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GATE 2019 EE – Question 52

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

The current $I$ flowing in the circuit shown below in amperes is ________.

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Correct answer: 0

Explanation

Each branch has a source-to-resistance ratio of $\frac{200}{50}=\frac{160}{40}=\frac{100}{25}=\frac{80}{20}=4$ A, so every branch behaves as a $4$ A Norton source in parallel with its resistor. With the polarities shown, two of these point one way and two the other, so the net source current into the common node is zero, the node voltage is zero, and no current flows through the $20\ \Omega$ resistor: $I=0$.