What is the output voltage v₀ (t) of the circuit shown in the figure?

Question

Question

This question was previously asked in

UPSC ESE Electronics & Communication 2022 Official Paper

Answer 1

i_s(t) = 3e̅^t. u(t) mA

i_s(t) = i_R+ i_C.

$i_{s} (t) = \frac{1}{R} v_{o} (t) + c \frac{d v o (t)}{d t}$

$c \frac{d v o (t)}{d t} + \frac{1}{R} v_{o} (t) = i_{s} (t)$

$\frac{d_{v} o (t)}{d t} + \frac{1}{R} v_{o} (t) = \frac{3 {̅ e}^{t} u (t)}{C}$

Applying Laplace,

$S V_{o} (s) + \frac{1}{R C} V_{o} (s) = \frac{3}{c} [\frac{1}{s + 1}]$

$v_{o} (s) [\frac{S R C + 1}{R C}] = \frac{3}{c} [\frac{1}{s + 1}]$

$v_{o} (s) = 3 R [\frac{1}{(S + 1) (S R C + 1)}]$ R = 10 k, C = 25μ f

$= \frac{30}{0.25} [\frac{1}{(s + 1) (s + 4)}]$

$120 = [\frac{A}{S + 1} \frac{B}{S + 4}]$

SA + 4A + BS + B = 1

S (A + B) + 4A + B = 1

A + B = 0, A = - B

4A + B = 1, 3A = 1, $A = \frac{1}{3}, B = \frac{- 1}{3}$

$v_{o} (s) = 120 [\frac{1}{3 (s + 1)} - \frac{1}{3 (s + 4)}]$

$40 [\frac{1}{(s + 1)} - \frac{1}{(s + 4)}]$

on Applying Inverse Laplace

v_o(t) = 40 [e̅^t. u(t) - e̅^ut u(t)]