Asked by Taylor
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
Answers
Answered by
bobpursley
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.
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.
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