: Y=1/Z. Substituting the value of Z from the question, we get Y = 1/4 = 0.25 = Y= 0.25 ohm-1. 8. The admittance of a system is 10 ohm-1, calculate its impedance. a) 10 ohm b) 0.1 ohm c) 1 ohm d) 1.1 ohm View Answer Answer: b Explanation: We know that: Z=1/Y. Z = 1/10 = 0.1 = Z = 0.1 ohm. 9. In A parallel circuit, with any number of impedances, The voltage across each impedance is? a) equal b) divided equally c) divided proportionaly d) zero View Answer Answer: a Explanation: In parallel circuits, the current across the circuits vary whereas the voltage remains the same. So, voltage across each impedance is equal in parallel circuit. 10. In a parallel circuit, current in each impedance is_____________ a) equal b) different c) zero d) infinite View Answer Answer: b Explanation: In parallel circuits, the current across the circuits vary whereas the voltage remains the same. So, current in each impedance is different. Basic Electrical Engineering Questions and Answers – Simple Parallel Circuits « Prev Next » This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Simple Parallel Circuits”. 1. From the given circuit, find the value of IR. a) 0 b) V/I c) V/R d) Cannot be determined View Answer Answer: c Explanation: In the given circuit, the voltage across the resistor is the same as the source voltage as they are connected in parallel. The current in the resistor is IR hence IR=V/R. 2. What is the relation between IR and V in the following circuit? a) IR leads V b) IR lags V c) IR and V are in phase d) No relation View Answer Answer: c Explanation: In the following circuit IR and V are in phase because IR is the current in the resistor and the current in the resistor is always in phase with the voltage across it. 3. What is the expression for the current in the inductor from the following circuit? a) V/I b) V/XL c) 0 d) Cannot be determined View Answer Answer: b Explanation: In the given circuit, the voltage across the inductor is the same as the source voltage as they are connected in parallel. The current in the inductor is IL hence IL=V/XL. 4. What is the phase relation between IL and V from the following circuit? a) IL lags V b) IL leads V c) IL and V are in phase d) No relation View Answer Answer: a Explanation: IL is the current across the inductor and we know that the current across the inductor always lags the voltage across it. Hence IL lags V. 5. Find the expression for the current I from the given circuit. a) I=IC b) I=IR c) I=IC+IR d) I=0 View Answer Answer: c Explanation: I is the total current in the circuit. Since this is a parallel connection, the total current in the circuit is equal to the sum of the currents in each branch of the circuit. Hence I=IC+IR. 6. Find the total current if IC=2A and IR=5A. a) 3A b) -3A c) 7A d) 10A View Answer Answer: c Explanation: I is the total current in the circuit. Since this is a parallel connection, the total current in the circuit is equal to the sum of the currents in each branch of the circuit. Hence I=IC+IR. I=2+5=7A. 7. Find the value of IR if I=10A and IC=8A. a) 5A b) 18A c) 12A d) 2A View Answer Answer: d Explanation: I is the total current in the circuit. Since this is a parallel connection, the total current in the circuit is equal to the sum of the currents in each branch of the circuit. Hence I=IC+IR. 10=8+IR = IR=2A. 8. Find the value of IL if IC=10A and IR=6A. a) 4A b) 18A c) 12A d) 2A View Answer Answer: a Explanation: I is the total current in the circuit. Since this is a parallel connection, the total current in the circuit is equal to the sum of the currents in each branch of the circuit. Hence I=IC+IR. 10=IC+6 = IC=4A. 9. What is the expression for the current in the capacitor from the following circuit? a) V/C b) V/I c) 0 d) V/XC View Answer Answer: d Explanation: In the given circuit, the voltage across the capacitor is the same as the source voltage as they are connected in parallel. The current in the capacitor is IC hence IC=V/XC. 10. What is the phase relation between IC and V from the following circuit? a) IC lags V b) IC leads V c) IC and V are in phase d) No relation View Answer Answer: b Explanation: IC is the current across the capacitor and we know that the current across the capacitor always leads the voltage across it. Hence IC leads V. Basic Electrical Engineering Questions and Answers – Parallel Impedance Circuits « Prev This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Parallel Impedance Circuits”. 1. In an impedance parallel network, the reactive component will ____________ the voltage by 90 degrees. a) Lead b) Lag c) Either lead or lag d) Depends on the circuit View Answer Answer: c Explanation: In an impedance parallel network the reactive component will either lead or lag the voltage by 90 degrees. 2. In an impedance parallel network, the reactive component will either lead or lag the voltage by _________ degrees. a) 0 b) 90 c) 45 d) 180 View Answer Answer: b Explanation: In an impedance parallel network the reactive component will either lead or lag the voltage by 90 degrees. 3. In an impedance parallel network, the reactive component will either lead or lag the ________ by 90 degrees. a) Voltage b) Current c) Either voltage or current d) Cannot be determined View Answer Answer: a Explanation: In an impedance parallel network the reactive component will either lead or lag the voltage by 90 degrees. 