The current (I) drawn from the battery in the | vedclass

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(B) To find the current drawn from the battery,we first determine the equivalent resistance $(R_{eq})$ of the circuit between points $A$ and $D$.$1$.

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$0.5$

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(B) To find the current drawn from the battery,we first determine the equivalent resistance $(R_{eq})$ of the circuit between points $A$ and $D$.$1$. The $5 \ \Omega$ resistor is between $A$ and $B$. The $10 \ \Omega$ resistor is between $A$ and $C$. The $10 \ \Omega$ resistor is between $B$ and $C$. This forms a delta connection between $A, B, C$. Converting the delta $(5 \ \Omega, 10 \ \Omega, 10 \ \Omega)$ to star,the equivalent resistances are $R_A = (5 	imes 10) / (5 + 10 + 10) = 2 \ \Omega$,$R_B = (5 	imes 10) / 25 = 2 \ \Omega$,and $R_C = (10 	imes 10) / 25 = 4 \ \Omega$.$2$. Now,the circuit simplifies: $R_A$ is in series with the parallel combination of $(R_B + 10 \ \Omega)$ and $(R_C + 20 \ \Omega)$.$3$. The parallel branch is between $B$ and $D$. The resistance of the upper branch is $2 + 10 = 12 \ \Omega$. The resistance of the lower branch is $4 + 20 = 24 \ \Omega$.$4$. The equivalent resistance of these two parallel branches is $R_p = (12 	imes 24) / (12 + 24) = 288 / 36 = 8 \ \Omega$.$5$. The total equivalent resistance $R_{eq} = R_A + R_p = 2 \ \Omega + 8 \ \Omega = 10 \ \Omega$.$6$. The current drawn from the battery is $I = V / R_{eq} = 5 \ V / 10 \ \Omega = 0.5 \ A$.

The current $(I)$ drawn from the battery in the given circuit is (in $A$)

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