Answer:
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
