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(C) The circuit consists of two parallel branches connected to a $10 \, V$ battery. Branch $1$ contains an inductor $(0.1 \, H)$ and a resistor $(10 \

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$1$

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(C) The circuit consists of two parallel branches connected to a $10 \, V$ battery. Branch $1$ contains an inductor $(0.1 \, H)$ and a resistor $(10 \, \Omega)$ in series. Branch $2$ contains a resistor $(10 \, \Omega)$. At $t = 0$ (just after closing the switch),the inductor acts as an open circuit due to the back $EMF$. Thus,no current flows through the first branch. The total current in the circuit is only through the second branch: $i_{min} = \frac{10 \, V}{10 \, \Omega} = 1 \, A$. At $t = \infty$ (long after closing the switch),the inductor acts as a short circuit (zero resistance). The total current is the sum of currents in both branches: $i_{branch1} = \frac{10 \, V}{10 \, \Omega} = 1 \, A$ and $i_{branch2} = \frac{10 \, V}{10 \, \Omega} = 1 \, A$. Thus,$i_{max} = 1 \, A + 1 \, A = 2 \, A$. The difference between the maximum and minimum current is $i_{max} - i_{min} = 2 \, A - 1 \, A = 1 \, A$.

In the circuit shown,the switch $S$ is initially open. The switch $S$ is closed at $t = 0$. The difference between the maximum and minimum current that can flow in the circuit is.....$A$.

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