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