InterviewSolution
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InterviewSolution
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Calculate the equivalent resistance of the network across the points A and B shown in figure. |
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Answer» Here in this diagram, say: R1 = 4Ω R2 = 8Ω R3 = 4Ω R4 = 8Ω When we want to find the resistance across points A and B then resistance R1 and R2 are connected in series and resistances R3 and R4 are also connected in series. Thus, R5 = R1 + R2 = 4Ω + 8Ω = 12Ω And, R6 = R3 + R4 = 4Ω + 8Ω = 12Ω Now, this R5 and R6 are connected to each other in parallel. Thus the equivalent resistance RAB across points A and B is: \(\frac{1}{R_{AB}}=\frac{1}{R_5}+\frac{1}{R_6}\) \(\frac{1}{R_{AB}}=\frac{1}{12}+\frac{1}{12}\) \(\frac{1}{R_{AB}}=\frac{2}{12}+\frac{1}{6}\) ∴ RAB = 6Ω |
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