In a Wheatstone's bridge, three resistances P,Q and R connected in the three arms and the fourth arm is formed by two resistances `S_1 and S_2` connected in parallel. The condition for the bridge to be balanced will be

Options

A. `(P)/(Q)=(R(S_(1)+S_(2)))/(S_(1)+S_(2))`
B. `(P)/(Q)=(R )/(S_(1)+S_(2))`
C. `(P)/(Q)=(R(S_(1)+S_(2)))/(S_(1)S_(2))`
D. `(P)/(Q)=(R(S_(1)+S_(2)))/(2S_(1)S_(2))`

Answer

C

Explanation

For the fourth arm, `(1)/(S)=(1)/(S_1)+(1)/(S_(2))`
`therefore S= (S_(1)S_(2))/(S_(1)+S_(2))`
`therefore " For balance " (P)/(Q)=(R )/(S) therefore (P)/(Q)=(R(S_(1)+S_(2)))/(S_(1)S_(2))`