What will be the current when the capacitor has acquired 1/4 of its maximum charge?

I = 0.625 Amp

Explanation:

Given:-

- The capacitance of the capacitor, C = 1.50 uF

- The capacitor charges through a resistance, R = 12.0 ohms

- The potential difference of battery, V = 10.0 V

Find:-

What will be the current when the capacitor has acquired 1/4 of its maximum charge?

Solution:-

- Note that the charge is increasing with time while the current across the capacitor decreases. But both obey the exponential equations.

- The charge (Q) obeys the equation:

Q = Q_max * (1 - e^ (-t/RC))

- While the current I obeys the relationship:

I = I_max * (e^ (-t/RC))

- When the charge hs taken 1/4 of its maximum value we can write:

0.25*Q_max = Q_max * (1 - e^ (-t/RC))

0.25 = (1 - e^ (-t/RC))

e^ (-t/RC) = 1 - 0.25 = 0.75

- Where the current across the capacitor at this time would be:

I = I_max * (e^ (-t/RC))

Where, I_max = V / R = (10 / 12)

I = (10 / 12) * (3 / 4)

I = 0.625 Amp