Question

A 6.00 μFcapacitor that is initially uncharged is connected in series with a 4400 Ωresistor and a 503 Vemf source with negligible internal resistance.

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

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

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

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

Part E: A long time after the circuit is completed (after many time constants), what are the values of the preceding four quantities?

Answer 1

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

At the instant, there is no voltage across the capacitor. Since it has no charge stored.

Capacitor acts like a straight wire. Then, we have

V_C = 0

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

Since no voltage drop occurs across the capacitor (V_C = 0).

The full battery voltage appear across the resistor. Then, we have

V_R =

V_R = 503 V

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

There is no charge stored on the capacitor. Then, we have

Q = 0

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

From an ohm's law, we get

I_R = V_R / R

I_R = [(503 V) / (4400 )]

I_R = 0.114 A

Part E : A long time after the circuit is completed, what are the values of the preceding four quantities?

After a long time, capacitor acts like a very large resistor. So, the voltage drop across the capacitor which will be given as -

V_C = 503 V

Since all the voltage drops across the capacitor after a very long time. Then, the voltage drop across the resistor which will be given as -

V_R = 0 V

After a long time is passed, the full battery voltage across the capacitor is 503 V.

Then, the charge stored on the capacitor which will be given by -

Q = C V_C

Q = [(6 x 10^-6 F) (503 V)]

Q = 3.01 x 10^-3 C

Capacitor acts like a break or a giant resistor. Then, the current through the resistor which will be given as -

From an ohm's law, we get

I_R = V_R / R

I_R = (0 V) / ()

I_R = 0 A