A 4.80 F capacitor that is initially uncharged is connected in series with a 7.7

Question

A 4.80 F capacitor that is initially uncharged is connected in series with a 7.70 k resistor and an emf source with E= 220 V negligible internal resistance.

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

B) Just after the circuit is completed, what is the voltage drop across the resistor?

C) Just after the circuit is completed, what is the charge on the capacitor?

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

Totally lost. Please help!

Explanation / Answer

here,

C = 4.80 uF = 4.8 * 10^-6 F
R = 7.70 k = 7700
E = 220 V

A) Capacitor acts like a straight wire so voltage drop across the capacitor is zero

B) As there is no Voltage drop across capacitor so, the voltage drop across the resistor will be 220 V

C) as we know

Charge = volatge * Capacitance

Q = 0 * C

Q = 0

so, Just after the circuit is completed,charge on the capacitor is Zero.

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

from ohms Law :
I = V/R = 220 / 7700

I = 0.0285 A

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