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BEE End-Sem
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Unit 03 • Unit 04 • Unit 05 • Unit 06



Unit No:- 03
1. Derive the expression of current drawn and power consumed by a circuit consisting of L connected
across 𝒗 = 𝑽𝒎 𝒔𝒊𝒏 𝝎𝒕 supply. Draw waveforms.

To derive the expression for current drawn and power consumed by a circuit consisting of an inductor (L)
connected across a sinusoidal voltage supply v = Vm sin(ωt), we need to consider the behavior of an
inductor in an AC circuit.

The relationship between the voltage and current in an inductor is given by the following equation:

di
v = L( )
dt
Where:

v is the voltage across the inductor,

L is the inductance of the inductor,

di/dt is the rate of change of current with respect to time.

By rearranging the equation, we can find an expression for the current flowing through the inductor:

di 1
= ( )v
dt L
Taking the derivative of both sides with respect to time:

d 2𝑖 1 dv
2
= ( )
dt L dt



Since v = Vm sin(ωt), we can find the derivative of v with respect to time:

dv
= 𝑉𝑚 ω cos(ωt)
dt
Substituting this back into the previous equation:

d2 i 1
2
= ( ) Vm ω cos(ωt)
dt L
https://msha.ke/btechnotes pg. 1

,Integrating both sides twice with respect to time, we obtain:

𝑉𝑚
i = ( ) sin(ωt) + A
ωL
Where:

i is the current flowing through the inductor,

A is the constant of integration.

The constant A represents the initial condition of the current, which depends on the circuit's initial state.

Now, let's consider the power consumed by the circuit. The instantaneous power consumed by an inductor
can be given by the equation:

p(t) = v(t) × i(t)

Substituting the values of v(t) and i(t):

𝑉𝑚
p(t) = (V𝑚 sin(ωt)) × [( ) sin(ωt) + A]
ωL
Simplifying the equation:
2
V𝑚
p(t) = ( ) sin2 (ωt) + A V𝑚 sin(ωt)
ωL

The average power consumed by the circuit over a complete cycle can be found by taking the time average of
the power equation. Since the average of sin2 (ωt) over a complete cycle is 1/2, and the average of sin(ωt)
over a complete cycle is 0, we can simplify the equation further:
2
Vm
Pavg = ( )
2ωL

Where:

Pavg is the average power consumed by the circuit.

Now, let's draw the waveforms:

1. Voltage waveform:

The voltage waveform across the inductor is given by v(t) = Vm sin(ωt). It will be a sinusoidal waveform,
oscillating between positive and negative values.

2. Current waveform:




https://msha.ke/btechnotes pg. 2

,The current waveform through the inductor is given by i(t) = (Vm/ωL) sin(ωt) + A. It will also be a sinusoidal
waveform, but it may have a phase shift or an initial offset (represented by the constant A) depending on the
initial conditions of the circuit.

3. Power waveform:

V2
m
The power waveform, p(t) = ( ωL ) sin2 (ωt) + A Vm sin(ωt), will have two components. The first
component represents the average power (Pavg ) and will be constant over time. The second component
represents the instantaneous power and will vary sinusoidally with the same frequency as the voltage and
current waveforms.

Note: The exact shape of the waveforms will depend on the specific values of Vm, ω, L, and the initial
conditions of the circuit. The above expressions provide a general understanding of the waveforms for an
inductor connected to a sinusoidal voltage supply

2. Derive the expression of current drawn and power consumed by a circuit consisting of C connected
across 𝒗 = 𝑽𝒎 𝒔𝒊𝒏 𝝎𝒕 supply. Draw waveforms.

To derive the expression of current drawn and power consumed by a circuit consisting of a capacitor (C)
connected across a voltage supply v(t) = Vm sin(ωt), we can analyze the behavior of the capacitor in an AC
circuit.



1. Expression of Current Drawn:

The current flowing through a capacitor in an AC circuit is given by the derivative of the capacitor voltage with
respect to time, multiplied by the capacitance (C):

d
i(t) = C × [v(t)]
dt
Given that v(t) = Vm sin(ωt), we can differentiate it with respect to time:

d d
[v(t)] = Vm × [sin(ωt)] = Vm × ω × cos(ωt)
dt dt
Substituting this into the expression for the current:

i(t) = C × Vm × ω × cos(ωt)

The current waveform is a sinusoidal waveform that leads the voltage waveform by 90 degrees in a capacitor.

2. Power Consumed:

In an AC circuit, the power consumed by a capacitor is given by the product of the voltage across the capacitor
and the current flowing through it:

https://msha.ke/btechnotes pg. 3

, p(t) = v(t) × i(t)

Substituting the expressions for voltage (v(t)) and current (i(t)):

p(t) = Vm × sin(ωt) × C × Vm × ω × cos(ωt)

Simplifying:
2
p(t) = Vm × C × ω × sin(ωt) × cos(ωt)

The power waveform is also a sinusoidal waveform, but it oscillates at twice the frequency of the voltage and
current waveforms.

Now, let's draw the waveforms:



- Voltage waveform (v(t)): A sinusoidal waveform with amplitude Vm , frequency ω, and no phase shift.

- Current waveform (i(t)): A sinusoidal waveform with amplitude Vm × C × ω, frequency ω, and leading the
voltage waveform by 90 degrees.
2
- Power waveform (p(t)): A sinusoidal waveform with amplitude Vm × C × ω, frequency 2ω (twice the
frequency of voltage and current waveforms), and oscillating between positive and negative values.

Note: The above waveforms represent the general behavior of a capacitor connected across an AC voltage
supply. The exact shapes and amplitudes will depend on the specific values of Vm , C, and ω.

3. Derive the expression of current drawn and power consumed by a circuit consisting of R & C connected
across 𝒗 = 𝑽𝒎 𝒔𝒊𝒏 𝝎𝒕 supply. Draw waveforms.

To derive the expression for current drawn and power consumed by a circuit consisting of a resistor (R) and
capacitor (C) connected across a sinusoidal voltage supply v = Vm sin(ωt), we can use the principles of AC
circuit analysis.

1. Current Expression:

The voltage across a resistor in an AC circuit is given by Ohm's law:

v = iR

Where:

v is the voltage across the resistor,

i is the current flowing through the resistor,

R is the resistance.

For a capacitor, the current is related to the rate of change of voltage with respect to time:
https://msha.ke/btechnotes pg. 4