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TS EAMCET · Physics · Capacitance

If a capacitor of capacitance \(100 \mu \mathrm{F}\) is charged at a steady rate of \(100 \mu \mathrm{C} \mathrm{s}^{-1}\), then the time taken to produce a potential difference of \(100 \mathrm{~V}\) between the capacitor plates is

  1. A \(50 \mathrm{~s}\)
  2. B \(200 \mathrm{~s}\)
  3. C \(150 \mathrm{~s}\)
  4. D \(100 \mathrm{~s}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(100 \mathrm{~s}\)

Step-by-step Solution

Detailed explanation

Total charge on capacitor, \[ \begin{aligned} & Q=C V \\ & =(100 \mu \mathrm{F})(100 \mathrm{~V}) \\ & =100 \times 10^{-6} \times 100=10^{-2} \mathrm{C} \end{aligned} \] Total charge \(=Q(\) Rate of charge \() \times(\) time taken \()\)…