# Three-Phase Circuits

 Question 1
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 1 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 2
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 2 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 3
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 3 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 4
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 4 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 4 questions to complete.