4. The reactive component in an impedance parallel circuit leads the voltage when the current _________ the voltage. a) Leads b) Lags c) Either leads or lags d) Cannot be determined View Answer Answer: a Explanation: The reactive component in an impedance parallel circuit leads the voltage when the current leads the voltage. 5. The active component in an impedance parallel circuit will __________ the voltage. a) Leads b) Lags c) Be in phase with d) Either leads or lags View Answer Answer: c Explanation: The active component in an impedance parallel network will always be in phase with the voltage in the circuit. 6. The phase difference between the active component of an impedance parallel circuit and the voltage in the network is __________ a) 0 b) 90 c) 180 d) 360 View Answer Answer: a Explanation: The active component in an impedance parallel network will always be in phase with the voltage in the circuit. Hence the phase difference is 0. 7. The quadrature component is also known as? a) Active component b) Reactive component c) Either active or reactive component d) Neither active nor reactive component View Answer Answer: b Explanation: The quadrature component is also known as the reactive component because the reactive component forms a quadrature with the voltage. 8. Find the expression for the current I from the given circuit. a) I=IL b) I=IR c) I=IL+IR d) I=0 View Answer Answer: c Explanation: I is the total current in the circuit. Since this is a parallel connection, the total current in the circuit is equal to the sum of the currents in each branch of the circuit. Hence I=IR+IL. 9. Find the value of IR if I=10A and IL=8A. a) 5A b) 18A c) 12A d) 2A View Answer Answer: d Explanation: We know that I=IR+IL. 10=IR+8 = IR=2A. 10. Find the total current if IL=2A and IR=8A. a) 3A b) -3A c) 7A d) 10A View Answer Answer: d Explanation: We know that I=IR+IL. I=8+2=10A. Basic Electrical Engineering Questions and Answers – Series Circuits Next » This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Series Circuits”. 1. Find the current in the circuit. a) 1 A b) 2 A c) 3 A d) 4 A View Answer Answer: b Explanation: I=V/R. Total resistance R = 20+40=60ohm. V=120V. I=120/60=2A. 2. In a series circuit, which of the parameters remain constant across all circuit elements such as resistor, capacitor and inductor etcetera? a) Voltage b) Current c) Both voltage and current d) Neither voltage nor current View Answer Answer: b Explanation: In a series circuit, the current across all elements remain the same and the total voltage of the circuit is the sum of the voltages across all the elements. 3. Voltage across the 60ohm resistor is______ a) 72V b) 0V c) 48V d) 120V View Answer Answer: b Explanation: The 60ohm resistance is shorted since current always choses the low resistance path. Voltage across short circuit is equal to zero, hence voltage across the resistor is 0. 4. Find the voltage across the 6 ohm resistor. a) 150V b) 181.6V c) 27.27V d) 54.48V View Answer Answer: c Explanation: Total current I=150/(6+12+15)=(150/33)V. V across 6 ohm = 6*I = 6*(150/33)V = 27.27V. 5. If there are two bulbs connected in series and one blows out, what happens to the other bulb? a) The other bulb continues to glow with the same brightness b) The other bulb stops glowing c) The other bulb glows with increased brightness d) The other bulb also burns out View Answer Answer: b Explanation: Since the two bulbs are connected in series, if the first bulb burns out there is a break in the circuit and hence the second bulb does not glow. 6. What is the value of x if the current in the circuit is 5A? a) 15 ohm b) 25 ohm c) 55 ohm d) 75 ohm View Answer Answer: a Explanation: Total voltage=sum of voltages across each resistor. =150=10*5+5*5+5*x. Solving the equation, we get x=15 ohm. 7. A voltage across a series resistor circuit is proportional to? a) The amount of time the circuit was on for b) The value of the resistance itself c) The value of the other resistances in the circuit d) The power in the circuit View Answer Answer: b Explanation: V=IR hence the voltage across a series resistor circuit is proportional to the value of the resistance. 8. Many resistors connected in series will? a) Divide the voltage proportionally among all the resistors b) Divide the current proportionally c) Increase the source voltage in proportion to the values of the resistors d) Reduce the power to zero View Answer Answer: a Explanation: In a series circuit, the current remains the same across all resistors hence the voltage divides proportionally among all resistors. 9. What is the voltage measured across a series short? a) Infinite b) Zero c) The value of the source voltage d) Null View Answer Answer: b Explanation: A short is just a wire. The potential difference between two points of a wire is zero hence the voltage measured is equal to zero. 10. What happens to the current in the series circuit if the resistance is doubled? a) It becomes half its original value b) It becomes double its original value c) It becomes zero d) It becomes infinity View Answer Answer: a Explanation: I=V/R. If R becomes 2R then I becomes I/2 i.e. half of its original value. Basic Electrical Engineering Questions and Answers – Parallel Networks « Prev Next » This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Parallel Networks”. 