GATE 2022 EE Q.21: solution and answer

GATE 2022 · EEQ.21 · 1 mark · MCQ

In the circuit shown below, a three-phase star-connected unbalanced load is connected to a balanced three-phase supply of $1003 V$ with phase sequence $A B C$ . The star connected load has $Z_{A} = 10 Ω$ and $Z_{B} = 20∠6 0^{\circ} Ω$ . The value of $Z_{C}$ in $Ω$ , for which the voltage difference across the nodes $n$ and $n^{'}$ is zero, is

GATE 2022 EE question 21: Left: a star-connected source with common node n (filled dot). — Figure from the GATE 2022 EE paper. It is part of the question.

(A)
$20∠ - 3 0^{\circ}$
(B)
$20∠3 0^{\circ}$
(C)
$20∠ - 6 0^{\circ}$
(D)
$20∠6 0^{\circ}$

Answer and solution

ANSWER

$Z_{C} = 20∠ - 6 0^{\circ} Ω$ , option (C).

Official answer: CSource: Official answer key, GATE 2022 EE (IIT Kharagpur)

Phase voltages of the balanced supply. $1003 V$ is the line voltage, so the phase voltage is $E_{p} = \frac{100 3}{3} = 100 V$ . For sequence $A B C$ , taking $E_{A}$ as reference: $E_{A} = 100∠ 0^{\circ} V$ , $E_{B} = 100∠ - 12 0^{\circ} V$ , $E_{C} = 100∠12 0^{\circ} V$ (rms).
Condition for zero neutral shift. By Millman's theorem, the voltage of the floating load neutral $n^{'}$ with respect to the source neutral $n$ is $V_{n^{'} n} = \frac{\frac{E _{A}}{Z _{A}} + \frac{E _{B}}{Z _{B}} + \frac{E _{C}}{Z _{C}}}{\frac{1}{Z _{A}} + \frac{1}{Z _{B}} + \frac{1}{Z _{C}}}$ so $V_{n^{'} n} = 0$ exactly when the numerator is zero. With $V_{n^{'} n} = 0$ , each impedance sees its own source phase voltage, so each line current is $I_{k} = E_{k} / Z_{k}$ . The numerator being zero is then just KCL at $n^{'}$ : $I_{A} + I_{B} + I_{C} = 0$ .
Phase A current: $I_{A} = \frac{100∠ 0 ^{\circ}}{10} = 10∠ 0^{\circ} A$ .
Phase B current: $I_{B} = \frac{100∠ - 12 0 ^{\circ}}{20∠6 0 ^{\circ}} = 5∠ - 18 0^{\circ} A = - 5 A$ .
KCL at $n^{'}$ : $I_{C} = - (I_{A} + I_{B}) = - (10 - 5) = - 5 A = 5∠18 0^{\circ} A$ .
Solve for the impedance: $Z_{C} = \frac{E _{C}}{I _{C}} = \frac{100∠12 0 ^{\circ}}{5∠18 0 ^{\circ}} = 20∠ - 6 0^{\circ} Ω = (10 - j 17.32) Ω$ This is a capacitive impedance.
Checks. (i) The supply magnitude cancels out of $\sum E_{k} / Z_{k} = 0$ , because all three $E_{k}$ have the same magnitude. Reading $1003 V$ as a phase voltage would give the same $Z_{C}$ . So would choosing a different reference angle. (ii) In admittance form, with $a = 1∠12 0^{\circ}$ the condition is $Y_{A} + a^{2} Y_{B} + a Y_{C} = 0$ . Here $Y_{A} = 0.1$ and $a^{2} Y_{B} = 0.05∠ - 18 0^{\circ} = - 0.05$ , so $a Y_{C} = - 0.05$ . That gives $Y_{C} = 0.05∠6 0^{\circ} S$ , i.e. $Z_{C} = 20∠ - 6 0^{\circ} Ω$ , which agrees. The answer is option (C).

KEY IDEA

In a three-wire star system, the load neutral sits at the source neutral potential exactly when the Millman numerator $\sum E_{k} / Z_{k}$ is zero. The same condition says the currents $E_{k} / Z_{k}$ satisfy KCL at $n^{'}$ . The load does not have to be balanced for this: it is enough that the three phase currents, each set by its own phase voltage, add to zero.

WATCH OUT FOR

Assuming $V_{n^{'} n} = 0$ needs a balanced load and picking $Z_{C} = Z_{B} = 20∠6 0^{\circ} Ω$ (option D). That cannot be right, since $Z_{A} = 10 Ω$ already differs from $Z_{B}$ . Two other slips: using $E_{C} = 100∠ - 12 0^{\circ}$ (the $A C B$ sequence) instead of $100∠12 0^{\circ}$ , and dropping the minus sign in $I_{C} = - (I_{A} + I_{B})$ , which gives $20∠12 0^{\circ}$ , not one of the options.

From the GATE 2022 EE paper. Every question of this paper · Download the full paper · All 11 Three-phase (EE) questions, with notes

GATE 2022 EE Q.21: solution and answer

Answer and solution

More Three-phase (EE) questions