A 4.80 F capacitor that is initially uncharged is connected in series with a 7.8

Question

A 4.80 F capacitor that is initially uncharged is connected in series with a 7.80 k resistor and an emf source with E= 125 V 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?

Explanation / Answer

Here ,

just after the circuit is closed , the capacitor will behave as a short circuit and it will provide zero resistance to current flowing through the circuit

part A)

as the capacitor is short circuited

the voltage drop across the capacitor is Zero

part B)

voltage across the resistance = 125 V

voltage across the resistance = 125 V

the voltage across the resistance is 125 V

c)

charge on the capacitor = C * V

charge on the capacitor = 4.8 * 0 C

charge on the capacitor is 0 C

d)

current in resistor = E/R

current in resistor = 125/7.8 mA

current in resistor = 16.03 mA

the current in resistor just after closing the circuit is 16.03 mA

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