1. If two bulbs are connected in parallel and one bulb blows out, what happens to the other bulb? a) The other bulb blows out as well b) The other bulb continues to glow with the same brightness c) The other bulb glows with increased brightness d) The other bulb stops glowing View Answer Answer: b Explanation: If one bulb blows out, it acts as an open circuit. Current does not flow in that branch but it continues to flow in the other branch of the parallel circuit. Hence the other bulb continues to glow. Also the voltage across other bulb remains the same due to which power delivered to it remains the same so it continues to glow with the same brightness. 2. Calculate the current across the 20 ohm resistor. a) 10A b) 20A c) 6.67A d) 36.67A View Answer Answer: a Explanation: I=V/R. Since in parallel circuit, voltage is same across all resistors. Hence across the 20 ohm resistor, V=200V so I=200/20=10A. 3. In a parallel circuit, with a number of resistors, the voltage across each resistor is ________ a) The same for all resistors b) Is divided equally among all resistors c) Is divided proportionally across all resistors d) Is zero for all resistors View Answer Answer: a Explanation: In parallel circuits, the current across the circuits vary whereas the voltage remains the same. 4. The current in each branch of a parallel circuit is proportional to _________ a) The amount of time the circuit is on for b) Proportional to the value of the resistors c) Equal in all branches d) Proportional to the power in the circuit View Answer Answer: b Explanation: I=V/R. In a parallel circuit, the voltage across each resistor is equal, hence the value of the current is proportional (inversely) to the value of the resistance. 5. Calculate the total current in the circuit. a) 20 A b) 10 A c) 11.43 A d) 15 A View Answer Answer: c Explanation: The 1 ohm and 2 ohm resistor are in series which is in parallel to the 3 ohm resistor. The equivalent of these resistances (3/2 ohm) is in series with the 4 ohm and 5 ohm resistor. Total R = 21/2 ohm. I=V/R=120/(21/2)=240/21=11.43 A. 6. The voltage across the open circuit is? a) 100V b) Infinity c) 90V d) 0V View Answer Answer: a Explanation: The voltage across all branches in a parallel circuit is the same as that of the source voltage. Hence the voltage across the 10 ohm resistor and the open circuit is the same=100V. 7. The voltage across the short is? a) 135V b) Infinity c) Zero d) 11.25V View Answer Answer: c Explanation: The voltage across a short is always equal to zero whether it is connected in series or parallel. 8. If the current through x ohm resistance in the circuit is 5A, find the value of x. a) 27 ohm b) 5 ohm c) 12 ohm d) 135 ohm View Answer Answer: a Explanation: R=V/I. In this circuit I=5A and V=135V. Therefore, R=135/5=27 ohm. 9. The currents in the three branches of a parallel circuit are 3A, 4A and 5A. What is the current leaving it? a) 0A b) Insufficient data provided c) The largest one among the three values d) 12A View Answer Answer: d Explanation: The total current leaving a node is the same as the current that enters it. Total I=I1+I2+I3=3+4+5=12A. 10. The total resistance between A and B are? a) 20 ohm b) 5 ohm c) 80 ohm d) 0 ohm View Answer Answer: b Explanation: The resistors are connected in parallel, hence the equivalent resistance = 1/(1/20+1/20+1/20+1/20)=5A. Basic Electrical Engineering Questions and Answers – Series Circuits and Parallel Networks « Prev Next » This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Series Circuits and Parallel Networks”. 1. It is preferable to connect bulbs in series or in parallel? a) Series b) Parallel c) Both series and parallel d) Neither series nor parallel View Answer Answer: b Explanation: Bulbs are connected in parallel so that even if one of the bulbs blow out, the others continue to get a current supply. 2. Calculate the total resistance between the points A and B. a) 7 ohm b) 0 ohm c) 7.67 ohm d) 0.48 ohm View Answer Answer: c Explanation: 1 ohm in parallel with 2 ohm give 2/3 ohm equivalent which is in series with 4 ohm and 3 ohm so total resistance between A and B = 4 + 2/3 + 3 = 23/3 = 7.67 ohm. 3. Calculate the equivalent resistance between A and B. a) 60 ohm b) 15 ohm c) 12 ohm d) 48 ohm View Answer Answer: c Explanation: 5 ohm and 15 ohm are connected in series to give 20 ohm.10ohm and 20 ohm are connected in series to give 30 ohm. Now both equivalent resistances (20ohm and 30 ohm) are in parallel to give equivalent resistance 20*30/(20+30) = 12 ohm. 4. Calculate the resistance between A and B. a) 3.56 ohm b) 7 ohm c) 14.26 ohm d) 29.69 ohm View Answer Answer: a Explanation: The 1 ohm, 2 ohm and 3 ohm resistors are connected in parallel. Its equivalent resistance is in series with the 4 ohm resistor and the parallel connection of the 5 ohm and 6 ohm resistor. The equivalent resistance of this combination is 80/11 ohm. This is in parallel with 7 ohm to give equivalent resistance between A and B is 3.56 ohm. 5. Batteries are generally connected in______ a) Series b) Parallel c) Either series or parallel d) Neither series nor parallel View Answer Answer: a Explanation: Batteries are generally connected in series so that we can obtain the desired voltage since voltages add up once they are connected in series. 6. In a _________ circuit, the total resistance is greater than the largest resistance in the circuit. a) Series b) Parallel c) Either series or parallel d) Neither series nor parallel View Answer Answer: a Explanation: In series circuits, the total resistance is the sum of all the resistance in the circuit, hence the total is greater than the largest resistance. 