26.41: A 4.60-F capacitor that is initially uncharged is con- nected in series w

Question

26.41: A 4.60-F capacitor that is initially uncharged is con- nected in series with a 7.50-k resistor and an emf source with £- 245 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)?

Explanation / Answer

a)

just after the circuit is completed , the charge stored in the capacitor is zero. hence the Voltage across the capacitor is also zero since

V = Q/C

b)

just as the circuit is completed , the Voltage across the capacitor is zero. hence the all the voltage of the battery appears across the resistor.

so Vr = Voltage across the resistor = E = emf of battery = 245 volts

c)

since the Voltage across the capacitor is zero , hence charge is given as

Q = C Vc

Q = C (0)

Q = 0

d)

using ohm's law , current through the resistor is given as

i = E/R

i = 245/(7500)

i = 0.033 A

e)

after long time ... the capacitor is fully charged and all the Voltage apppears across it and no Voltage appears across the resistor

a)

Vc = Voltage across the capacitor = E = 245 Volts

b)

Vr = Voltage drop across the resistor = 0 Volts

c)

Q = CVc = (4.60 x 10-6) (245) = 0.001127 C

d)

i = 0 since Vr = 0

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