Question

1. A 4.60 pF capacitor that is initially uncharged is connected in series with a 7.50 kΩ resistor and an emf source with E = 125 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)?

Answer 1

Answer:

a) Voltage across the capacitor just after connected is ZERO

b) Voltage drop across the resistor is 125 V

c) Charge on the capacitor is ZERO

d) current through resistor is 0.0167 A

e) After long time capacitor gets complete charge so complete voltage of the battery will be across the capacitor plates V = 125 V

voltage drop across the resistor is ZERO now

charge on the capacitor is given as

$Q = 5.75 \times 10^{-10} C$

current through the resistance must be ZERO

Explanation:

When capacitor is initially uncharged then just after it is connected into the circuit it will pass all the charge through it

a)

Voltage across the capacitor just after connected is ZERO as there is no charge on it

b)

V = i R

so complete voltage of the battery is across the resistor

so we have

V = 125 V

c)

Charge on the capacitor is ZERO as just after connection charging not done

d)

current through resistor is given as

$V = i R$

$125 = i (7500)$

$i = 0.0167 A$

e)

After long time capacitor gets complete charge so complete voltage of the battery will be across the capacitor plates

so we have

$V = 125 V$

voltage drop across the resistor is ZERO now

charge on the capacitor is given as

$Q = CV$

$Q= (4.60 pF)(125)$

$Q = 5.75 \times 10^{-10} C$

current through the resistance must be ZERO

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Topic : Capacitor charging circuit

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