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OpenStudy (anonymous):

A capacitor of 2.0μF is connected to a battery of 2.0V through a resistance of 10kΩ. What is the initial current in the circuit and the current after 0.02s?

OpenStudy (anonymous):

Using \[C = \frac{ Q }{ V } = \frac{ I \Delta t}{ V}\] Got \[\Delta t = RC = 10 \times 10^3 \times 2.0 \times 10^{-6} = 0.02 s\]

OpenStudy (anonymous):

Hi @ganeshie8. Pls assist me if you got the time. Thx :)

OpenStudy (anonymous):

In my list of options, 0.2μA and 0.074mA only seemed feasible even though I don't understand why they have wide variations in negative powers of ten.

ganeshie8 (ganeshie8):

\(i(t) = \frac{Vin}{R} e^{\frac{-t}{RC}}\)

ganeshie8 (ganeshie8):

for initial current, plugin t = 0 for current after 0.02s, plugin t = 0.02

OpenStudy (anonymous):

Thanks @ganeshie8 (:

OpenStudy (anonymous):

However, my answer for t=0 is \[2 \times 10^{-4} A\]Quite different by power of 2 from the answer.

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