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(B) The given circuit is a Wheatstone bridge. The condition for a balanced Wheatstone bridge is $\frac{P}{Q} = \frac{R}{S}$.Here,$P = 1 \, \Omega$,$Q

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$4$

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(B) The given circuit is a Wheatstone bridge. The condition for a balanced Wheatstone bridge is $\frac{P}{Q} = \frac{R}{S}$.Here,$P = 1 \, \Omega$,$Q = 3 \, \Omega$,$R = 7 \, \Omega$,and $S = 21 \, \Omega$.Since $\frac{1}{3} = \frac{7}{21}$,the bridge is balanced.In a balanced Wheatstone bridge,no current flows through the central resistor $(10 \, \Omega)$. Therefore,the equivalent resistance $R_{eq}$ is the series combination of $(1 \, \Omega + 3 \, \Omega)$ in parallel with $(7 \, \Omega + 21 \, \Omega)$.$R_{eq} = \frac{(1+3) 	imes (7+21)}{(1+3) + (7+21)} = \frac{4 	imes 28}{4 + 28} = \frac{112}{32} = 3.5 \, \Omega$.The voltage of the battery is $V = I 	imes R_{eq} = 4 \, A 	imes 3.5 \, \Omega = 14 \, V$.When the $10 \, \Omega$ resistor is replaced by a $20 \, \Omega$ resistor,the bridge remains balanced because the ratio $\frac{P}{Q} = \frac{R}{S}$ is still satisfied.Thus,no current flows through the $20 \, \Omega$ resistor,and the equivalent resistance of the circuit remains $3.5 \, \Omega$.The new current drawn from the battery is $I' = \frac{V}{R_{eq}} = \frac{14 \, V}{3.5 \, \Omega} = 4 \, A$.

In the circuit shown in the figure,the current drawn from the battery is $4 \, A$. If the $10 \, \Omega$ resistor is replaced by a $20 \, \Omega$ resistor,then the current drawn from the circuit will be .............. $A$.

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