Ask Question
1 June, 09:16

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

+3
Answers (1)
  1. 1 June, 13:13
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “What will be the current when the capacitor has acquired 1/4 of its maximum charge? ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers