Four resistances of 15, Omega, 12, Omega, 4, | vedclass

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(B) Let the resistance to be connected in parallel with the $10\, \Omega$ resistor be $R_p$.For a balanced Wheatstone bridge, the ratio of resistances

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$10$

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(B) Let the resistance to be connected in parallel with the $10\, \Omega$ resistor be $R_p$.For a balanced Wheatstone bridge, the ratio of resistances in adjacent arms must be equal: $\frac{P}{Q} = \frac{R}{S}$.Here, the resistances are $15\, \Omega, 12\, \Omega, 4\, \Omega$ and the effective resistance of $10\, \Omega$ in parallel with $R_p$, which is $R_{eff} = \frac{10 R_p}{10 + R_p}$.Setting the bridge balance condition: $15 	imes 4 = 12 	imes R_{eff}$.$60 = 12 	imes \frac{10 R_p}{10 + R_p}$.$5 = \frac{10 R_p}{10 + R_p}$.$5(10 + R_p) = 10 R_p$.$50 + 5 R_p = 10 R_p$.$5 R_p = 50$.$R_p = 10\, \Omega$.

Four resistances of $15\, \Omega, 12\, \Omega, 4\, \Omega$ and $10\, \Omega$ are connected in cyclic order to form a Wheatstone bridge network. The resistance that is to be connected in parallel with the resistance of $10\, \Omega$ to balance the network is ................. $\Omega$.

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