GATE 2018 EE Q.8: solution and answer

GATE 2018 · EEQ.8 · 1 mark · MCQ

In the figure, the voltages are $v_{1} (t) = 100 cos (ω t)$ , $v_{2} (t) = 100 cos (ω t + π /18)$ and $v_{3} (t) = 100 cos (ω t + π /36)$ . The circuit is in sinusoidal steady state, and $R << ω L$ . $P_{1}$ , $P_{2}$ and $P_{3}$ are the average power outputs. Which one of the following statements is true?

GATE 2018 EE question 8: Three ideal voltage sources share a common bottom (ground) wire. — Figure from the GATE 2018 EE paper. It is part of the question.

(A)
$P_{1} = P_{2} = P_{3} = 0$
(B)
$P_{1} < 0, P_{2} > 0, P_{3} > 0$
(C)
$P_{1} < 0, P_{2} > 0, P_{3} < 0$
(D)
$P_{1} > 0, P_{2} < 0, P_{3} > 0$

Answer and solution

ANSWER

Option (C): $P_{1} < 0$ , $P_{2} > 0$ , $P_{3} < 0$ . The middle source $v_{2}$ leads both of the others, so it supplies real power to $v_{1}$ and to $v_{3}$ .

Official answer: CSource: Official answer key, GATE 2018 EE (IIT Guwahati)

Phasors (rms): $π /18 = 1 0^{\circ}$ and $π /36 = 5^{\circ}$ , so $V_{1} = 70.71∠ 0^{\circ} V$ , $V_{2} = 70.71∠1 0^{\circ} V$ , $V_{3} = 70.71∠ 5^{\circ} V$ . Write $V = 100/ 2 = 70.71 V$ , so $V^{2} = 5000 V^{2}$ . All three magnitudes are equal; only the phase angles differ.
The bottom terminals of all three sources are joined, so each outer source is connected straight to the top of $v_{2}$ through one series branch $Z = R + j X$ , where $X = ω L$ . The order of $R$ and $L$ inside a branch makes no difference. Because $R ≪ ω L$ , each branch is almost a pure reactance.
Power output of $v_{1}$ : the current leaving its $+$ terminal is $I_{1} = (V_{1} - V_{2}) / Z$ , so $P_{1} = Re (V_{1} I_{1}^{*}) = Re [\frac{V ^{2} - V _{1} V _{2}^{*}}{Z ^{*}}]$ . Put $V_{1} V_{2}^{*} = V^{2} ∠ (- δ)$ , where $δ = 1 0^{\circ}$ is the angle by which $V_{2}$ leads $V_{1}$ , and $\frac{1}{Z ^{*}} = \frac{R + j X}{∣ Z ∣ ^{2}}$ : $P_{1} = \frac{V ^{2}}{∣ Z ∣ ^{2}} Re [(1 - cos δ + j sin δ) (R + j X)] = \frac{V ^{2} [ R ( 1 - cos δ ) - X sin δ ]}{∣ Z ∣ ^{2}}$
Sign of $P_{1}$ : $P_{1} < 0$ exactly when $R (1 - cos δ) < X sin δ$ , that is when $\frac{R}{X} < \frac{sin δ}{1 - cos δ} = cot \frac{δ}{2} = cot 5^{\circ} \approx 11.4$ . Since $R ≪ ω L$ , this certainly holds. Dropping $R$ (so $∣ Z ∣^{2} \approx X^{2}$ ) gives $P_{1} \approx - \frac{V ^{2} sin 1 0 ^{\circ}}{ω L} = - \frac{868}{ω L} W < 0$ . So $v_{1}$ absorbs power.
Power output of $v_{3}$ : it is connected only to $v_{2}$ , through the same $Z$ . With $I_{3} = (V_{3} - V_{2}) / Z$ leaving its $+$ terminal, and $V_{2}$ leading $V_{3}$ by $δ = 1 0^{\circ} - 5^{\circ} = 5^{\circ}$ , the same algebra gives $P_{3} = \frac{V ^{2} [ R ( 1 - cos 5 ^{\circ} ) - X sin 5 ^{\circ} ]}{∣ Z ∣ ^{2}}$ . This is negative because $R / X < cot 2. 5^{\circ} \approx 22.9$ . Approximately, $P_{3} \approx - \frac{V ^{2} sin 5 ^{\circ}}{ω L} = - \frac{436}{ω L} W < 0$ . So $v_{3}$ also absorbs power.
Power output of $v_{2}$ : the current leaving its $+$ terminal is $\frac{V _{2} - V _{1}}{Z} + \frac{V _{2} - V _{3}}{Z}$ . Each branch now contributes $+ X sin δ$ instead of $- X sin δ$ : $P_{2} = \frac{V ^{2} [ R ( 1 - cos 1 0 ^{\circ} ) + X sin 1 0 ^{\circ} ]}{∣ Z ∣ ^{2}} + \frac{V ^{2} [ R ( 1 - cos 5 ^{\circ} ) + X sin 5 ^{\circ} ]}{∣ Z ∣ ^{2}} > 0$ since every term is positive. Approximately, $P_{2} \approx \frac{1304}{ω L} W$ . Check: $P_{1} + P_{2} + P_{3} = \frac{2 R V ^{2} [ ( 1 - cos 1 0 ^{\circ} ) + ( 1 - cos 5 ^{\circ} ) ]}{∣ Z ∣ ^{2}}$ . That is exactly the loss $∣ I_{1} ∣^{2} R + ∣ I_{3} ∣^{2} R$ in the two resistors, so energy balances.
So $P_{1} < 0$ , $P_{2} > 0$ and $P_{3} < 0$ , which is option (C). This matches the rule of thumb for a nearly lossless inductive link: real power flows from the leading voltage to the lagging one. $v_{2}$ leads both of its neighbours, so it supplies both.

KEY IDEA

When sources of equal magnitude are joined by a mostly inductive branch ( $R ≪ ω L$ ), real power flows from the source whose voltage leads to the one that lags, $P \approx \frac{V ^{2} sin δ}{X}$ . Compare each source only with the source it is actually wired to. Here $v_{1}$ and $v_{3}$ are each wired to $v_{2}$ , and $v_{2}$ leads both of them.

WATCH OUT FOR

Picking (A) because all three magnitudes are equal, when it is the phase difference that drives real power through a reactance. Another mistake is picking (B): comparing $v_{3}$ with $v_{1}$ (it leads $v_{1}$ by $5^{\circ}$ ) and calling it a supplier. But $v_{3}$ is connected only to $v_{2}$ , which leads it by $5^{\circ}$ , so $v_{3}$ absorbs power.

From the GATE 2018 EE paper. Every question of this paper · Download the full paper · All 34 AC, power & resonance questions, with notes

GATE 2018 EE Q.8: solution and answer

Answer and solution

More AC, power & resonance questions