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
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!
1 answer