A 460mu F capacitor that is initially uncharged is connected

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 are (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? (e) A long time after the circuit is completed (after many time constants) what are the values of the quantities in parts (a)-(d)?

Solution

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

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

c) The charge on capacitor = 0

d) I = V_R / R = 245 / 7500 = 0.0327 A

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

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

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