The initial three-phase voltage phasors ($\bar V_A$, $\bar V_B$, and $\bar V_C$) at a bus of a power network are as shown in Case-1. Due to a disturbance, the bus voltage phasors changed in phase by a small angle $\Delta\theta$, and the magnitudes remained the same as depicted in Case-2.
Which one of the following statements is correct about the zero sequence components?

Question

A single-phase full-controlled thyristor converter bridge is used for regenerative braking of a separately excited DC motor with the following specifications:

Rated armature voltage = 210 V
Rated armature current = 10 A
Rated speed = 1200 rpm
Armature resistance = 1 Ω
Input to the converter bridge = 240 V at 50 Hz

Assume that the motor is running at 600 rpm and the armature terminals are suitably reversed for regenerative braking.

If the armature current of the motor is to be maintained at the rated value, the triggering angle of the converter bridge (in degrees) should be ____________ (rounded off to two decimal places).

Motor and converter parameters:

Parameter	Value
Rated armature voltage	210 V
Rated armature current	10 A
Rated speed	1200 rpm
Armature resistance	1 Ω
Input to converter bridge	240 V at 50 Hz

The armature of the DC motor is fed from the full-controlled bridge, and the field current is maintained constant.

Answer 1

Instead of factoring the rotation out algebraically, picture what happens to the phasor sum geometrically.

Draw the three Case-1 phasors tip to tail (or just added as vectors) to get a resultant vector, which, divided by 3, is the zero sequence phasor $\bar V_0$ for Case-1.

Now, rotating every one of the three original phasors by the same angle $\Delta\theta$ is exactly the same as rotating the entire rigid picture, phasors and all, by $\Delta\theta$ about the origin. Since vector addition commutes with rotation (rotating each vector and then adding gives the same result as adding then rotating the sum), the resultant vector for Case-2 is just the Case-1 resultant rotated by $\Delta\theta$.

A rotation never changes a vector's length, only its direction. So the length of the Case-2 resultant (and hence the magnitude of $\bar V_0'$ after dividing by 3) is identical to the Case-1 resultant's length (and $\bar V_0$'s magnitude), while its direction (phase angle) has shifted by exactly $\Delta\theta$.

So the two zero sequence phasors have the same magnitude but point in different directions, differing by $\Delta\theta$ in phase.

\[ \boxed{\text{Same magnitude, different phase angles}} \]

Answer 2

Instead of factoring the rotation out algebraically, picture what happens to the phasor sum geometrically.

Draw the three Case-1 phasors tip to tail (or just added as vectors) to get a resultant vector, which, divided by 3, is the zero sequence phasor $\bar V_0$ for Case-1.

Now, rotating every one of the three original phasors by the same angle $\Delta\theta$ is exactly the same as rotating the entire rigid picture, phasors and all, by $\Delta\theta$ about the origin. Since vector addition commutes with rotation (rotating each vector and then adding gives the same result as adding then rotating the sum), the resultant vector for Case-2 is just the Case-1 resultant rotated by $\Delta\theta$.

A rotation never changes a vector's length, only its direction. So the length of the Case-2 resultant (and hence the magnitude of $\bar V_0'$ after dividing by 3) is identical to the Case-1 resultant's length (and $\bar V_0$'s magnitude), while its direction (phase angle) has shifted by exactly $\Delta\theta$.

So the two zero sequence phasors have the same magnitude but point in different directions, differing by $\Delta\theta$ in phase.

\[ \boxed{\text{Same magnitude, different phase angles}} \]

\(\omega\) (rad/s)	Magnitude (dB)	Phase (degrees)
0.5	-7	-40
1.0	-10	-80
2.0	-18	-130
10.0	-40	-200

Show Hint

The Correct Option is D

Solution and Explanation

Questions Asked in GATE EE exam