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In a Wheatstone bridge, three resistance...

In a Wheatstone bridge, three resistance, P, Q and R are connected in the three arms and fourth arm is formed by two resistance `s_(1) and s_(2)` connected in parallel. The condition for the bridge to be balanced for ratio `((p)/(Q))` is (given `R=10Omega, S_(1)=6Omega, S_(2)=3Omega`)

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The correct Answer is:
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If S is the equivalent resistance of `s_(1) and s_(2)` in parallel, then `S=(S_(1)S_(2))/(S_(1)+S_(2))`. For Wheat stone bridge to be balanced `(P)/(Q)=(R)/(S)or" "(P)/(Q)=(R)/(S_(1)S_(2)//(S_(1)+S_(2)))=(R(S_(1)+S_(2)))/(S_(1)S_(2))`
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