GATE 2022 IN Q.52: solution and answer

GATE 2022 · INQ.52 · 2 marks · NAT

In the circuit shown, the load is driven by a sinusoidal ac voltage source $V_{1} = 100∠ 0^{\circ} V$ at 50 Hz. Given $R_{1} = 20 Ω$ , $C_{1} = (\frac{1000}{π}) μ F$ , $L_{1} = (\frac{20}{π}) mH$ and $R_{2} = 4 Ω$ , the power factor is ________ (round off to one decimal place)

GATE 2022 IN question 52: A single-loop circuit. — Figure from the GATE 2022 IN paper. It is part of the question.

Answer and solution

ANSWER

The load impedance is $Z = 8 - j 6 Ω$ ( $∣ Z ∣ = 10 Ω$ ), so the power factor is $8/10 = 0.8$ (leading).

Official answer: 0.8Source: Official answer key, GATE 2022 IN (IIT Kharagpur)

Angular frequency: $ω = 2 π f = 2 π (50) = 100 π rad/s$ .
Capacitive reactance: $X_{C} = \frac{1}{ω C _{1}} = \frac{1}{100 π \cdot \frac{1000}{π} \times 1 0 ^{- 6}} = \frac{1}{0.1} = 10 Ω$ so the capacitor's impedance is $- j 10 Ω$ .
Inductive reactance: $X_{L} = ω L_{1} = 100 π \cdot \frac{20}{π} \times 1 0^{- 3} = 2 Ω$ so the inductor's impedance is $j 2 Ω$ .
$R_{1}$ and $C_{1}$ are in parallel (product over sum): $Z_{1} = \frac{20 ( - j 10 )}{20 - j 10} = \frac{- j 200 ( 20 + j 10 )}{2 0 ^{2} + 1 0 ^{2}} = \frac{2000 - j 4000}{500} = 4 - j 8 Ω$
Everything to the right of the marked terminals is the load: $Z_{1}$ , $L_{1}$ and $R_{2}$ in series. $Z = Z_{1} + j X_{L} + R_{2} = (4 - j 8) + j 2 + 4 = 8 - j 6 Ω, ∣ Z ∣ = 8^{2} + 6^{2} = 10 Ω$
Power factor $= cos θ = \frac{Re ( Z )}{∣ Z ∣} = \frac{8}{10} = 0.8$ . It is leading because the net reactance, $- 6 Ω$ , is capacitive: the current $I = \frac{100∠ 0 ^{\circ}}{10∠ - 36.8 7 ^{\circ}} = 10∠36.8 7^{\circ} A$ leads the voltage. The source voltage magnitude does not affect the power factor.

KEY IDEA

Power factor is $cos$ of the load impedance angle, i.e. $Re (Z) /∣ Z ∣$ . The unusual values ( $C_{1} = 1000/ π μ F$ , $L_{1} = 20/ π mH$ ) are chosen so the $π$ cancels against $ω = 100 π$ . That gives clean reactances ( $X_{C} = 10 Ω$ , $X_{L} = 2 Ω$ ). Reduce $R_{1} ∥ C_{1}$ to one series impedance, then add the series elements.

WATCH OUT FOR

Adding $R_{1}$ and $X_{C}$ as if they were in series, which gives $Z = 24 - j 8$ and a power factor of about $0.95$ . The same error happens when $Re (Z)$ is taken as $R_{1} + R_{2} = 24 Ω$ instead of using the real part of the parallel combination ( $4 Ω$ ) plus $R_{2}$ . A second slip is forgetting the $1 0^{- 6}$ or $1 0^{- 3}$ unit factors on $C_{1}$ and $L_{1}$ .

From the GATE 2022 IN paper. Every question of this paper · Download the full paper · All 34 AC, power & resonance questions, with notes

GATE 2022 IN Q.52: solution and answer

Answer and solution

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