ISRO EC 2025 Q.3: solution and answer

ISRO 2025 · ECQ.3 · 1 mark · MCQ

In Miller's theorem, if a capacitor C is connected between the input and output of an amplifier with gain A, the equivalent input capacitance is :

(A)
C
(B)
(1 + A) C
(C)
(1 – A) C
(D)
C / (1 + A)

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Answer and solution

ANSWER

The equivalent input capacitance is $(1 - A) C$ , option (c).

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

Let $A = V_{o} / V_{i}$ be the voltage gain, and let a capacitor $C$ (impedance $Z = \frac{1}{j ω C}$ ) be connected between the input node and the output node.
Current leaving the input node through the bridging capacitor: $I = \frac{V _{i} - V _{o}}{Z} = \frac{V _{i} ( 1 - A )}{Z}$ .
Miller's theorem replaces the bridge by an impedance $Z_{1}$ from the input node to ground that draws the same current: $Z_{1} = \frac{V _{i}}{I} = \frac{Z}{1 - A}$ .
With $Z = \frac{1}{j ω C}$ this gives $Z_{1} = \frac{1}{j ω C ( 1 - A )}$ , which is a capacitor of value $C_{in} = (1 - A) C .$
For an inverting amplifier ( $A < 0$ ) this becomes $(1 + ∣ A ∣) C$ , the familiar multiplication of capacitance. Option (b), $(1 + A) C$ , matches only if $A$ is taken as a magnitude for an inverting stage, not for the gain $A$ as written.
Option (a) ignores the gain, and option (d) does not follow from the theorem. The equivalent input capacitance is $(1 - A) C$ , option (c).

KEY IDEA

Miller's theorem: a bridging impedance $Z$ appears at the input as $Z / (1 - A)$ , so a bridging capacitor $C$ appears as $(1 - A) C$ , where $A = V_{o} / V_{i}$ is signed.

WATCH OUT FOR

Writing $(1 + A) C$ from memory of the inverting-amplifier case. That form holds only when $A$ is a magnitude for an inverting stage. With $A$ as the signed gain, the factor is $(1 - A)$ .

From the ISRO EC 2025 paper. Every question of this paper · The full paper (PDF, ISRO) · All 2 ISRO Amplifiers questions, with notes