7. In a ____________ circuit, the total resistance is smaller than the smallest resistance in the circuit. a) Series b) Parallel c) Either series or parallel d) Neither series nor parallel View Answer Answer: b Explanation: in a parallel circuit, the equivalent resistance=1/sum of the reciprocals of all the resistances in the circuit. Hence it is smaller than the smallest resistance in the circuit. 8. Which is the most cost efficient connection? a) Series b) Parallel c) Either series or parallel d) Neither series nor parallel View Answer Answer: a Explanation: The advantage of series-connections is that they share the supply voltage, hence cheap low voltage appliances may be used. 9. Calculate the equivalent resistance between A and B. a) 2 ohm b) 4 ohm c) 6 ohm d) 8 ohm View Answer Answer: b Explanation: R=((2+3)||5)+1.5)||4. The 2 and the 3 ohm resistor are in series. The equivalent of these two resistors is in parallel with the 5 ohm resistor. The equivalent of these three resistances is in series with the 1.5 ohm resistor. Finally, the equivalent of these resistances is in parallel with the 4 ohm resistor. 10. Calculate the equivalent resistance between A and B. a) 6.67 ohm b) 46.67 ohm c) 26.67 ohm d) 10.67 ohm View Answer Answer: a Explanation: R=20||20||20=6.67 ohm. The three 20 ohm resistors are in parallel and re-sistance is measured across this terminal. Basic Electrical Engineering Questions and Answers – Kirchhoff’s Current Law « Prev Next » This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Kirchhoff’s Current Law”. 1. Find the value of v if v1=20V and value of current source is 6A. a) 10V b) 12V c) 14V d) 16V View Answer Answer: b Explanation: The current through the 10 ohm resistor=v1/10=2A.Applying KCL at node 1: i5=i10+i2. i2=6-2=4A. Thus the drop in the 2 ohm resistor = 4×2 = 8V. v1=20V; hence v2=20-v across 2 ohm resistor=20-8=12V v2=v since they are connected in parallel. v=12V. 2. Calculate the current A. a) 5A b) 10A c) 15A d) 20A View Answer Answer: c Explanation: KCl states that the total current leaving the junction is equal to the current entering it. In this case, the current entering the junction is 5A+10A=15A. 3. Calculate the current across the 20 ohm resistor. a) 20A b) 1A c) 0.67A d) 0.33A View Answer Answer: d Explanation: Assume lower terminal of 20 ohm at 0V and upper terminal at V volt and applying KCL, we get V/10 +V/20 =1. V=20/3V So current through 20 ohm = V/20 = (20/3)/20 =1/3=0.33V. 4. Calculate the value of I3, if I1= 2A and I2=3A. a) -5A b) 5A c) 1A d) -1A View Answer Answer: a Explanation: According to KCl, I1+I2+I3=0. Hence I3=-(I1+I2)=-5A. 5. Find the value of i2, i4 and i5 if i1=3A, i3=1A and i6=1A. a) 2,-1,2 b) 4,-2,4 c) 2,1,2 d) 4,2,4 View Answer Answer: a Explanation: At junction a: i1-i3-i2=0. i2=2A. At junction b: i4+i2-i6=0. i4=-1A. At junction c: i3-i5-i4=0. i5=2A. 6. What is the value of current if a 50C charge flows in a conductor over a period of 5 seconds? a) 5A b) 10A c) 15A d) 20A View Answer Answer: b Explanation: Current=Charge/Time. Here charge = 50c and time = 5seconds, so current = 50/5 = 10A. 7. KCL deals with the conservation of? a) Momentum b) Mass c) Potential Energy d) Charge View Answer Answer: d Explanation: KCL states that the amount of charge entering a junction is equal to the amount of charge leaving it, hence it is the conservation of charge. 8. KCL is applied at _________ a) Loop b) Node c) Both loop and node d) Neither loop nor node View Answer Answer: b Explanation: KCL states that the amount of charge leaving a node is equal to the amount of charge entering it, hence it is applied at nodes. 9. KCL can be applied for __________ a) Planar networks b) Non-planar networks c) Both planar and non-planar d) Neither planar nor non-planar View Answer Answer: c Explanation: KCL is applied for different nodes of a network whether it is planar or non-planar. 10. What is the value of the current I? a) 8A b) 7A c) 6A d) 5A View Answer Answer: a Explanation: At the junction, I-2+3-4-5=0. Hence I=8A. Basic Electrical Engineering Questions and Answers – Kirchhoff’s Voltage Law « Prev Next » This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Kirchhoff’s Voltage Law”. 1. Calculate the value of V1 and V2. a) 4V, 6V b) 5V, 6V c) 6V, 7V d) 7V, 8V View Answer Answer: a Explanation: Using KVL, 12-V1-8=0. V1= 4V. 8-V2-2=0. V2=6V. 2. KVL deals with the conservation of? a) Mass b) Momentum c) Charge d) Energy View Answer Answer: d Explanation: KVL states that the sum of the potential energy and taken with the right sign is equal to zero, hence it is the conservation of energy since energy doesn’t enter or leave the system. 3. Calculate the voltage across the 10 ohm resistor. a) 12V b) 4V c) 10V d) 0V View Answer Answer: b Explanation: Total resistance = 5+10+15 = 30 ohm. Current in the circuit is 12/30 A. Voltage across 10 ohm resistor is 10*(12/30) = 4V. 4. Find the value of the currents I1 and I2. a) 0.3, 0.1 b) -0.1, -0.3 c) -0.3, -0.1 d) 0.1, 0.2 View Answer Answer: d Explanation: Using KVL in loop 1, 10-100 i1=0. i1=0.1A Using KVL in outer loop, -100i2+20=0 i2=0.2A. 5. The sum of the voltages over any closed loop is equal to __________ a) 0V b) Infinity c) 1V d) 2V View Answer Answer: a Explanation: According to KVL, the sum of the voltage over any closed loop is equal to 0. 