Question

A 4.60-mu F capacitor that is initially uncharged is connected in series with a 7.50-k ohm resistor and an emf source with epsilon = 245 V and negligible internal resistance. Just after the circuit is completed, what arc (a) the voltage drop across the capacitor; (b) the voltage drop across the resistor; (c) the charge on the capacitor; (d) the current through the resistor? (c) A long time after the circuit is completed (after many time constants) what are the values of the quantities in parts (a)-(d)? [(a) V_c = 0, (b) V_R = 245V, (c) q = 0, (d) i = 0.0327 A, (e) V_c = 245V, V_R = 0, q = 1.13 times 10^-3C, i = 0]

Answer 1

Capacity = C = 4.60 uF

Resistance = R = 7.50 kΩ = 7500 Ω   

emf of source = E=245 V negligible internal resistance.

A)Just after the circuit is completed, the voltage drop across the capacitor is zero

Capacitor acts like a straight wire, thus V = 0 V

B) Just after the circuit is completed, the voltage drop across the resistor is 245 V

Since no voltage drop occurs across the capacitor, all the voltage drops across the resistor: Vr = 245 V

C) just after the circuit is completed, the charge on the capacitor is Q = 0

D)Just after the circuit is completed, the current through the resistor is

I = (Vapplied - Vcap)/R = 245 / 7500 = 0.0327 A


E) A long time after the circuit is completed the valuesof the quantities in parts (a)-(d) are

A) After long time,the voltage drop across the capacitor is 245 V

After a long time, capacitor acts like a very large resistor (or break). So voltage across the capacitor becomes Vc = 245 V

B) After long time,the voltage drop across the resistor is zero

Since all the voltage drops across the capacitor after a very long time, Vr = 0 V

C) the charge on the capacitor is Q = C*V = 4.60*10^-6*245 = 1.127*10^-3 coulombs

D) After long time,the current through the resistor is zero

Again, capacitor acts line a break or a giant resistor, using ohm’s law with R = infinity, I = 0 A