Discussion
For the given circuit, determine I(t) in Laplace domain i.e. I(s).
Answer: A
When the switch closes at \(t = 0\), a 10 V step is applied to the series combination of the 1 ohm resistor and the 1 F capacitor, with no initial charge. In the Laplace domain the source is \(10/s\) and the impedance is \(R + 1/(sC) = 1 + 1/s\).
\[ \begin{aligned} I(s) = \frac{10/s}{1 + 1/s} \\ &= \frac{10/s}{(s+1)/s} = \frac{10}{s + 1} \end{aligned} \]
In the time domain this is \(i(t) = 10e^{-t}\), the current of the charging capacitor, which starts at 10 A and falls to zero. The option \(10/s\) would be the current in a purely inductive path, and \(10/(s-1)\) would grow without limit. The answer is option A.
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