Three-Phase Circuits

Question 1
In the circuit shown below, a three-phase star-connected unbalanced load is connected to a balanced three-phase supply of 100\sqrt{3} with phase sequence ABC. The star connected load has Z_A=10\Omega and Z_B=20\angle 60^{\circ}. The value of Z_C in \Omega, for which the voltage difference across the nodes nand n' is zero, is

A
20\angle -30^{\circ}
B
20\angle 30^{\circ}
C
20\angle -60^{\circ}
D
20\angle 60^{\circ}
GATE EE 2022   Electric Circuits
Question 1 Explanation: 
Given: n & n' are at same potential, therefore,
I_{nn'}=0
I_{A}+I_{B}+I_{C}=0
\frac{E_A}{Z_A}+\frac{E_B}{Z_B}+\frac{E_C}{Z_C}=0
where ,
E_A=100\angle 0^{\circ}V \text{ and }Z_A=10\Omega
E_B=100\angle -120^{\circ}V \text{ and }Z_B=20\angle 60^{\circ}
E_C=100\angle 120^{\circ}V \text{ and }Z_C=?
\therefore \frac{100\angle 0^{\circ}}{10}+\frac{100\angle -120^{\circ}}{20\angle 60^{\circ} }+\frac{100\angle 120^{\circ}}{Z_C}=0
\Rightarrow Z_C=20\angle -60^{\circ} \Omega
Question 2
A three-phase balanced voltage is applied to the load shown. The phase sequence is \text{RYB}. The ratio \frac{\left | I_{B} \right |}{|I_{R}|} is ____________.

A
1
B
2
C
3
D
4
GATE EE 2021   Electric Circuits
Question 2 Explanation: 


\begin{aligned} I_{R} &=\frac{V_{R B}}{-j 10}=\frac{V_{L} \angle-60^{\circ}}{-j 10}=\frac{V_{L}}{10} \angle 30^{\circ} \\ I_{Y} &=\frac{V_{Y B}}{j 10}=\frac{V_{L} \angle-120^{\circ}}{-j 10}=\frac{V_{L}}{10} \angle 150^{\circ} \\ \text { and }\qquad \qquad I_{B}&=-\left(I_{R}+I_{Y}\right)=\frac{V_{L}}{10} \angle-90^{\circ} \\ \therefore \qquad \qquad \left|\frac{I_{B}}{I_{R}}\right|&=1 \end{aligned}
Question 3
In the given network V_{1}=100\angle 0^{\circ}V, V_{2}=-120\angle 0^{\circ} V, V_{3}=+120\angle 0^{\circ} V. The phasor current i (in Ampere) is
A
173.2\angle -60^{\circ}
B
173.2\angle 120^{\circ}
C
100.0\angle -60^{\circ}
D
100.0\angle 120^{\circ}
GATE EE 2015-SET-2   Electric Circuits
Question 3 Explanation: 
'i' is taken as input,

Apply KCL at node
Node voltage is 100\angle 120^{\circ}
i=\frac{100\angle 120^{\circ}-100\angle 0^{\circ}}{-j1} +\frac{100\angle 120^{\circ}-100\angle -120^{\circ}}{j1}
i=173.2\angle -60^{\circ}A
Question 4
The line A to neutral voltage is 10 \angle 15^{\circ}V for a balanced three phase star connected load with phase sequence ABC . The voltage of line B with respect to line C is given by
A
10 \sqrt{3}\angle 105^{\circ} V
B
10 \angle 105^{\circ} V
C
10 \sqrt{3}\angle -75^{\circ} V
D
-10 \sqrt{3}\angle 90^{\circ} V
GATE EE 2014-SET-3   Electric Circuits
Question 4 Explanation: 
Given,
V_{AN}=10\angle 15^{\circ}volt
As the system is balanced and phase sequence is ABC, therefore,
V_{AN}=10\angle 15^{\circ}
V_{BN}=10\angle 135^{\circ}V and V_{CN}=10\angle 255^{\circ}V
\therefore voltage of line w.r.t. line C is
V_{BC}=V_{BN}-V_{CN}
V_{BC}=10\angle 255^{\circ}-10\angle 135^{\circ}
V_{BC}=10\sqrt{3}\angle -75^{\circ} Volt
Question 5
A two-phase load draws the following phase currents : i_{1}(t)=I_{m}sin(\omega t-\phi _{1}), i_{2}(t)=I_{m}cos(\omega t-\phi _{2}). These currents are balanced if \phi _{1} is equal to.
A
-\phi _{2}
B
\phi _{2}
C
(\pi/2-\phi _{2})
D
(\pi/2+\phi _{2})
GATE EE 2012   Electric Circuits
Question 5 Explanation: 
In two phase, current are balanced if phase difference is 90^{\circ}.
i_1(t)=I_m \sin (\omega t-\phi _1)
i_2(t)=I_m \sin (\omega t-\phi _2+90^{\circ})
Phase difference is 90^{\circ},
\omega t-\phi _2+90^{\circ}-(\omega t-\phi _1)=90^{\circ}
\phi _1-\phi _2=0
\phi _1=\phi _2
Question 6
For the three-phase circuit shown in the figure, the ratio of the currents I_{R}: I_{Y}: I_{B} is given by
A
1:1:\sqrt{3}
B
1:1:2
C
1:1:0
D
1:1:\sqrt{3/2}
GATE EE 2005   Electric Circuits
Question 6 Explanation: 


Assuming phase sequence to be RYB
Taking V_{RY} as the reference,
V_{RY}=V\angle 0^{\circ}
V_{YB}=V\angle -120^{\circ}
V_{BR}=V\angle -240^{\circ}
I_R=\frac{V_{RB}}{R_1}=-\frac{V_{BR}}{R_1} =-\frac{V\angle -240^{\circ}}{R_1}=\frac{V}{R_1}\angle -60^{\circ}
I_Y=\frac{V_{YB}}{R_1} =-\frac{V\angle -120^{\circ}}{R_1}
Using KCL,
I_R+I_Y+I_B=0
\frac{V}{R_1}\angle -60^{\circ}+\frac{V}{R_1}\angle -120^{\circ} I_B=0
\Rightarrow \; I_B=\sqrt{3}\frac{V}{R_1}\angle 90^{\circ}
So, I_R:I_Y:I_B=\frac{V}{R_1}:\frac{V}{R_1}:\sqrt{3}\frac{V}{R_1}
\;\;\; =1:1:\sqrt{3}
There are 6 questions to complete.