6. What is the basic law that has to be followed in order to analyze the circuit? a) Newton’s laws b) Faraday’s laws c) Ampere’s laws d) Kirchhoff’s law View Answer Answer: d Explanation: Kirchhoff’s laws, namely Kirchhoff’s Current Law and Kirchhoff’s Voltage law are the basic laws in order to analyze a circuit. 7. Every____________ is a ____________ but every __________ is not a __________ a) Mesh, loop, loop, mesh b) Loop, mesh, mesh, loop c) Loop, mesh, loop, mesh d) Mesh, loop, mesh, loop View Answer Answer: a Explanation: According to Kirchhoff’s Voltage Law, Every mesh is a loop but every loop is not a mesh. Mesh is a special case of loop which is planar. 8. What is the voltage across the 5 ohm resistor if current source has current of 17/3 A? a) 2.32V b) 5.21V c) 6.67V d) 8.96V View Answer Answer: b Explanation: Assuming i1 and i2 be the currents in loop 1 and 2 respectively. In loop 1, 4+2i1+3(i117/3)+4(i1-i2)+5=0 In loop 2, i2(4+1+5)-4i1-5=0 =-4i1+10i2=5. Solving these equations simultaneously i2=1.041A and i1=1.352A V=i2*5= 5.21V. 9. Calculate VAB. a) 3.5V b) 12V c) 9.5V d) 6.5V View Answer Answer: a Explanation: For branch A: VAC=15*20/(25+15)=7.5V For branch B: VBC= 10*20/(10+40)=4V Applying KVL to loop ABC: VAB+VBC+VCA=0 VAB=3.5V. 10. KVL is applied in ____________ a) Mesh analysis b) Nodal analysis c) Both mesh and nodal d) Neither mesh nor nodal View Answer Answer: a Explanation: Mesh analysis helps us to utilize the different voltages in the circuit as well as the IR products in the circuit which is nothing but KVL Basic Electrical Engineering Questions and Answers – Power « Prev Next » This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Power”. 1. Which of the following is not an expression power? a) P=VI b) P=I2R c) P=V2/R d) P=I/R View Answer Answer d Explanation: Power is the product of voltage and current. Writing I in terms of V, we get P=V2/R and writing V in terms of I, we get P=I2r. 2. Which of the following statements are true? a) Power is proportional to voltage only b) Power is proportional to current only c) Power is neither proportional to voltage nor to the current d) Power is proportional to both the voltage and current View Answer Answer: d Explanation: Power is proportional to both voltage and current. 3. A 250V bulb passes a current of 0.3A. Calculate the power in the lamp. a) 75W b) 50W c) 25W d) 90W View Answer Answer: a Explanation: Here, V = 250v and I = 0.3A. P=VI. Which implies that, P=250*0.3=75W. 4. Kilowatt-hour(kWh) is a unit of? a) Current b) Power c) Energy d) Resistance View Answer Answer: c Explanation: Power is the energy per unit time. That is, P=E/t. If the unit of power in kW and the unit of time is an hour, then the unit of energy=unit of power*unit of time=kWh. 5. Calculate the power in the 20 ohm resistance. a) 2000kW b) 2kW c) 200kW d) 2W View Answer Answer: b Explanation: Here V = 200v and Resistance( R) = 20ohm. P=V2/R= 2002/20=2000W=2kW. 6. A current of 5A flows in a resistor of 2 ohms. Calculate the energy dissipated in 300 seconds in the resistor. a) 15kJ b) 15000kJ c) 1500J d) 15J View Answer Answer: a Explanation: P=I2R =52*2=50W. E= Pt=50*300=15000J=15kJ. 7. Calculate the power across each 20 ohm resistance. a) 1000W, 1000W b) 500W, 500W c) 1000kW, 1000kW d) 500kW, 500kW View Answer Answer: b Explanation: This is a series connected circuit hence the current across each resistance is the same. To find current: I=V/R=200/20=5A. 2 2 To find power: P=I2R=52*20=500W. Since both the resistors have a resistance of 20 ohms, the power across both is the same. 8. Calculate the power across each 10 ohm resistance. a) 1000kW, 1000kW b) 1kW, 1kW c) 100W, 100W d) 100kW, 100kW View Answer Answer: b Explanation: This is parallel connected circuit, hence the voltage across each of the resistors is the same. P =(V2)/R=(1002)/10 = 1000W=1kW. Since both the resistors receive the same amount of voltage, the power in both is the same. 9. Calculate the work done in a resistor of 20 ohm carrying 5A of current in 3 hours. a) 1.5J b) 15J c) 1.5kWh d) 15kWh View Answer Answer: c Explanation: To find power: P=I2R=52*20=500W=0.5kW. To find Work done: W=Pt=0.5*3=1.5kWh. 10. The SI unit of power is? a) kW(kilo-watt) b) J/s(joules per second) c) Ws(watt-second) d) J/h(joules per hour View Answer Answer: b Explanation: Power = energy/time SI unit of power = SI unit of energy/SI unit of time = joule/second. Basic Electrical Engineering Questions and Answers – Energy « Prev Next » This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Energy”. 1. Which among the following is a unit for electrical energy? a) V(volt) b) kWh(kilowatt-hour) c) Ohm d) C(coloumb) View Answer Answer: b Explanation: Kilowatt is a unit of power and hour is a unit of time. Energy is the product of power and time, hence the unit for Energy is kWh. 2. A bulb has a power of 200W. What is the energy dissipated by it in 5 minutes? a) 60J b) 1000J c) 60kJ d) 1kJ View Answer Answer: c Explanation: Here, Power = 200w and time = 5min. E=Pt = E= 200*5= 1000Wmin=60000Ws= 60000J= 60kJ. 3. Out of the following, which one is not a source of electrical energy? a) Solar cell b) Battery c) Potentiometer d) Generator View Answer Answer: c Explanation: Solar cell converts light energy to electrical energy. Battery converts chemical energy to electrical energy. Generator generates electrical energy using electromagnetic induction. A potentiometer is an instrument used for measuring voltage and consumes electrical energy instead of generating it. 4. Calculate the energy dissipated by the circuit