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GATE 2016 EE – Question 65

Electric Circuits · Network solution methods: KCL, KVL, Node and Mesh analysis · 2 marks · Numerical answer

In the circuit shown below, the node voltage $V_A$ is ________ V.

Node A has a 5 Ω resistor to ground, a 5 A current source into the node from ground, and a middle branch with a 5 Ω resistor in series with a dependent voltage source of $10 I_1$ (its + terminal at the ground side). A 5 Ω resistor carrying current $I_1$ leaves node A and goes through another 5 Ω resistor to a 10 V source, whose + terminal is at the top.

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Show answer and explanation

Correct answer: 11.25 to 11.50

Explanation

The right-hand branch gives $I_1 = \frac{V_A - 10}{10}$. The middle branch carries $\frac{V_A + 10I_1}{5}$, because its dependent source puts the lower end at $-10I_1$ relative to ground. KCL at node A gives $5 = \frac{V_A}{5} + \frac{V_A + 10I_1}{5} + I_1 = 0.4V_A + 3I_1$. Substituting $I_1$ gives $5 = 0.4V_A + 0.3V_A - 3$, so $0.7V_A = 8$ and $V_A = 11.43$ V.