A capacitor of capacitance 30µF is charged by a constant current of 10mA. If initially, the capacitor was uncharged what is the time taken for the potential difference across the capacitor to reach 300V ?
Correct answer: A. 0.9sec
- A. 0.9sec
- B. 15 sec
- C. 1.5 x 10^5sec
- D. 0.9 x 10^2sec
Explanation
The required charge is Q = CV = (30 x 10^-6 F)(300 V) = 0.009 C. With constant current I = 10 mA = 0.010 A, the charging time is t = Q/I = 0.009 C/0.010 A = 0.9 s. The likely wrong choice is d, caused by a power-of-ten error in converting microfarads or milliamperes.
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Capacitors store electric charge and energy in an electric field between conductors. Work includes capacitance, the relation Q = CV, dielectric materials, charging and discharging, energy formulas, and combinations of capacitors in series and parallel. Electric potential difference is needed to understand why charge and stored energy change.
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