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The current in the above network is

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  1. |t - 1 + e-t| u(t)
  2. |t2 - 1 + e-t| u(t)
  3. |t + 1 + e-t| u(t)
  4. |t - 1 - e-t| u(t)

Answer (Detailed Solution Below)

Option 1 : |t - 1 + e-t| u(t)
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Detailed Solution

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V(t) = t u(t)

The circuit can be drawn in the Laplace domain as follows:-

Z(s) = s + 1

By taking inverse Laplace transform, we get

i(t) = -u(t) + t u(t) + e-t u(t)

= (t – 1 + e-t) u(t)

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