Posted by **sonia** on Saturday, November 12, 2011 at 6:07pm.

The magnetic field in the region shown below is increasing by 5 T/s. The circuit shown (with R=25 , C=55 µF, dimensions 1 m by 2 m) is inserted into the region as shown. The capacitor is initially uncharged.

1. How long will it take the capacitor to charge to 25% of its final charge?

Please help I tried T=RC but not right formula!

- no picture -
**Damon**, Saturday, November 12, 2011 at 8:09pm
I do not have a picture of your circuit so find this hard to follow. However for an RC circuit at initial voltage V when the switch is closed:

V = i R + (1/C) integral i dt

with i = 0 at t = 0

0 = R di/dt + i/C

0 = RC di/dt + i

let i = Io (e^kt)

di/dt = k Io e^kt

0 = RC kIo e^kt + i

so

i = -RC k Io e^kt = Io e^kt

so k = -1/RC

so if in form i = Io e^-t/T

T indeed = RC

Now

The voltage around this loop = rate of change of magnetic flux

E = 5*2square meters = 10 volts

so we put a ten volt force on our RC circuit

i = Io e^-t/RC

Initially all the voltage appears on the resistor because there is no charge yet on the capacitor

Io = E/R = 10/R

so

i = (10/R)e^-t/RC

charge on capacitor Q= integral i dt

= (10 /R) (-RC e^-t/RC + k)

when t = 0, Q = 0 so

k = RC

so

Q = 10 C (1 - e^-t/RC)

Q final = 10 C

when does Q = (10/4)C ?

.25 = 1 - e^-t/RC

e^-t/RC = .75

-t/RC = ln .75 = -.2877

so

t = .2877 R C

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