Show that the magnitude of the charge on a capacitor given by Q =Q (1-e ) and Q

Question

Show that the magnitude of the charge on a capacitor given by Q =Q (1-e ) and Q = Q e for charging and discharging, respectively, What is the voltage across a capacitor after a time of constants when (a) charging from zero voltage and (b) discharging from a fully charged condition? With V = V e ,it mathematically takes an infinite time for a capacitor in an RC circuit to discharge. Practically, how many time constants does it take for a capacitor to discharge to less than 1% of its initial voltage?

Explanation / Answer

R and C are in series and V is the EMf in the cicuit

applying loop law

-q/C -iR +V =0

i = dq/dt

dq/dt = V/R -q/RC

dq/(V-q/C) = dt/R

integrating both sides we get

ln(V-q/C) = -t/RC

V-q/C = exp(-t/RC + c ) , c is the integration const.

when t=0 q =0 , charging

c = ln(V)  

q = VC (1 -exp(-t/RC) , VC = qo

q = qo (1 - exp(-t/RC) )

while discharging at t=0 q = qo

hence q = qo exp(-t/RC)

2. a while charging

q = q0 (1-exp (-2)) = 0.86 qo

V = q/C = 0.86qo/C = 0.86Vo

while discharging

V = q/C = qo *0.135 /C = 0.135Vo

3. V/Vo = 0.01   ( dischrged to 1 % )

              = exp(-t/RC)

-t/RC = ln(0.01)

t = 4.6 RC , approx. 5 time constants it owuld discharge to less than 1%

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