Question

1. Find and plot the voltage volt) in the circuit shown. 18V+ 2. Find and plot the current io(t) in the circuit shown t-0 12 mA 300 ?F o(t)

Answer 1

t = 0 V_c(0) = 0 volts therefore V_0(t) = 0 volts @ t = 0 switch is connected to supply as t rightarrow infinity capacital reaches steady state and acts as open circuit therefore V_c (infinity) = 18 * 6 k \ 18 k V_c(infinity) = 6 volts & V_0 (infinity) = V_c(infinity) * 4000/6000 v_0 = 4 volts & time contant T = Re * C = 4000 * 100 * 10 ^ - 6 t = 0 middot 4 sec therefore t greaterthan 0 V_c(t) = V - c(infinity) - [v_c (infinity) - v_c(0)]e ^ - t \ t = 6 - [6 - 0]e ^ - t \ t v_c(t) = 6 - 6 e - t \ 0 middot 4 v _ c(t) = 6 - 6 e ^ - 2 middot 5 t volts & v _ 0 (t) = v_0 (infinity) - [v_0 (infinity) = v_0(0)] e ^ - t \ t v_0(t) = 4 - 4 e ^2 \r\nfor t < 0 switch is closed poxition for long time hence capacitor reaches its study state and acts as open circuits i_0(0) = 12 * 10 ^ - 3 * 6000 \ 9000 i_0(0) = 8 m amps v_c(0) = 8 * 10 ^ - 3 * 2000 = 16 volts at t = 0 switch is opend as t tightarrow infinity capacitor fully discharges & V_c (infinity) = 0 volts & i_0 (infinity) = 0 amps therefore t = RC = 2000 * 300 * 10^ - 6 therefore t greaterthan 0 v_c(t) = v_c (infinity) - [v_c (infinity) - v_c(0)]e ^ - t \ T = 0 - [0 - 16]e ^ - t 10 middot 6 v_c(t) = 16 e ^ - 1 middot 667t & i_0(t) = i_0(infinity) - [i - 0(infity) - i_0(0)]e ^ - t \ t = 0 - [0 - 8 * 10 ^ - 3]e ^ - t \ 0 middot 6 i_0(t) = 8 e - 1 middot 667t m amps (or)