What is the equivalent resistance between points A and C in the circuit shown in fig.?

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$2 R$

$\frac{4 R}{3}$

$\frac{5 R}{12}$

$\frac{7 R}{5}$

answer is D.

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Detailed Solution

In the loop ABDA,

$I_{1} R_{1} + (2 I_{1} - I) (R_{1}) - (I - I_{1}) (R_{2}) = 0$

$I_{1} [3 R_{1} + R_{2}] - I [R_{1} + R_{2}] = 0 :$

$I_{1} = I \frac{(R_{1} + R_{2})}{3 R_{1} + R_{2}} - - - - - (1)$

In the loop ABCEA,

$I_{1} R_{1} + (I - I_{1}) (R_{2}) - E = 0$

$I_{1} [R_{1} - R_{2}] + {IR}_{2} = E$

$\Rightarrow I \frac{(R_{1} + R_{2})}{3 R_{1} + R_{2}} (R_{1} - R_{2}) + {IR}_{2} =$

$\Rightarrow I [\frac{(R_{1}^{2} - R_{2}^{2}) + R_{2} (3 R_{1} + R_{2})}{3 R_{1} + R_{2}}] = E$

$\Rightarrow E = I [R_{1} \frac{(R_{1} + 3 R_{2})}{3 R_{1} + R_{2}}]$

$∴ R_{eff} = \frac{R_{1} (R_{1} + 3 R_{2})}{3 R_{1} + R_{2}} = \frac{R [R + 3 (2 R)]}{3 R_{1} + 2 R} = \frac{7 R}{5} .$

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