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
- A \(50 \mathrm{~s}\)
- B \(200 \mathrm{~s}\)
- C \(150 \mathrm{~s}\)
- D \(100 \mathrm{~s}\)
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 \()\)…
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