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(A) Let the potential at $A$ be $V_A = 6\,V$ and at $C$ be $V_C = 0\,V$.The circuit can be redrawn as a Wheatstone bridge where the potential at $B$ i

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(A) Let the potential at $A$ be $V_A = 6\,V$ and at $C$ be $V_C = 0\,V$.The circuit can be redrawn as a Wheatstone bridge where the potential at $B$ is determined by the potential divider rule in the branch $ABC$:$V_B = V_A - I_1 	imes 5\,\Omega = 6 - \left(\frac{6}{5+15}ight) 	imes 5 = 6 - \left(\frac{6}{20}ight) 	imes 5 = 6 - 1.5 = 4.5\,V$.The potential at $D$ is determined by the potential divider rule in the branch $ADC$:$V_D = V_A - I_2 	imes 10\,\Omega = 6 - \left(\frac{6}{10+30}ight) 	imes 10 = 6 - \left(\frac{6}{40}ight) 	imes 10 = 6 - 1.5 = 4.5\,V$.The potential difference between $B$ and $D$ is $V_B - V_D = 4.5\,V - 4.5\,V = 0\,V$.

Four resistors are connected as shown in the figure. $A$ $6\,V$ battery of negligible resistance is connected across terminals $A$ and $C$. The potential difference across terminals $B$ and $D$ will be ............. $V$.

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