ISRO EC 2025 Q.47: solution and answer

ISRO 2025 · ECQ.47 · 1 mark · MCQ

The network of a system is shown below. Signal flow graph for the system is

ISRO EC 2025 question 47: Circuit: source (+ at top) drives node v1. — Figure from the ISRO EC 2025 paper. It is part of the question.

(A)
Forward gains 1, R3, R2, R4, 1; feedback gains -1/R1, -R3, -1/R2
(B)
Forward gains 1/R1, R3, 1/R2, R4, 1; feedback gains -1/R1, -R3, -1/R2
(C)
Forward gains 1, R3, R2, R4, 1; feedback gains -1/R1, -1/R3, -1/R2
(D)
Forward gains 1/R1, R3, 1/R2, R4, 1; feedback gains -1/R1, -1/R3, -R2

Answer and solution

ANSWER

The correct signal flow graph is option (B).

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

Take both mesh currents $i_{1}$ and $i_{2}$ as flowing left to right through $R_{1}$ and $R_{2}$ . Write one equation per variable. Each current is written in terms of voltages and each voltage in terms of currents. The nodes are $v_{1}, i_{1}, v_{2}, i_{2}, v_{3}$ in that order.
The current through $R_{1}$ is $i_{1}$ , so by Ohm's law $i_{1} = \frac{v _{1} - v _{2}}{R _{1}} = \frac{1}{R _{1}} v_{1} - \frac{1}{R _{1}} v_{2}$ . This gives a branch $v_{1} \to i_{1}$ of gain $\frac{1}{R _{1}}$ and a branch $v_{2} \to i_{1}$ of gain $- \frac{1}{R _{1}}$ .
The voltage $v_{2}$ is across $R_{3}$ , which carries $i_{1} - i_{2}$ . So $v_{2} = R_{3} (i_{1} - i_{2}) = R_{3} i_{1} - R_{3} i_{2}$ . This gives a branch $i_{1} \to v_{2}$ of gain $R_{3}$ and a branch $i_{2} \to v_{2}$ of gain $- R_{3}$ .
The current through $R_{2}$ is $i_{2}$ , so $i_{2} = \frac{v _{2} - v _{3}}{R _{2}} = \frac{1}{R _{2}} v_{2} - \frac{1}{R _{2}} v_{3}$ . This gives a branch $v_{2} \to i_{2}$ of gain $\frac{1}{R _{2}}$ and a branch $v_{3} \to i_{2}$ of gain $- \frac{1}{R _{2}}$ .
The voltage $v_{3}$ is across $R_{4}$ , which carries $i_{2}$ . So $v_{3} = R_{4} i_{2}$ , a branch $i_{2} \to v_{3}$ of gain $R_{4}$ . The output is $v_{o} = v_{3}$ , which adds a unity branch to the output node.
The forward path therefore has gains $\frac{1}{R _{1}}, R_{3}, \frac{1}{R _{2}}, R_{4}, 1$ . The feedback branches are $- \frac{1}{R _{1}}$ (from $v_{2}$ to $i_{1}$ ), $- R_{3}$ (from $i_{2}$ to $v_{2}$ ) and $- \frac{1}{R _{2}}$ (from $v_{3}$ to $i_{2}$ ). This is option (b).
Option (a) has gain $1$ at the input and $R_{2}$ in place of $\frac{1}{R _{2}}$ . Option (c) has those same two errors and also $- \frac{1}{R _{3}}$ in place of $- R_{3}$ . Option (d) has the right forward path, but its feedback gains $- \frac{1}{R _{3}}$ and $- R_{2}$ are inverted.

KEY IDEA

Build the signal flow graph in alternating form. Each current is a conductance times a voltage difference, so its branches have gain $1/ R$ . Each node voltage is a resistance times a current, so its branches have gain $R$ .

WATCH OUT FOR

Using $R$ where $1/ R$ belongs and the reverse. A current equation takes $1/ R_{1}$ and $1/ R_{2}$ . A voltage equation takes $R_{3}$ and $R_{4}$ . Another slip is giving the input branch gain 1 when it should be $1/ R_{1}$ .

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