NETWORKS (ELECTRIC CIRCUITS) · THEOREMS

GATE 2022 IN, question 54: Theorems — solution and answer

GATE 2022 · INQ.54 · 2 marks · NAT Checked against the official key

Given Circuit A with currents $I_{1}$ and $I_{2}$ as shown, the current $I_{3}$ in Circuit B (in amperes), is ________ (round off to one decimal place)

GATE 2022 IN question 54, as printed: Two separate ladder circuits, A and B, each with its own bottom rail (the reference node). — As printed in the GATE 2022 IN paper — the figure is part of the question.

Answer and worked solution

ANSWER

$I_{3} = 0.0 A$ . The 8 V source alone gives $+ 1 A$ (Circuit A at half the voltage), and the 48 V source alone gives $- 1 A$ (Circuit A reversed by reciprocity, at 3 times the voltage and pushing the other way). The two cancel.

Official key: 0.0 to 0.0 · Official answer key printed with the GATE 2022 IN paper (IIT Kharagpur)

Superposition. Circuit B is linear, so $I_{3} = I_{3}^{'} + I_{3}^{''}$ . Here $I_{3}^{'}$ is the current with only the 8 V source acting (48 V replaced by a short) and $I_{3}^{''}$ is the current with only the 48 V source acting (8 V replaced by a short). Each $R_{0}$ stays in its branch, so the passive network in both cases is exactly the resistor network of Circuit A.
8 V acting alone (homogeneity). With the 48 V shorted, the right branch is just $R_{0}$ , so the network is exactly Circuit A driven by 8 V instead of 16 V. Every current scales by the same factor: $I_{3}^{'} = \frac{8}{16} \times 2 = 1 A$
Reciprocity theorem. In Circuit A, 16 V in the left branch pushes current up that branch (in the $I_{1}$ direction) and produces $\frac{1}{3} A$ downward in the right branch, which is just $R_{0}$ . By reciprocity, 16 V placed in series with the right-hand $R_{0}$ and oriented to push current downward in that branch (+ terminal at the bottom) would produce $\frac{1}{3} A$ in the left branch, now just $R_{0}$ , flowing in the $I_{1}$ direction, which is the $I_{3}$ direction.
48 V acting alone. The 48 V source has its + terminal on top, so it pushes current upward in the right branch, opposite to the orientation used in step 3. Scale linearly and reverse the sign: $I_{3}^{''} = - \frac{48}{16} \times \frac{1}{3} = - 1 A$
Add the two parts. $I_{3} = I_{3}^{'} + I_{3}^{''} = 1 - 1 = 0 A$
Check by finding $R_{0}$ from Circuit A. Call the voltages at the three top nodes $V_{1}, V_{2}, V_{3}$ . With $I_{2} = \frac{1}{3}$ : $V_{3} = \frac{R _{0}}{3}$ and $V_{2} = V_{3} + 8 \cdot \frac{1}{3} = \frac{R _{0} + 8}{3}$ . The 4 Ω current is $\frac{1}{3} + \frac{V _{2}}{6}$ , so $V_{1} = V_{2} + 4 (\frac{1}{3} + \frac{V _{2}}{6}) = \frac{5 R _{0} + 52}{9}$ . KVL in the left branch gives $V_{1} = 16 - 2 R_{0}$ . Solving gives $R_{0} = 4 Ω$ and $V_{1} = 8 V$ . KCL at node 1 confirms it: $\frac{8}{8} + \frac{1}{3} + \frac{4}{6} = 2 A$ .
Nodal analysis on Circuit B with $R_{0} = 4 Ω$ . $node 1: \frac{V _{1} - 8}{4} + \frac{V _{1}}{8} + \frac{V _{1} - V _{2}}{4} = 0 node 2: \frac{V _{2} - V _{1}}{4} + \frac{V _{2}}{6} + \frac{V _{2} - V _{3}}{8} = 0 node 3: \frac{V _{3} - V _{2}}{8} + \frac{V _{3} - 48}{4} = 0$ The solution is $V_{1} = 8 V$ , $V_{2} = 12 V$ , $V_{3} = 36 V$ (node 1: $0 + 1 - 1 = 0$ ; node 2: $1 + 2 - 3 = 0$ ; node 3: $3 - 3 = 0$ ). Then $I_{3} = \frac{8 - V _{1}}{4} = 0 A$ , which matches step 5. To one decimal place, $I_{3} = 0.0 A$ .

KEY IDEA

Circuit B has the same resistor network as Circuit A, so you don't need $R_{0}$ . Use superposition. With only the left source acting, Circuit A is scaled by homogeneity: 8 V instead of 16 V gives half of $I_{1}$ . With only the right source acting, apply reciprocity to Circuit A: 16 V in the left branch gives $\frac{1}{3}$ A in the right branch, so 16 V in the right branch gives $\frac{1}{3}$ A in the left branch. Then scale by 48/16 and fix the sign from the source's polarity.

COMMON MISTAKE

The usual mistake is the sign of the 48 V contribution. Reciprocity needs the moved source to push current in the same sense as the reference direction of the branch whose current is measured, here downward through the right-hand $R_{0}$ . The 48 V source has + on top and pushes current upward, so its contribution is $- 1$ A, not $+ 1$ A. Getting this wrong gives 2.0 A. Two other slips: forgetting that the 8 V source gives only half of $I_{1}$ (1 A, not 2 A), and making algebra errors while solving for $R_{0}$ when it isn't needed.

From the GATE 2022 IN paper. Every question of this paper · Download the full paper · All 16 Theorems questions, with notes

GATE 2022 IN, question 54: Theorems — solution and answer

Answer and worked solution

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