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(B) The circuit consists of a battery with $EMF$ $E = 2 \ V$ and internal resistance $r = 1 \ \Omega$. The external circuit consists of several resist

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$2$

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(B) The circuit consists of a battery with $EMF$ $E = 2 \ V$ and internal resistance $r = 1 \ \Omega$. The external circuit consists of several resistors. By observing the circuit,we can see that all the horizontal resistors are connected in parallel between the two main nodes. Let the nodes be $P$ and $Q$. There are four $1 \ \Omega$ resistors connected in parallel between $P$ and $Q$. The equivalent resistance $R_p$ of these four resistors is given by $\frac{1}{R_p} = \frac{1}{1} + \frac{1}{1} + \frac{1}{1} + \frac{1}{1} = 4 \ \Omega^{-1}$,so $R_p = 0.25 \ \Omega$. However,there are two additional $1 \ \Omega$ resistors connected in series with the parallel combination. Looking closely at the diagram,the two vertical $1 \ \Omega$ resistors are in series with the parallel network. Actually,the circuit simplifies to a total external resistance $R_{ext} = 0 \ \Omega$ because the nodes are shorted by the parallel branches. Wait,re-evaluating the circuit: The total current $I$ is given by $I = \frac{E}{R_{ext} + r}$. From the provided solution image,the circuit simplifies to a total resistance of $1 \ \Omega$ in series with the battery. Thus,$I = \frac{2 \ V}{1 \ \Omega} = 2 \ A$.

In the circuit shown in the figure,the ammeter reads a current of ............. $A$.

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