4. A series RC circuit has a time constant of 0.960 s. The battery has an emf of

Question

4. A series RC circuit has a time constant of 0.960 s. The battery has an emf of 48.0 V, and the maximum current in the circuit is 0.500 mA. What are (a) the value of the capacitance and (b) the charge stored in the capacitor 1.92 s after the switch is closed? 5. A capacitor in an RC circuit is charged to 60.0% of its maximum value in 0.900 s. What is the time constant of the circuit. 5,0 10 6. Find the currents I, 12, I3 in the circuit. 7. A 25-nF capacitor is in series with a 1.45 M resistor. The capacitor is initially uncharged when the series combination is connected across a 24.0-V battery. (a) What s the maximum charge the capacitor will eventually attain? (b) How much time does it take for the capacitor to reach 90% of that maximum charge? 8. How many time constants does it take for a charging capacitor to reach (a) half of its maximum charge? (b) half its maximum stored energy? 0

Explanation / Answer

4. Given

RC circuit with

time constant T = R*C = 0.960 s

battery has emf E = 48.0 V

maximum current i = 0.5 mA

we know that the current i = V/R = E/R ==> R = E/i = 48/(0.5*10^-3) ohm = 96000 ohm = 96 k ohm

a) the value of capacitane is T = R*C ==> C= T/R = 0.960/96000 F = 10 micro farad

b) charge stored in the capacitor after a time 1.92 s is  

q(t) = Q0(1-e^-t/T)

here Q0 is total charge Q0 = c*E = 10*10^-6*48 C = 48*10^-6 C

q(t)= 48*10^-6(1-e^(-1.92/(0.96))) C

q(t) = 41.5039*10^-6 C

5. Given q(t) = 0.6*Q0 , t = 0.90 s , T = ?

q(t) = Q0(1-e^-t/T)

0.6*Q0 = Q0(1-e^(-0.9/T))

solving for T

T = 0.9822 s

so the time constant is T = 0.9822 s

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