GATE 2022 EC Q.14: solution and answer

GATE 2022 · ECQ.14 · 1 mark · MCQ

Consider the circuit shown in the figure. The current $I$ flowing through the $10 Ω$ resistor is ________.

GATE 2022 EC question 14: Two sub-circuits joined by a single 10 Ω resistor. — Figure from the GATE 2022 EC paper. It is part of the question.

(A)
1 A
(B)
0 A
(C)
0.1 A
(D)
-0.1 A

Answer and solution

ANSWER

The current through the 10 Ω resistor is I = 0 A, option (B). The right-hand loop connects to the rest of the circuit through that one resistor and has no return path, so KCL forces the current to zero.

Official answer: BSource: Official answer key, GATE 2022 EC (IIT Kharagpur)

Look at how the parts are connected. The right loop (the $2 Ω$ resistor, the right-hand $3 V$ source and the $1 Ω$ resistor) has its own bottom wire. That wire is not joined to the left bottom wire or to ground. The only thing linking the right loop to the rest of the circuit is the $10 Ω$ resistor.
KCL on a closed surface drawn around the whole right loop. Exactly one conductor crosses that surface: the $10 Ω$ branch. The net current leaving a closed surface must be zero, so $I = 0$ .
Check with the node voltages. Take the left bottom wire as ground. With $I = 0$ , the left loop is a plain series circuit: $i = \frac{3}{1 + 2} = 1 A$ , so $V_{A} = 1 \times 2 = 2 V$ .
In the right loop, $i^{'} = \frac{3}{2 + 1} = 1 A$ , flowing out of the $+$ terminal, through the $2 Ω$ resistor into node B and down through the $1 Ω$ resistor. So node B sits $1 \times 1 = 1 V$ above the right bottom wire. Because $I = 0$ , there is no drop across the $10 Ω$ resistor, so $V_{B} = V_{A} = 2 V$ and the floating right bottom wire sits at $1 V$ . Everything is consistent: the voltage across the $10 Ω$ resistor is $0$ .
Therefore $I = 0 A$ , which is option (B).

KEY IDEA

Draw a closed surface around a subcircuit that is joined to the rest of the network by only one wire. KCL says the net current crossing any closed surface is zero, so the current in that single wire must be zero. A current needs a complete loop, and here there is none. The voltages on each side don't matter.

WATCH OUT FOR

Assuming the two bottom wires are both ground. Students then work out the open-circuit node voltages (V_A = 2 V, V_B = 1 V relative to each bottom wire) and write I = (2 − 1)/10 = 0.1 A, which is option (C), or −0.1 A, option (D), if they flip the direction. This is wrong for two reasons. The right bottom wire is floating, with no common return path. And even if a common return existed, the source resistances would load the circuit and give 1/(10 + 4/3) ≈ 0.088 A, not 0.1 A.

From the GATE 2022 EC paper. Every question of this paper · Download the full paper · All 25 Laws & node/mesh questions, with notes

GATE 2022 EC Q.14: solution and answer

Answer and solution

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