in 50 seconds. a) 50kJ b) 50J c) 100j d) 100kJ View Answer Answer: a Explanation: Here V = 100 and R = 10. Power in the circuit= V2/R = 1002/10 = 1000W. Energy= Pt= 1000*50 = 50000J = 50kJ. 5. Which among the following is an expression for energy? a) V2It b) V2Rt c) V2t/R d) V2t2/R View Answer Answer: c Explanation: Expression for power = VI, substituting I from ohm’s law we can write, P=V2/R. Energy is the product of power and time, hence E=Pt = V2t/R. 6. Calculate the energy in the 10 ohm resistance in 10 seconds. a) 400J b) 40kJ c) 4000J d) 4kJ View Answer Answer: b Explanation: Since the resistors are connected in parallel, the voltage across both the resistors are the same, hence we can use the expression P=V2/R. P=2002/10= 4000W. E=Pt = 4000*10=40000Ws = 40000J = 40kJ. 7. A battery converts___________ a) Electrical energy to chemical energy b) Chemical energy to electrical energy c) Mechanical energy to electrical energy d) Chemical energy to mechanical energy View Answer Answer: b Explanation: A battery is a device in which the chemical elements within the battery react with each other to produce electrical energy. 8. A current of 2A flows in a wire offering a resistance of 10ohm. Calculate the energy dissipated by the wire in 0.5 hours. a) 72Wh b) 72kJ c) 7200J d) 72kJh View Answer Answer b Explanation: Here I (current) = 2A and Resistance(R) = 10ohm. Power = I2R = 22*10=40. Energy = Pt = 40*0.5*60*60 = 72000J=72kJ. 9. Calculate the energy in the 5 ohm resistor in 20 seconds. a) 21.5kJ b) 2.15kJ c) 2.15J d) 21.5kJ View Answer Answer: a Explanation: The current in the circuit is equal to the current in the 5 ohm resistor since it a series connected circuit, hence I=220/(5+10)=14.67A. P=I2R = 14.672*5=1075.8W. E=Pt = 1075.8*20 = 21516J=21.5kJ. 10. Practically, if 10kJ of energy is supplied to a device, how much energy will the device give back? a) Equal to10kJ b) Less than 10kJ c) More than 10kJ d) Zero View Answer Answer: b Explanation: Practically, if 10kJ of energy is supplied to a system, it returns less than the supplied energy because, some of the energy is lost as heat energy, sound energy etc. Basic Electrical Engineering Questions and Answers – Resistivity « Prev Next » This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Resistivity”. 1. Materials which easily allow the passage of electric current are known as ______ a) Insulators b) Conductors c) Dielectrics d) Semi-conductors View Answer Answer: b Explanation: Conductors are materials(mostly metals), which freely allow the passage of electrons through it. If electrons can flow freely through a material, it implies that even current can flow freely through that material since current is the rate of flow of electrons. 2. A wire of length 2m and another wire of length 5m are made up of the same material and have the same area of cross section, which wire has higher resistance? a) Both have equal resistance b) The 2m wire has higher resistance c) The 5m wire has higher resistance d) The value of resistance cannot be determined from the given data View Answer Answer: c Explanation: Resistance is directly proportional to the length of the wire, hence as the length of the wire increases, resistance increases. 3. A wire having an area of cross section = 10sqm and another wire having an area of cross section= 15sqm, have the same length and are made up of the same material. Which wire has more resistance? a) Both have equal resistance b) The 10sqm wire has higher resistance c) The 15sqm wire has higher resistance d) The value of resistance cannot be determined from the given data View Answer Answer: b Explanation: Resistance is inversely proportional to the area of cross-section. As an area of crosssection increases, resistance decreases. Hence the 10sqm wire has a higher resistance than the 15sqm wire. 4. Which of the following statements are true with regard to resistance? a) Resistance is directly proportional to a length of the wire b) Resistance is directly proportional to an area of cross section of the wire c) Resistance is inversely proportional to the length of the wire d) Resistance is inversely proportional to the resistivity of the wire View Answer Answer: a Explanation: The expression for resistance is: Resistance=Resistivity*length of wire/ area of cross section of the wire. Hence resistance is directly proportional to length. 5. A wire has the same resistance as the one given in the figure. Calculate its resistivity if the length of the wire is 10m and its area of cross section is 2m. a) 16 ohm-metre b) 8 ohm-metre c) 16 kiloohm-metre d) 8 kiloohm-metre View Answer Answer: b Explanation: From the given circuit, R=V/I = 200/5 = 40ohm. Resistivity= Resistance*Area of cross section/ Length of the wire. Resistivity= 40*2/10= 8 ohm-metre. 6. Which, among the following is a unit for resistivity? a) ohm/metre b) ohm/metre2 c) ohm-metre d) ohm-metre2 View Answer Answer: c Explanation: Resistivity = Resistance* Length/area of cross section. Unit of resistivity = ohm*(m2)/m = ohm-m. 7. What is the resistivity of Copper? a) 1.59*10-8ohm-m b) 2.7*10-8ohm-m c) 7.3*10-8ohm-m d) 5.35*10-8ohm-m View Answer Answer: a Explanation: Resistivity is a material property. Different materials have different resistivity. Resistivity of copper is 1.72*10-8 ohm-m. 8. Calculate the ratio of the resistivity of 2 wires having the same length and same resistance with area of cross section 2m2 and 5m2 respectively. a) 5:7 b) 2:7 c) 2:5 d) 7:5 View Answer Answer: c Explanation: Resistivity = R*A/L Since resistance and length of the two wires are same so resistivity is directly proportional to area of cross section. Ratio of area is 2:5 so the ratio of resistivity is also 2:5. 