The Wheatstone bridge shown in the diagram is | vedclass

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(A) For a balanced Wheatstone bridge,the condition is given by $\frac{R_1}{L} = \frac{R_3}{K}$,which implies $\frac{R_3}{R_1} = \frac{K}{L}$.Since the

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$\frac{K}{L}$

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(A) For a balanced Wheatstone bridge,the condition is given by $\frac{R_1}{L} = \frac{R_3}{K}$,which implies $\frac{R_3}{R_1} = \frac{K}{L}$.Since the bridge is balanced,no current flows through the galvanometer. Thus,$R_1$ and $R_3$ are in series with the same current $I_1$ flowing through them.The power dissipated by a resistor is given by $P = I^2 R$.Therefore,the power dissipated by $R_3$ is $P_3 = I_1^2 R_3$ and the power dissipated by $R_1$ is $P_1 = I_1^2 R_1$.The ratio of the powers is $\frac{P_3}{P_1} = \frac{I_1^2 R_3}{I_1^2 R_1} = \frac{R_3}{R_1}$.Substituting the condition from the balanced bridge,we get $\frac{P_3}{P_1} = \frac{K}{L}$.

The Wheatstone bridge shown in the diagram is balanced. If $P_3$ is the power dissipated by $R_3$ and $P_1$ is the power dissipated by $R_1$,then the ratio $\frac{P_3}{P_1}$ is

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