For the circuit shown the switch has been open for a very lo

For the circuit shown, the switch has been open for a very long time before closing at t = 0. Find i(0^+), v(0^+), i(infinity), v(infinity), di/dt|_i = 0, and dv/dt|_i=0^+. Given: L = 0.2 H, C = 20 mu F, and V_s = 30 V.

Solution

Ans)

For t<0 the circuit is in steady state ,inductor acts as short circuit and capacitor acts as open circuit ,using this we can find the initial current and voltages

i(0^-)=Vs/(50+100+200)=30/350=85.71 mA

i(0^-)=85.71 mA

As inductor current cannot change instantaneously

i(0⁺)=i(0^-)=85.71 mA

i(0⁺)=85.71 mA

--------------------------------

Voltage across 200 ohm resistor is equal to voltage across capacitor,so

v(0⁺)=i(0⁺)*200=85.71m*200=17.143 V

v(0⁺)=17.143 V

----------------------------------

For t>0 the steady state currents are when switch is closed 100 ohm resistor is shorted,so

i(infinity)=Vs/(50+200)=30/250=120 mA

i(infinity)=120 mA

------------------------

v(infinity)=i(infinity)*200=120m *200=24 V

v(infinity)=24 V

-----------------------

The voltage across inductor for t=0+ is

V_L=Vs-i(0⁺)*(50+200)=30-85.714m*(250)=30-21.428=8.571 V

V_L=8.571 V

Now using the relation V_L=L di/dt

di/dt(0+)=V_L/L=8.571/0.2=42.857 A/s

di/dt(0+)=42.857 A/s

------------------------------------------

As current through the capacitor is zero for t=0+

dv/dt=i/C=0