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In the circuit given below, the current in the resistance 20 Ω(far end) is _________(a) 8.43 A(b) 5.67 A(c) 1.43 A(d) 2.47 AI got this question during an interview.Asked question is from Advanced Problems on Reciprocity Theorem topic in portion Useful Theorems in Circuit Analysis of Network Theory

Answer»

Right ANSWER is (c) 1.43 A

Explanation: EQUIVALENT Resistance REQ = 20 + [30 || (20 + (20||20))]

= 20 + [30 || (20 + \(\frac{20×20}{20+20}\))]

= 20 + [30 || (20+10)]

= 20 + [30 || 30]

= 20 + \(\frac{30 × 30}{30+30}\)

= 20 + 15 = 35 Ω

The current DRAWN by the circuit = \(\frac{200}{35}\) = 5.71 A

Now, by USING current division rule, we get, I2Ω = 1.43 A.



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