9. Which of the following statements are true with regard to resistivity? a) Resistance depends on the temperature b) Resistance does not depend on the temperature c) Resistivity depend on the length d) Resistivity depend on area of cross section View Answer Answer: a Explanation: Resistivity is material property. It depends only on temperature. For the same material with different length and area, resistivity remains the same until temperature remains constant. 10. The reciprocal of resistivity is________ a) Conductance b) Resistance c) Conductivity d) Impedance View Answer Answer: c Explanation: The expression for resistivity is = RA/l. The expression for conductivity = Cl/A; C=1/R = Conductivity = l/(AR) = 1/resistivity. Hence, conductivity is the reciprocal of resistivity. Basic Electrical Engineering Questions and Answers – Temperature Coefficient of Resistance « Prev Next » This set of Basic Electrical Engineering Interview Questions and Answers focuses on “Temperature Coefficient of Resistance”. 1. The resistance of pure metals ___________ a) Increases with an increase in temperature b) Decreases with an increase in temperature c) Remains the same with an increase in temperature d) Becomes zero with an increase in temperature View Answer Answer: a Explanation: In a conductor, the valence band and conduction band overlap each other, there is an excess of electrons in the conduction band. When the temperature increases, there is an overcrowding of electrons in the conduction band hence reducing the mobility and hence resistance increases. 2. The resistance of insulators __________ a) Increases with an increase in temperature b) Decreases with an increase in temperature c) Remains the same with an increase in temperature d) Becomes zero with an increase in temperature View Answer Answer: b Explanation: In the case of an insulator, the energy gap between the conduction band and the valence band is very large. When the temperature is increased, the electrons move from the conduction band to the valence band and hence it starts conducting. When conductance increases, resistance decreases, since C=1/R. Thus, when the temperature increases, resistance decreases in insulators. 3. Which of the following statements are true about metals? a) Metals have a positive temperature coefficient b) Metals have a negative temperature coefficient c) Metals have zero temperature coefficient d) Metals have infinite temperature coefficient View Answer Answer: a Explanation: The resistance of metals increases with an increase in temperature thus, it has a positive temperature coefficient. 4. Which of the following statements are true about insulators? a) Insulators have a positive temperature coefficient b) Insulators have a negative temperature coefficient c) Insulators have zero temperature coefficient d) Insulators have infinite temperature coefficient View Answer Answer: b Explanation: Insulators have a negative temperature coefficient because as temperature increases, the resistance of insulators decreases. 5. What is the unit of temperature coefficient? a) ohm/centigrade b) ohm-centigrade c) centigrade-1 d) centigrade View Answer Answer: c Explanation: R=Reff[1+temp. coeff(T-Teff)]. From the given expression: (R/Reff-1)/(T-Teff) = temp. coeff. Hence, the unit is the reciprocal of that of temperature = centigrade-1. 6. A copper coil has a resistance of 200 ohms when its mean temperature is 0 degree centigrade. Calculate the resistance of the coil when its mean temperature is 80 degree centigrade. Temperature coefficient of copper is 0. centigrade-1 a) 264.65 ohm b) 264.65 kilo-ohm c) 286.65 ohm d) 286.65 kilo-ohm View Answer Answer: a Explanation: R=R0(1+α dT) = 200(1+ 0.*80) = 264.65 ohm. 7. The temperature of a coil cannot be measured by which of the following methods? a) Thermometer b) Increase in resistance of the coil c) Thermo-junctions embedded in the coil d) Calorimeter View Answer Answer: d Explanation: Calorimeter measures the amount of heat and not the temperature of the coil. The temperature of a coil is mainly measured by thermometer. Resistance of coil increase with an increase in temperature of coil so we can measure temperature using this method. Another method is the formation of thermocouple inside coil due to high temperature at one end and low temperature at other ends. 8. The rise or fall in resistance with the rise in temperature depends on ________ a) The property of the conductor material b) The current in the metal c) Property of material as well current in that material d) Does not depend on any factor View Answer Answer: a Explanation: The rise or fall in resistance with a rise in temperature depends upon the property of the material. Hence it rises with temperature in metals and falls with temperature in insulators and semiconductors. 9. If the temperature is increased in semi-conductors such that the resistance incessantly falls, it is termed as _______ a) Avalanche breakdown b) Zener breakdown c) Thermal runway d) Avalanche runway View Answer Answer c Explanation: When the temperature keeps increasing, the resistance keeps falling continuously and hence the current to increase. This causes the heat in the semi-conductor to rise. This causes the temperature to increase further and the resistance to further decrease. This process continues and until there is sufficient heat to destroy the structure of the semi-conductor completely. This is known as a thermal runway. 