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Four resistances of $15 Ω, 12 Ω, 4 Ω and 10 Ω$ respectively in cyclic order to form Wheatstone’s network. The resistance that is to be connected in parallel with resistance of $10 Ω$ to balance the network is… $Ω$

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answer is 10.

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Detailed Solution

balance condition, $\frac{P}{Q} = \frac{R}{S}$

$\frac{15}{12} = \frac{(\frac{10 x}{10 + x})}{4} = \frac{10 x}{4 (10 + x)}$

$10 + x = 2 x$

$x = 10 Ω$

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