GATE 2022 EE Q.55: solution and answer

GATE 2022 · EEQ.55 · 2 marks · NAT

In the circuit shown below, the magnitude of the voltage $V_{1}$ in volts, across the $8 k Ω$ resistor is ______. (round off to nearest integer)

GATE 2022 EE question 55: Two meshes that share a bottom ground wire. — Figure from the GATE 2022 EE paper. It is part of the question.

Answer and solution

ANSWER

$∣ V_{1} ∣ = 100 V$ .

Official answer: 98 to 102Source: Official answer key, GATE 2022 EE (IIT Kharagpur)

Look at the topology first. The middle vertical wire is a plain short from the node after the $0.5 V_{1}$ source to the bottom wire, so that node sits at ground potential. The circuit splits into two loops that affect each other only through the dependent sources.
Right loop. The dependent current source is in series with the $8 k Ω$ resistor, so it forces current $I$ left to right and then down through the resistor, entering at the $+$ end of $V_{1}$ . So $V_{1} = 8000 I$ .
Left loop (KVL, clockwise, closing through the short). Going up through the source is a rise of $75 V$ . The current $I$ gives a drop of $2000 I$ across the $2 k Ω$ resistor, and passing through the CCVS from $+$ to $-$ is a drop of $0.5 V_{1}$ : $75 - 2000 I - 0.5 V_{1} = 0$
Substitute $V_{1} = 8000 I$ : $75 = 2000 I + 0.5 (8000 I) = 6000 I \Rightarrow I = 12.5 mA$
Then $V_{1} = 8000 \times 12.5 \times 1 0^{- 3} = 100 V$ , so $∣ V_{1} ∣ = 100 V$ .

KEY IDEA

The short in the middle holds the node after the CCVS at ground, so the circuit splits into two loops. Only the dependent sources connect them. The right loop is a series path driven by the current source, which gives $V_{1} = 8000 I$ . Put that into the left loop's KVL and you have one equation in $I$ .

WATCH OUT FOR

One mistake is to treat the middle wire as if it carried the $8 k Ω$ current, or to think the short "kills" the right loop. The short only fixes a node voltage. The current source still pushes $I$ through the $8 k Ω$ resistor. A second mistake is to get the CCVS polarity wrong, adding $0.5 V_{1}$ as a rise instead of a drop when the loop is traversed from $+$ to $-$ . That gives $75 = 2000 I - 4000 I$ , so $I$ comes out negative and $∣ V_{1} ∣ = 300 V$ .

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

GATE 2022 EE Q.55: solution and answer

Answer and solution

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