The circuit shown below is connected long enough so that the

The circuit shown below is connected long enough so that the capacitor carries a constant charge, Compute the charge on the capacitor, How long after the source elf is removed from the circuit does it take for the charge on the capacitor to drop to one quarter of its original value? R1 4.00 k ohm R2 2.00 k ohm R3 3.00 k ohm R4 3.00 k ohm Cal 40.0 mu F e m f 72.0 V

Solution

a) The formula is :

V = Q/C => Q = CV = 40*10^-6*72= 2880*10^-6 C.

b) The formula for time (t) is:

t = RC

For quater drop C= 40/4 = 10 micro Farad.

Since, R1 and R3 are parallel,

R(parallel1) = R1*R3/(R1+R3) = 1.7 K(ohm).

Then, R2 and R4 are parallel,

R(parallel2) = R2*R4/(R2+R4) = 1.2 K(ohm).

Finally, R(parallel1) is in series with R(parallel2):

R (series) = R(parallel1)+R(parallel2) = 2.9 K(ohm).

Hence, t=2.9 K(ohm)/10V = 290 sec.