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A capacitor with a capacitance of C = 29.2 μF is slowly charged by a constant...

A capacitor with a capacitance of C = 29.2 μF is slowly charged by a constant current of I = 43.7 nA. How long does it take to charge the capacitor to a voltage of V = 25.9 V?

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Answer #1

The current is given by, (q is the charge)

The charge in the capacitor can be written as,

Substituting, we can write,

Since it is given that the current is constant and it is slowly charged we can approximate the derivative as,

The current and capacitance is given. The change in voltage is 25.9 V. Substituting we get,

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