GATE 2023 IN Q.46: solution and answer

GATE 2023 · INQ.46 · 2 marks · NAT

In the circuit shown, $ω = 100 π$ rad/s, $R_{1} = R_{2} = 2.2 Ω$ and $L = 7$ mH. The capacitance $C$ for which $Y_{in}$ is purely real is ________ mF (rounded off to two decimal places).

GATE 2023 IN question 46: An AC voltage source V_0 cos (ω t) sits on the left vertical wire between a top rail and a bottom rail. — Figure from the GATE 2023 IN paper. It is part of the question.

Answer and solution

ANSWER

$C \approx 1.45 mF$ .

Official answer: 1.40 to 1.50Source: Official answer key, GATE 2023 IN (IIT Kanpur)

Reactances: $X_{L} = ω L = 100 π \times 7 \times 1 0^{- 3} = 2.199 Ω$ and $X_{C} = \frac{1}{ω C}$ . Write $R = R_{1} = R_{2} = 2.2 Ω$ .
Admittances of the two parallel branches: $Y_{1} = \frac{1}{R + j X _{L}} = \frac{R - j X _{L}}{R ^{2} + X _{L}^{2}}, Y_{2} = \frac{1}{R - j X _{C}} = \frac{R + j X _{C}}{R ^{2} + X _{C}^{2}}$
Parallel admittances add, so $Y_{in} = Y_{1} + Y_{2}$ . For $Y_{in}$ to be purely real its imaginary part must be zero: $\frac{X _{L}}{R ^{2} + X _{L}^{2}} = \frac{X _{C}}{R ^{2} + X _{C}^{2}}$
Cross-multiply, $X_{L} (R^{2} + X_{C}^{2}) = X_{C} (R^{2} + X_{L}^{2})$ , and rearrange into a quadratic in $X_{C}$ : $X_{L} X_{C}^{2} - (R^{2} + X_{L}^{2}) X_{C} + R^{2} X_{L} = 0 \Rightarrow (X_{C} - X_{L}) (X_{L} X_{C} - R^{2}) = 0$
Root 1, $X_{C} = X_{L}$ : $C = \frac{1}{ω ^{2} L} = \frac{1}{( 100 π ) ^{2} \times 7 \times 1 0 ^{- 3}} = \frac{1}{690.87} = 1.4474 \times 1 0^{- 3} F$ .
Root 2, $X_{C} = R^{2} / X_{L}$ : $C = \frac{X _{L}}{ω R ^{2}} = \frac{L}{R ^{2}} = \frac{7 \times 1 0 ^{- 3}}{4.84} = 1.4463 \times 1 0^{- 3} F$ .
Because $X_{L} = 2.199 Ω$ is almost exactly equal to $R = 2.2 Ω$ , the two roots are nearly equal. Substituting either one back gives a real $Y_{in}$ (about $0.455 S$ ), and both round to the same value: $C \approx 1.45 mF$ .

KEY IDEA

Work with admittances, because parallel admittances simply add. Set the imaginary part of $Y_{1} + Y_{2}$ to zero. The condition factorises into $X_{C} = X_{L}$ or $X_{C} X_{L} = R^{2}$ (that is, $C = L / R^{2}$ ). Here $ω L \approx R$ , so the network is almost the classic constant-resistance circuit ( $R_{1} = R_{2} = L / C$ ). That is why the two roots nearly coincide at about 1.45 mF.

WATCH OUT FOR

Setting $ω L = 1/ (ω C)$ straight away, as if this were a series or pure-LC resonance, without deriving the condition. That happens to land in the range here, but only because of the numbers. A second common slip is writing $Y_{2} = 1/ (R + j X_{C})$ with the wrong sign on the capacitive reactance. That makes the two imaginary parts add instead of cancel, so no real solution exists.

From the GATE 2023 IN paper. Every question of this paper · Download the full paper · All 34 AC, power & resonance questions, with notes

GATE 2023 IN Q.46: solution and answer

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