Question

Find the voltage drop across 6Ω resistor in the circuit shown below.

(a) 6.58

(b) 7.58

(c) 8.58

(d) 9.58

The question was posed to me in an international level competition.

Query is from Norton’s Theorem topic in chapter Useful Theorems in Circuit Analysis of Network Theory

Answer 1

Correct answer is (c) 8.58

For explanation I would say: The voltage across the 6Ω resistor is V = 1.43×6 = 8.58V. So the current and voltage have the same values both in the original circuit and Norton’s equivalent circuit.