In the case of a Wheatstone bridge,the balanc | vedclass

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(D) In a balanced Wheatstone bridge,the galvanometer shows zero deflection,meaning no current flows through the galvanometer branch $(BD)$.This implie

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$P / Q = R / S$

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(D) In a balanced Wheatstone bridge,the galvanometer shows zero deflection,meaning no current flows through the galvanometer branch $(BD)$.This implies that the potential at point $B$ is equal to the potential at point $D$ $(V_B = V_D)$.Therefore,the potential difference across $AB$ is equal to the potential difference across $AD$:$I_1 P = I_2 R$  --- $(i)$Similarly,the potential difference across $BC$ is equal to the potential difference across $DC$:$I_1 Q = I_2 S$  --- $(ii)$Dividing equation $(i)$ by equation $(ii)$:$\frac{I_1 P}{I_1 Q} = \frac{I_2 R}{I_2 S}$$\frac{P}{Q} = \frac{R}{S}$

In the case of a Wheatstone bridge,the balanced condition is.....

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