In this circuit, VS = 350V, RS = 1000 ohms, C = 330 uF, and RL = 10 ohms. The sw

Question

In this circuit, VS = 350V, RS = 1000 ohms, C = 330 uF, and RL = 10 ohms. The switch has been in the position shown for a very long time.

At t = 0 the switch is flipped to the left.

What is the voltage on the capacitor at t = 0.7, in Volts?

What is the energy stored in the capacitor at t = 0.7 seconds, in Joules?

How much energy was dissipated in the resistor RS between t = 0 and t = 0.7, in Joules?

At t = 0.7 the switch is flipped to the right.

Just after the flip, what is the current through the resistor RL, in Amperes?

What is the voltage VL, in Volts at time t = 0.71 seconds?

Explanation / Answer

AT LEFT FLIP voltage across capacitor: Vc(t) = Vs(1 - e^(-t/RsC) ) Vc(t) = 350(1 - e^(-t/0.330) ) Vc(t=0.7) = 308.04 V energy stored = (1/2)CVc^2 = (1/2)(330*10^-6)*(308.04)^2 J = 15.6566259 J voltage across Rs: Vrs = Vs - Vc Vrs(t) = Vs*e^-(t/RC) Vrs(t) = 350*e^(-t/0.33) now, energy dissipated: E = integral(Vrs^2/Rs)dt from t=0 to 0.7 = integral(350^2/1000)*(e^(-2t/0.33))dt from t=0 to 0.7 = (350^2/1000)*(1-e^(-2*0.7/0.33))/(2/0.33) = 19.92199170124481 J AFTER RIGHT SWITCH potential across capacitor doesn't chnage instantly so, Vc(0.7+) = 308.04 V => iL(0.7+)= Vc/RL = 308.04/10 = 30.804 A the discharging equation will be : Vc(t) = 308.04*e^(-t/C*RL) = 308.04*e^(-(t-0.7)/0.0033) Vc(0.71) = 308.04*e^-(0.01/0.0033) = 14.88V

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