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A capacitor is connected to a 20 \mathrm{~V} battery through a resistance of 10 \mathrm{\Omega}. It is found that the potential difference across the capacitor rises to 2 \mathrm{~V}$ in $1 \mu \mathrm{s}. The capacitance of the capacitor is___________ \mu \mathrm{F}.
Given \: \: \ln \left(\frac{10}{9}\right)=0.105
Option: 1 0.95
Option: 2 9.52
Option: 3 1.85
Option: 4 0.105

Answers (1)

best_answer


By charging of capacitor condition, q_{t}= CV\left ( 1-e^{\frac{-t}{t}} \right )
\tau = RC= 10C
q_{t}= CV_{t}= CV\left ( 1-e^{\frac{-t}{ t}} \right )
\frac{1}{10}= 1-e^{-10^{\frac{-6}{t}}}
\frac{9}{10}= e^{-10^{\frac{-6}{10C}}}
ln\left ( \frac{9}{10} \right )= \frac{-10^{-6}}{10C}
C= \frac{10^{-6}}{1\cdot 05}
C= 0\cdot 95\times 10^{-6}F
C= 0\cdot 95\mu F
The correct option is (1)


 

Posted by

vishal kumar

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