ISRO EC 2025 Q.2: solution and answer

ISRO 2025 · ECQ.2 · 1 mark · MCQ

A source with internal resistance Rs = 50 Ω is to be matched to a load RL = 200 Ω using a lossless LC network at a frequency of 10 MHz. Which of the following configurations will provide proper impedance matching and high-pass filtering?

(A)
A series inductor followed by a parallel capacitor
(B)
A parallel inductor followed by a series capacitor
(C)
A series capacitor followed by a parallel inductor
(D)
A parallel capacitor followed by a series inductor

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Answer and solution

ANSWER

The correct choice is (C): a series capacitor followed by a parallel inductor.

Official answer: CSource: ISRO's final answer key of 15 May 2026

Read the options from the source towards the load. A two-element L-match has one series element and one shunt element, and the shunt element must sit across the larger resistance. Here $R_{L} = 200 Ω > R_{s} = 50 Ω$ , so the shunt element goes on the load side and the series element on the source side.
Options (B) and (D) put a shunt element on the source side, across the smaller resistance. A shunt element across $50 Ω$ can only lower the resistance seen from the source, so it cannot match $50 Ω$ to $200 Ω$ . Only (A) and (C) have the series element first and the shunt element second.
Quality factor of the match: $Q = \frac{R _{L}}{R _{s}} - 1 = \frac{200}{50} - 1 = 3 \approx 1.73$ .
Between (A) and (C), decide by the filtering requirement. A series capacitor blocks low frequencies, and a shunt inductor shorts low frequencies to ground, so series C with shunt L is high-pass. Series L with shunt C, as in (A), passes low frequencies and attenuates high ones, so it is low-pass.
Element values at $10 MHz$ ( $ω = 2 π \times 1 0^{7} rad/s$ ): the shunt inductor has $X_{L} = \frac{R _{L}}{Q} \approx 115.5 Ω$ , so $L = \frac{X _{L}}{ω} \approx 1.84 μ H$ . The series capacitor has $X_{C} = Q R_{s} \approx 86.6 Ω$ , so $C = \frac{1}{ω X _{C}} \approx 184 pF$ .
Verify the match. The load $200 Ω$ in parallel with $j X_{L}$ has $G = 0.005 S$ and $B = 1/115.5 \approx 0.00866 S$ . Its equivalent series resistance is $\frac{G}{G ^{2} + B ^{2}} = \frac{0.005}{2.5 \times 1 0 ^{- 5} + 7.5 \times 1 0 ^{- 5}} = 50 Ω$ , and its series reactance is $\frac{B}{G ^{2} + B ^{2}} = 86.6 Ω$ (inductive). The series capacitor contributes $- j 86.6 Ω$ and cancels it, so the source sees a purely resistive $50 Ω$ .
Hence a series capacitor followed by a parallel inductor gives both the impedance match and the high-pass response: option (C).

KEY IDEA

In an L-match the shunt element goes across the larger resistance and the series element on the smaller side, with $Q = R_{high} / R_{low} - 1$ . For a high-pass match, use a series capacitor and a shunt inductor.

WATCH OUT FOR

Putting the shunt element on the source (low-resistance) side, as in (B) and (D), or choosing series L with shunt C in (A), which matches but is a low-pass section.

From the ISRO EC 2025 paper. Every question of this paper · The full paper (PDF, ISRO) · All 4 ISRO AC, power & resonance questions, with notes

ISRO EC 2025 Q.2: solution and answer

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