The capacitance of a cylindrical capacitor is given by 65.0 µF. After being conn

Question

The capacitance of a cylindrical capacitor is given by 65.0 µF. After being connected to a D.C. power source the capacitor is fully charged to 210.0 µC. You disconnect the capacitor from the battery. You now take two resistors in parallel and connect the capacitor in parallel with the resistors. The resistance of the resistors are (10.0 and 48.0 ). (a) What is the time constant for this circuit? s (b) What is the time when the capacitor has lost half of its charge? s (c) What is the current through the capacitor at that time? A

Explanation / Answer

a)Net resistance=(10*48/58)=8.276 ohm

Capacitance=65*10^-6 F

time costant=RC=8.276*65*10^-6=5.38*10^-4

Q=210*10^-6*e-t/(8.276*65*10^-6)

at half charge

(210*10^-6)/2=210*10^-6*e-t/(8.276*65*10^-6)

ln2=t/(8.276*65*10^-6)

t=0.00037=370 microsecond

c)Q=210*10^-6*e-t/(8.276*65*10^-6)

dQ/dt=I=-0.39*e-t/(8.276*65*10^-6)

I=0.39-0.196 ampere=0.194ampere

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