Question

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 A

I 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 1

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.