# 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 $n$and $n'$ 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.