Two 6.0 μF capacitors, two 2.2 kΩ resistors, and a 14.5 V source are connected in series. Starting from the uncharged state, how long does it take for the current to drop from its initial value to 1.40 mA

Answer 1

RC= 4.4*3.0E-6

I=14.5/4.4 * e^t/RC

take the log (e) of each side.

ln 1.40=ln 14.5 - ln4.4 + t/RC
solve for t.