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
173.2\angle -60^{\circ} | |
173.2\angle 120^{\circ} | |
100.0\angle -60^{\circ} | |
100.0\angle 120^{\circ} |
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
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
10 \sqrt{3}\angle 105^{\circ} V | |
10 \angle 105^{\circ} V | |
10 \sqrt{3}\angle -75^{\circ} V | |
-10 \sqrt{3}\angle 90^{\circ} V |
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
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.
-\phi _{2} | |
\phi _{2} | |
(\pi/2-\phi _{2}) | |
(\pi/2+\phi _{2}) |
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
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
1:1:\sqrt{3} | |
1:1:2 | |
1:1:0 | |
1:1:\sqrt{3/2} |
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.
