Question

A 14.7 μF capacitor is charged to a potential of 55.0 V and then discharged through a 75.0 Ω resistor.

(a) How long after discharge begins does it take for the capacitor to lose 90.0% of the following?

(i) its initial charge: _______s

(ii) its initial energy: _______s

(b) What is the current through the resistor at both times in part (a)?

(i) at t_{charge: ________}A

(ii) at t_{energy: ________}A

Answer 1

a) time constant=R*C=1.1025 ms

initial voltage on capacitor=55 volts
initial charge=C*V=0.8085 mC

then voltage across capacitor as a function of time t
is given as

v(t)=55*exp(-t/1.1025) ) volts where t is in ms
charge on the capacitor is given by q(t)=V(t)*C=0.8085*exp(-t/1.1025) mC, where t is ms

i)let at time t , the charge is 10% of initial charge (it has lost 90% )

==> 0.1*0.8085=0.8085*exp(-t/1.1025)

==>t=2.5386 ms

ii) energy in capacitor is given by 0.5*C*V^2

so for energy to be 0.1 times initial energy, voltage will be sqrt(0.1)=0.3162 times initial voltage

hence 0.3162*55=55*exp(-t/1.1025)

==>t=1.2693 ms

b)
as only capacitor and resistor are connected in series, voltage across resistor=-voltage across capacitor

hence to find the magnitude of current (ignoring the direction),

current=voltage across capacitor/resistance

i) when charge is 10% of initial charge, voltage is 10% of initial voltage (as charge=C*V, hence directly proportional)

then voltage=55*0.1=5.5 volts

current=5.5/75=0.0733 A

ii) voltage=0.3162*55=17.391 volts

then current=voltage/resistance=0.23188 A