Professional Electrical Engineering Subjects
Test Problems – Day 1
Answers with Solutions
1. At a 115kV substation, the PT ratio is 1000 and the CT ratio is 1200/5. The
potentialgoing into the wattmeter is 115volts. What is the MW indicated when
the wattmeter reads 800 watts?
A. 150 MW B. 192 MW C. 15 MW D. 19.2 MW
Solution:
Given:
V = 115 kV
aCT = 1200/5
aVT = 1000/1
Vw = 115 V
Prd = 800 W
Formula,
PMW = aVT aCT Prd
Substitute the given,
1200
PMW = (1000) ( 5 ) (800 W)
PMW = 192 MW
2. A 36 Ω resistor is connected in parallel with an unknown resistor R. Their
combination is them connected in series with a 12 Ω resistor. Find the value
of R such that the power drawn by the parallel combination of 36 Ω and R is
equal to the power inthe 12 Ω resistor.
A. 6 Ω B. 18 Ω C. 24 Ω D. 48 Ω
Solution:
Given:
R1 = 36 Ω
R 2 = 12 Ω
Formula,
1 1 −1
For parallel circuit, R eq = (R + R)
1
If Peq = P𝟏𝟐𝛀 , then, 𝑹𝒆𝒒 = 𝟏𝟐 𝛀
1 1 −1
12 Ω = (36 Ω + R)
R = 18 Ω
, 3. For two alternators operating in parallel, if the load shared by one of the,
to beincreased, as field excitation is
A. to be strengthened keeping input torque the same
B. to be weakened keeping input torque the same
C. to be kept constant but input torque should be increased
D. to be kept constant but input torque should be decreased
4. The impedance triangle is similar to the triangle with the
resistancephasor in place of the .
A. Current, resistor current C. voltage, impedance
B. Current, resistor voltage D. voltage, resistor voltage
5. What is the IEEE device function number 50?
A. AC time overcurrent relay C. AC circuit breaker
B. instantaneous overcurrent relay D. Differential protective relay
6. In long transmission line, at no-load the receiving end voltage as
compared to the sending end voltage is
I. higher II. lower III. Remains the same
A. I only B. II only
C. Either I or II D. III only
7. A 40μF capacitor is supplied with a voltage of v(t) = 100 sin 377t. What
is the current at t=0?
A. 0 B. 1.51 A C. 3.02 A D. 1.07 A
Solution:
Given:
C = 40 μF
v(t) = 100 sin 377t V
Formula,
dV
i(t) = C
dt
Substitute the given,
dV
= (377)(100 cos 377t) = 37700 cos 377t
dt
i(t) = (40 μF)(37700 cos 377t) = 1.508 cos 377t A
At t=0,
i(0) = 1.508 cos 377(0)
i(0) = 1.508 A
Test Problems – Day 1
Answers with Solutions
1. At a 115kV substation, the PT ratio is 1000 and the CT ratio is 1200/5. The
potentialgoing into the wattmeter is 115volts. What is the MW indicated when
the wattmeter reads 800 watts?
A. 150 MW B. 192 MW C. 15 MW D. 19.2 MW
Solution:
Given:
V = 115 kV
aCT = 1200/5
aVT = 1000/1
Vw = 115 V
Prd = 800 W
Formula,
PMW = aVT aCT Prd
Substitute the given,
1200
PMW = (1000) ( 5 ) (800 W)
PMW = 192 MW
2. A 36 Ω resistor is connected in parallel with an unknown resistor R. Their
combination is them connected in series with a 12 Ω resistor. Find the value
of R such that the power drawn by the parallel combination of 36 Ω and R is
equal to the power inthe 12 Ω resistor.
A. 6 Ω B. 18 Ω C. 24 Ω D. 48 Ω
Solution:
Given:
R1 = 36 Ω
R 2 = 12 Ω
Formula,
1 1 −1
For parallel circuit, R eq = (R + R)
1
If Peq = P𝟏𝟐𝛀 , then, 𝑹𝒆𝒒 = 𝟏𝟐 𝛀
1 1 −1
12 Ω = (36 Ω + R)
R = 18 Ω
, 3. For two alternators operating in parallel, if the load shared by one of the,
to beincreased, as field excitation is
A. to be strengthened keeping input torque the same
B. to be weakened keeping input torque the same
C. to be kept constant but input torque should be increased
D. to be kept constant but input torque should be decreased
4. The impedance triangle is similar to the triangle with the
resistancephasor in place of the .
A. Current, resistor current C. voltage, impedance
B. Current, resistor voltage D. voltage, resistor voltage
5. What is the IEEE device function number 50?
A. AC time overcurrent relay C. AC circuit breaker
B. instantaneous overcurrent relay D. Differential protective relay
6. In long transmission line, at no-load the receiving end voltage as
compared to the sending end voltage is
I. higher II. lower III. Remains the same
A. I only B. II only
C. Either I or II D. III only
7. A 40μF capacitor is supplied with a voltage of v(t) = 100 sin 377t. What
is the current at t=0?
A. 0 B. 1.51 A C. 3.02 A D. 1.07 A
Solution:
Given:
C = 40 μF
v(t) = 100 sin 377t V
Formula,
dV
i(t) = C
dt
Substitute the given,
dV
= (377)(100 cos 377t) = 37700 cos 377t
dt
i(t) = (40 μF)(37700 cos 377t) = 1.508 cos 377t A
At t=0,
i(0) = 1.508 cos 377(0)
i(0) = 1.508 A
