Compute the average and effective values of the square voltage wave shown in the following figure.

For t<0,  the switch is open.

Capacitor acts open circuit to DC.

Capacitor is charging.

Voltage across the Capacitor, 

   v(0^-)= {(12||4)×V} ÷ { (12||4)+6} = 8V

At t=0, switch is opened.

Since voltage across capacitor cannot change instantaneously,

Hence v(0)=v(0^-)= 8V

At t>0, capacitor is discharging, 

R_th = (12||4) = 3 Ohm

C= 1/6F

Time constant, Π=R_thC = 0.5

Voltage across the capacitor,

v(t) = v(0)e^-t/Π

        = 8e^-t/0.5 V (Ans.)

9 months after

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The initial energy stored in the Capacitor,

W=(1/2)CV_o² 

    = 0.5×0.167×8² 

    = 5.333J (Ans)

9 months after

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