GATE 2025 EC Q.57: solution and answer

GATE 2025 · ECQ.57 · 2 marks · MSQ

All the diodes in the circuit given below are ideal.

Which of the following plots is/are correct when $V_{I}$ (in Volts) is swept from $- M$ to $M$ ?

GATE 2025 EC question 57: A voltage source V_I with its + terminal at the top and its − terminal at the bottom. — Figure from the GATE 2025 EC paper. It is part of the question.

(A)
Figure — plot of $V_{O}$ (vertical axis) against $V_{I}$ (horizontal axis). It is V-shaped: $V_{O} = M$ at $V_{I} = - M$ , it falls linearly to 0 at the origin, and it rises linearly back to $V_{O} = M$ at $V_{I} = M$ . Dashed lines mark the level M above $V_{I} = \pm M$ .
(B)
Figure — plot of $V_{O}$ against $V_{I}$ . It is a straight line through the origin from $(- M, - M)$ to $(M, M)$ . Dashed lines mark $V_{O} = \pm M$ at $V_{I} = \pm M$ .
(C)
Figure — plot of $I_{I}$ against $V_{I}$ . It is V-shaped: $I_{I} = \frac{M}{R}$ at $V_{I} = - M$ , it falls linearly to 0 at the origin, and it rises linearly back to $\frac{M}{R}$ at $V_{I} = M$ .
(D)
Figure — plot of $I_{I}$ against $V_{I}$ . It is a straight line through the origin from $(- M, - \frac{M}{R})$ to $(M, \frac{M}{R})$ . Dashed lines mark $I_{I} = \pm \frac{M}{R}$ at $V_{I} = \pm M$ .

Answer and solution

ANSWER

(A) and (D). With ideal diodes the bridge gives $V_{O} = ∣ V_{I} ∣$ , a V-shape reaching M at both ends. The source current is $I_{I} = V_{I} / R$ , a straight line from $- \frac{M}{R}$ to $\frac{M}{R}$ , because the source always sees just the resistance R.

Official answer: A, DSource: Official answer key, GATE 2025 EC (IIT Roorkee)

Read the circuit from the diode directions. Call the top node $T$ , the bottom node $B$ , the left end of R $P$ (the $-$ of $V_{O}$ ) and the right end $Q$ (the $+$ of $V_{O}$ ). The diodes $T \to Q$ and $B \to Q$ both have their cathodes at $Q$ , and the diodes $P \to T$ and $P \to B$ both have their anodes at $P$ . This is a full-wave bridge rectifier: the input is across $T$ – $B$ , the output is across $P$ – $Q$ , and $V_{O} = V_{Q} - V_{P}$ . Ideal diode means zero drop when conducting and an open circuit when reverse biased.
Case $V_{I} > 0$ ( $T$ positive with respect to $B$ ). Take $D_{T Q}$ and $D_{P B}$ as conducting (short circuits). Then $V_{Q} = V_{T}$ and $V_{P} = V_{B}$ . Check the other two: $D_{P T}$ has anode at $V_{B}$ and cathode at $V_{T}$ , and $D_{B Q}$ has anode at $V_{B}$ and cathode at $V_{Q} = V_{T}$ , so each is reverse biased by $V_{I}$ . The assumption is consistent. Current flows $T \to Q \to R \to P \to B$ , and $V_{O} = V_{T} - V_{B} = V_{I}, I_{I} = \frac{V _{I}}{R} > 0.$
Case $V_{I} < 0$ ( $B$ positive with respect to $T$ ). Now $D_{B Q}$ and $D_{P T}$ conduct, and $D_{T Q}$ and $D_{P B}$ are each reverse biased by $∣ V_{I} ∣$ . Current flows $B \to Q \to R \to P \to T$ and returns to the source's $+$ terminal along the top wire. Now $V_{Q} = V_{B}$ and $V_{P} = V_{T}$ , so $V_{O} = V_{B} - V_{T} = - V_{I} = ∣ V_{I} ∣.$
Direction of the source current for $V_{I} < 0$ : the current leaves the bridge at $T$ and flows back into the source's $+$ terminal, which is opposite to the marked $I_{I}$ arrow. Its size is $\frac{∣ V _{I} ∣}{R}$ , so $I_{I} = - \frac{∣ V _{I} ∣}{R} = \frac{V _{I}}{R} < 0.$
Output voltage: in both cases $V_{O} = ∣ V_{I} ∣$ . This is a V-shape through the origin that reaches $M$ at both $V_{I} = - M$ and $V_{I} = M$ . So (A) is correct and (B) is wrong: (B) would need $V_{O}$ to go negative, but the current through R always flows from $Q$ to $P$ , so $V_{O} \geq 0$ .
Input current: $I_{I} = \frac{V _{I}}{R}$ for every $V_{I}$ . The bridge reverses the connection of R when $V_{I}$ changes sign, but the source still sees a plain resistance R, so the plot is a straight line from $(- M, - \frac{M}{R})$ to $(M, \frac{M}{R})$ . (D) is correct and (C) is wrong.
Correct options: (A) and (D).

KEY IDEA

A full-wave bridge rectifies what the LOAD sees, not what the SOURCE sees. The current through R always flows the same way ( $Q \to P$ ), so $V_{O} = ∣ V_{I} ∣$ . The current at the source terminals still reverses with $V_{I}$ . With ideal diodes the source sees the resistance R connected one way round or the other, so $I_{I} = V_{I} / R$ is linear and odd.

WATCH OUT FOR

Picking (C) by assuming that since the output is rectified, the input current is rectified too. The load current $V_{O} / R = ∣ V_{I} ∣/ R$ is always positive, but for $V_{I} < 0$ the source current flows against the marked $I_{I}$ arrow, so $I_{I}$ is negative. Picking (B) is the other slip: it treats the bridge as a straight-through connection and forgets that the two diodes conducting for $V_{I} < 0$ swap which end of R connects to which terminal.

From the GATE 2025 EC paper. Every question of this paper · Download the full paper · All 25 Diode circuits questions, with notes

GATE 2025 EC Q.57: solution and answer

Answer and solution

More Diode circuits questions