Four resistance of $15 Ω, 12 Ω, 4 Ω a n d 10 Ω$ respectively in cyclic order to form Wheatstone’s network. The resistance to be connected parallel to $10 Ω$ to balance the network is

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$20 Ω$

$10 Ω$

$30 Ω$

$40 Ω$

answer is B.

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Detailed Solution

$\frac{P}{Q} = \frac{R}{S}$

$\begin{array}{l} \frac{15}{12} = \frac{10 R^{1}}{\frac{10 + r^{1}}{4}} \Rightarrow 5 = \frac{10 R^{1}}{\begin{array}{l} 10 \\ r^{1} \end{array}} \\ = 50 + 5 R^{1} = 10 R^{1} \\ R^{1} = \frac{50}{5} = 10 Ω \end{array}$

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