Current in the capacitor is given by

Here charge is given by
![q = CE[1 - e^{-\frac{t}{RC}}]](http://img.homeworklib.com/questions/cf235f30-0364-11ec-a61b-95b887dec2fd.png?x-oss-process=image/resize,w_560)
now current is given by

![i = \frac{d}{dt} (CE[1 - e^{-\frac{t}{RC}}])](http://img.homeworklib.com/questions/d00d9e10-0364-11ec-b96c-77a4560c5944.png?x-oss-process=image/resize,w_560)


PART B)
Now for inductor we have
![i = \frac{9V}{R}[1 - e^{-\frac{Rt}{L}}]](http://img.homeworklib.com/questions/d178c910-0364-11ec-baf6-0fc1c12a1ab3.png?x-oss-process=image/resize,w_560)
So option (d) is correct
12. An uncharged capacihor (capacitonce C) is connected in series of the followin 13. An inductor...
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