A 4.15-muF capacitor that is initially uncharged is connected in series with a 7

Question

A 4.15-muF capacitor that is initially uncharged is connected in series with a 7.55-kOhm resistor and an emf source with epsilon = 140 V and negligible internal resistance. (a) Just after the circuit is completed, what is the voltage drop across the capacitor? V (b) Just after the circuit is completed, what is the voltage drop across the resistor? V (c) Just after the circuit is completed, what is the charge on the capacitor? C (d) Just after the circuit is completed, what is the current through the resistor? A (e) A long time after the circuit is completed (after many time constants) what are the values of the quantities in parts (a) - (d)? V_c = V V_R = V q = C I = A

Explanation / Answer

Given: capacitance of the capacitor = 4.15 uF

resistance = 7.55 kohm

volatge of emf source = 140 V

a) voltage drop across the capacitor

initially there is no charge on the capacitor plates

hence voltage drop = 0

b) volatge drop across the resistor

= voltage of the source = 140 V

c) charge on the capacitor initially

=0

d) current through the resistor

= V/R = 140 / 7550 = 0.018 A

e) volatge equation for capacitor

V = Vs(1-e^-t/RC)

after a long time

V = Vs = 140 V

Vr = 0

q = CV = 4.15*10^-6*140 = 581 uC

since voltage through the resistor is 0 , current flowing through the circuit will be zero after a long time

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