In a wheat stone's bridge, three reststance \(P, Q\) and \(R\) connected in three arm a and the fourth arm is formed by two resistances \(\mathrm{S}_{1}\) and \(\mathrm{S}_{2}\) connected in parallel The condition for bridge to be balanced will be. (A) $(\mathrm{P} / \mathrm{Q})=\left\\{\mathrm{R} /\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right)\right\\}$ (B) $(\mathrm{P} / \mathrm{Q})=\left\\{(2 \mathrm{R}) /\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right)\right\\}$ (C) $(\mathrm{P} / \mathrm{Q})=\left[\left\\{\mathrm{R}\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right)\right\\} /\left\\{\mathrm{S}_{1} \mathrm{~S}_{2}\right\\}\right]$ (D) $(\mathrm{P} / \mathrm{Q})=\left[\left\\{\mathrm{R}\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right)\right\\} /\left\\{2 \mathrm{~S}_{1} \mathrm{~S}_{2}\right\\}\right]$

Step by step solution

01

Recall Wheatstone Bridge principle

A Wheatstone bridge is balanced when the ratio of the resistances in the bridge is equal, that is, \(\frac{P}{Q} = \frac{R}{S}\), where S is the equivalent resistance of the parallel combination of S1 and S2.

02

Calculate equivalent resistance of S1 and S2 in parallel

The equivalent resistance S of two resistances connected in parallel can be found using the formula: \[\frac{1}{S} = \frac{1}{S_1}+\frac{1}{S_2}\] We can solve this equation for S to find the equivalent resistance: \[S = \frac{S_1S_2}{S_1+S_2}\]

03

Substitute equivalent resistance into Wheatstone Bridge principle

Now, substitute the equivalent resistance S into the Wheatstone bridge balance condition: \(\frac{P}{Q} = \frac{R}{S}\) Substitute S with the equivalent resistance calculated in the previous step: \[\frac{P}{Q} = \frac{R}{\frac{S_1S_2}{S_1+S_2}}\]

04

Simplify the expression

Finally, simplify the expression by multiplying both sides by \((S_1+S_2)\): \[\frac{P}{Q} = \frac{R(S_1+S_2)}{S_1S_2}\] This expression shows the condition for the bridge to be balanced, which matches option (C).

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