In the circuit shown, epsilon = 6 V, R = 150 k Ohm, C_1 = 200 mu F, C_2 = 100 mu
ID: 1549588 • Letter: I
Question
In the circuit shown, epsilon = 6 V, R = 150 k Ohm, C_1 = 200 mu F, C_2 = 100 mu F, and C_3 = 400 mu F. (1 mu F = 10^-6 F.) The switch S is initially open and there is no charge on any of the capacitors. a) What is the equivalent capacitance between points A and B? b) The switch is now closed. What is the current through the resistor just after the switch is closed? c) What is the current through the resistor 5 seconds after the switch is closed? d) We wait a longtime after the switch is closed. What is the voltage across capacitor C_1? e) What is the total potential energy stored in the capacitors at this time?Explanation / Answer
a) C2 , C3 in series will equivallent to Ceq1 = C2*C3/(C2+C3) = 100*400/(100+400) = 80 micro farad
now Ceq1 , C1 in parallel will equivallent to Ceq(AB) = Ceq1 + C1 = 80 + 200 = 280 micro farad
b) just after the switch is closed capacitor will act as a short circuit
=> i = E/R = 6/(150*10^3) = 40*10^-6 A
c) current in the RC circuit
i(t) = i*e^(-t/RCeq)
after t = 5 sec
i(t) = 40*10^-6*e^(-5/(150*10^3*280*10^-6)) = 35.51*10^-6 A
d) after long time voltage across C1 will be equal to E volts
=> voltage across Ceq1 will be E volts
=> voltage across C3 will be V = C2*E/(C2+C3) = 100*6/500 = 1.2 volts
e) total potential energy stored in the capacitor C3 at this time
PE = 0.5*C3*V^2 = 0.5*400*10^-6*(1.2)^2 = 288*10^-6 J