Correct Option: A

Given i (t) = – 20 e^{– 2t}
V (t) = – 20 e^{– 2t} × 1 = – 20 e^{– 2t}

iC = C	dV (t)
	dt

= 2.	d	(– 20 e– 2t)
	dt

or i_C = 80 e^{– 2t}
I = i (t) + i _C = – 20 e^{– 2t} + 80 e^{– 2t}
= 60 e^{– 2t}
V_s = V_R + V_L + V (t)

= 1 × 60 e– 2t + L	d	i + V (t)
	dt

= 1 × 60 e– 2t +		1		d	60 e– 2t – 20 e– t
		4		dt

= 60 e_{– 2t} – 15 × 2 e_{– 2t} – 20 e_{– 2t}
= 60 e_{– 2t} – 30e_{– 2t} – 20 e_{– 2t}
= 10e_{– 2t}