A 5.20 F capacitor that is initially uncharged is connected in series with a 7.0

Question

A 5.20 F capacitor that is initially uncharged is connected in series with a 7.00 k resistor and an emf source with E= 145 V negligible internal resistance.

(E) A long time after the circuit is completed (after many time constants) what is the voltage drop across the capacitor?

(F) A long time after the circuit is completed (after many time constants) what is the voltage drop across the resistor?

(G) A long time after the circuit is completed (after many time constants) what is the charge on the capacitor?

(H)A long time after the circuit is completed (after many time constants) what is the current through the resistor?

Explanation / Answer

After a long time, Capacitor will be Fully charged and will behave like an open circut.
Therefore, there will be no current flowinf in the circut.

(E)
  Voltage drop across the capacitor, = 145 v
(F)
Voltage drop across the Resistor , I*R = 0.
(G)
Charge on the Capacitor, Q = C*V
Q = 5.2 uF * 145
Q = 754 uC
Charge on the Capacitor, Q = 754 uC
(H)
As the capacitor will behave like open circut,
Current through the resistor, = 0 Amp

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