This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Three-Phase Unbalanced Circuits”.
1. If the system is a three-wire system, the currents flowing towards the load in the three lines must add to ___ at any given instant.
a) 1
b) 2
c) 3
d) zero
View Answer
Explanation: If the system is a three-wire system, the currents flowing towards the load in the three lines must add to zero at any given instant.
2. The three impedances Z1 = 20∠30⁰Ω, Z2 = 40∠60⁰Ω, Z3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Determine the phase current IR.
a) (17.32-j10) A
b) (-17.32-j10) A
c) (17.32+j10) A
d) (-17.32+j10) A
View Answer
Explanation: Taking VRY = V∠0⁰ as a reference phasor, and assuming RYB phase sequence, we have VRY = 400∠0⁰V Z1 = 20∠30⁰Ω = (17.32+j10)Ω IR = (400∠0o)/(20∠30o) = (17.32-j10) A.
3. The three impedances Z1 = 20∠30⁰Ω, Z2 = 40∠60⁰Ω, Z3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the phase current IY.
a) (10-j0) A
b) (10+j0) A
c) (-10+j0) A
d) (-10-j0) A
View Answer
Explanation: The voltage VYB is VYB = 400∠-120⁰V. The impedance Z2 is Z2 = 40∠60⁰Ω => IY = (400∠-120o)/(40∠60o)=(-10+j0)A.
4. The three impedances Z1 = 20∠30⁰Ω, Z2 = 40∠60⁰Ω, Z3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the phase current IB.
a) (34.64-j20) A
b) (34.64+j20) A
c) (-34.64+j20) A
d) (-34.64-j20) A
View Answer
Explanation: The voltage VBR is VBR = 400∠-240⁰V. The impedance Z3 is Z3 = 10∠-90⁰Ω => IB = (400∠240o)/(10∠-90o)=(-34.64-j20)A.
5. The three impedances Z1 = 20∠30⁰Ω, Z2 = 40∠60⁰Ω, Z3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the line current I1.
a) (-51.96-j10) A
b) (-51.96+j10) A
c) (51.96+j10) A
d) (51.96+j10) A
View Answer
Explanation: The line current I1 is the difference of IR and IB. So the line current I1 is I1 = IR – IB = (51.96+j10) A.
6. The three impedances Z1 = 20∠30⁰Ω, Z2 = 40∠60⁰Ω, Z3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the line current I2.
a) (-27.32+j10) A
b) (27.32+j10) A
c) (-27.32-j10) A
d) (27.32-j10) A
View Answer
Explanation: The line current I2 is the difference of IY and IR. So the line current I2 is I2 = IY – IR = (-27.32+j10) A.
7. The three impedances Z1 = 20∠30⁰Ω, Z2 = 40∠60⁰Ω, Z3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the line current I3.
a) (24.646+j20) A
b) (-24.646+j20) A
c) (-24.646-j20) A
d) (24.646-j20) A
View Answer
Explanation: The line current I3 is the difference of IB and IY. So the line current I3 is I3 = IB – IY = (-24.646-j20) A.
8. The three impedances Z1 = 20∠30⁰Ω, Z2 = 40∠60⁰Ω, Z3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the power in the R phase.
a) 6628
b) 6728
c) 6828
d) 6928
View Answer
Explanation: The term power is defined as the product of square of current and the impedance. So the power in the R phase = 202 x 17.32 = 6928W.
9. The three impedances Z1 = 20∠30⁰Ω, Z2 = 40∠60⁰Ω, Z3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the power in the Y phase.
a) 1000
b) 2000
c) 3000
d) 4000
View Answer
Explanation: The term power is defined as the product of square of current and the impedance. So the power in the Y phase = 102 x 20 = 2000W.
10. The three impedances Z1 = 20∠30⁰Ω, Z2 = 40∠60⁰Ω, Z3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the power in the B phase.
a) 0
b) 1
c) 3
d) 2
View Answer
Explanation: The term power is defined as the product of square of current and the impedance. So the power in the B phase = 402 x 0 = 0W.
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