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

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Dear Student, 
                      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 S₁ and S₂ must be equal to the ratio of P to Q.
   
                                                                $\frac{P}{Q}= \frac{R}{S} [Here S = \frac{S_1\times S_2}{S_1+S_2}]$
So, The correct option is (c). P/Q = R (S₁S₂) / (S₁+ S₂)
Regards