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In a wheatstone bridge 3 resistances P,Q& R connected in three arms and the fourth arm is formed by 2 resistances S1 and S2 connected in parallel what is the condition for the bridge to be balanced

a P/Q=R/S1+S2

b P/Q=2R/S1=S2

c P/Q=r(S1+S2)/S1S2

d P/q=R(S1=S2)/2S1S2

As we know to balance the wheatstone bridge, condition is that P / Q = R / S (resistance applied externally).

In the given figure attached it is clearly seen that the ratio of Resistance R to the effective resistance(S) of parallel combination of S

_{1}

_{ }and S

_{2}must be equal to the ratio of P to Q.

$\frac{P}{Q}=\frac{R}{S}[HereS=\frac{{S}_{1}\times {S}_{2}}{{S}_{1}+{S}_{2}}]$

So, The correct option is (c). P/Q = R (S

_{1}S

_{2}) / (S

_{1}+ S

_{2})

Regards

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