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(D) Case $1$: When the key $(K)$ is open,the circuit consists of the cell $(12 \ V)$ and one resistor of $40 \ \Omega$ in series with the ammeter. The

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$1: 2$

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(D) Case $1$: When the key $(K)$ is open,the circuit consists of the cell $(12 \ V)$ and one resistor of $40 \ \Omega$ in series with the ammeter. The current $i_1$ is given by Ohm's law: $i_1 = V / R = 12 / 40 = 0.3 \ A$.Case $2$: When the key $(K)$ is closed,the two $40 \ \Omega$ resistors are connected in parallel. The equivalent resistance $R_{eq}$ is: $1 / R_{eq} = 1 / 40 + 1 / 40 = 2 / 40 = 1 / 20$,so $R_{eq} = 20 \ \Omega$.The current $i_2$ is: $i_2 = V / R_{eq} = 12 / 20 = 0.6 \ A$.Therefore,the ratio $i_1: i_2 = 0.3: 0.6 = 1: 2$.

In the given circuit,the internal resistance of the cell is zero. If $i_1$ and $i_2$ are the readings of the ammeter when the key $(K)$ is opened and closed respectively,then $i_1: i_2=$

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