10. Materials having resistance almost equal to zero is _______ a) Semi-conductor b) Conductor c) Superconductors d) Insulators View Answer Answer: c Explanation: When the temperature of the material falls to absolute zero, the resistance falls to zero and hence there are no I2R losses. Since resistance is zero, conductance is almost infinity and hence these materials are known as superconductors. Basic Electrical Engineering Questions and Answers – Kirchhoff’s Laws and Network Solutions « Prev Next » This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Kirchhoff’s Laws and Network Solutions”. 1. Find the value of I1, I2 and I3. a) -0.566A, 1.29A, -1.91A b) -1.29A, -0.566A, 1.91A c) 1.29A, -0.566A, -1.91A d) 1.91A, 0.566A, 1.29A View Answer Answer: c Explanation: Using the matrix method: Matrix(3,-2,0) (I1)=(5) (-2,9,-4) (I2)=(0) (0,-4,9) (I3)=(-15) Solving this matrix equation, we get I1 = 1.29A, I2 = -0.566A and I3 = -1.91A. 2. Find the value of V, if the value of I3= 0A. a) 1.739 V b) 6.5 V c) 4.5V d)2.739V View Answer Answer: a Explanation: 5-3I1+2I2=0, 9I2-2I1=0, -4I2+V=0 On solving,V=1.739V. 3. Find the value of R if the power in the circuit is 1000W. a) 10 ohm b) 9 ohm c) 8 ohm d) 7 ohm View Answer Answer: c Explanation: To find the value of I: VI=P =100I=1000 = I=10A. Voltage across the 2 ohm resistor = 20V. Voltage across the R resistor = 100-20= 80V. R=V/I = R=80/10 = 8A. 4. Find the current in the 4 ohm resistor. a) 5A b) 0A c) 2.2A d) 20A View Answer Answer: b Explanation: The 4 ohm resistor gets shorted since current always prefers the low resistance path. All the current flows to the branch which is connected in parallel to the 4 ohm branch, hence no current flows in the 4 ohm resistance. 5. Nodal analysis is generally used to determine______ a) Voltage b) Current c) Resistance d) Power View Answer Answer: a Explanation: Nodal analysis uses Kirchhoff’s Current Law to find all the node voltages. Hence it is a method used to determine the voltage. 6. Mesh analysis is generally used to determine_________ a) Voltage b) Current c) Resistance d) Power View Answer Answer: b Explanation: Mesh analysis uses Kirchhoff’s Voltage Law to find all the mesh currents. Hence it is a method used to determine current. 7. What is the current in the circuit? a) 0A b) 15A c) 5A d) 10A View Answer Answer: a Explanation: If we move in the clockwise direction, we get the total voltage to be equal to: -10-20+30 = 0V. Since I=V/R = 0/4=0, I=0A. 8. Does the 15A source have any effect on the circuit? a) Yes b) No c) Cannot be determined d) Yes, only when the 10V source is removed View Answer Answer: b Explanation: The 15A current source has a lower resistance path associated with it and hence it keeps moving in that particular loop. It does not leave that loop and enter the circuit, hence the circuit is not affected by it. 9. KVL is associated with____________ a) Mesh analysis b) Nodal analysis c) Both mesh and nodal d) Neither mesh nor nodal View Answer Answer: a Explanation: KVL employs mesh analysis to find the different mesh currents by finding the IR products in each mesh. 10. KCL is associated with_________ a) Mesh analysis b) Nodal analysis c) Both mesh and nodal d) Neither mesh nor nodal View Answer Answer: b Explanation: KCL employs nodal analysis to find the different node voltages by finding the value if a current in each branch. Sanfoundry Global Education & Learning Series – Basic Electrical Engineering. Basic Electrical Engineering Questions and Answers – Mesh Analysis « Prev Next » This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Mesh Analysis”. 1. Find the value of the currents I1, I2 and I3 flowing clockwise in the first, second and third mesh respectively. a) 1.54A, -0.189A, -1.195A b) 2.34A, -3.53A, -2.23A c) 4.33A, 0.55A, 6.02A d) -1.18A, -1.17A, -1.16A View Answer Answer: a Explanation: The three mesh equations are: -3I1+2I2-5=0 2I1-9I2+4I3=0 4I2-9I3-10=0 Solving the equations, we get I1= 1.54A, I2=-0.189 and I3= -1.195A. 2. Find the value of the currents I1 and I2 flowing clockwise in the first and second mesh respectively. a) 0.96A, 1.73A b) 0.96A, -1.73A c) -0.96A, 1.73A d) -0.96A, -1.73A View Answer Answer: b Explanation: The two mesh equations are: 5I1-3I2=10 -3I1+7I2=-15 Solving the equations simultaneously, we get I1=0.96A and I2=-1.73A. 3. Find the value of V if the current in the 3 ohm resistor=0. a) 3.5V b) 6.5V c) 7.5V d) 8.5V View Answer Answer: c Explanation: Taking the mesh currents in the three meshes as I1, I2 and I3, the mesh equations are: 3I1+0I2+0V=5 -2I1-4I2+0V=0 0I1+9I2+V=0 Solving these equations simultaneously and taking the value of I2=0, we get V=7.5V. 4. Find the value of V1 if the current through the 1 ohm resistor=0A. a) 83.33V b) 78.89V c) 87.87V d) 33.33V View Answer Answer: a Explanation: Taking I1, I2 and I3 as the currents in the three meshes and taking I3=0 since it is the current across the 1 ohm resistor, the three mesh equations are: 15I1-5I2=V1 -5I1+10I2=0 3I2=10 Solving these equations simultaneously we get V1= 83.33V. 5. Calculate the mesh currents I1 and I2 flowing in the first and second meshes respectively. a) 1.75A, 1.25A b) 0.5A, 2.5A c) 2.3A, 0.3A d)