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(A) In the given circuit,the points $A$ and $B$ are short-circuited by the wire $AB$. This means the potential difference across the $4 \ \Omega$ and

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Power supplied by the battery is $300 \ W$

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(A) In the given circuit,the points $A$ and $B$ are short-circuited by the wire $AB$. This means the potential difference across the $4 \ \Omega$ and $12 \ \Omega$ resistors is zero,so no current flows through them. The circuit effectively consists of only the $3 \ \Omega$ resistor connected in series with the $30 \ V$ battery. The total resistance of the circuit is $R = 3 \ \Omega$. The current flowing in the circuit is $I = \frac{V}{R} = \frac{30 \ V}{3 \ \Omega} = 10 \ A$. The power supplied by the battery is $P = V 	imes I = 30 \ V 	imes 10 \ A = 300 \ W$. Since the current in the wire $AB$ is the total current of the circuit,it is $10 \ A$,not zero. Therefore,only option $A$ is correct.

Considering the following circuit,select the correct alternative.

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