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A parallel combination of 0.1 \mathrm{M} \Omega  resistor and a 10 \mu \mathrm{F}  capacitor is connected across a 1.5 \mathrm{~V} source of negligible resistance. The time required for the capacitor to get charged up to 0.75 \mathrm{V}  is approximately (in seconds)
 

Option: 1

\infty


Option: 2

\mathrm{ \log _e 2}


Option: 3

\mathrm{\log _{10} 2}


Option: 4

Zero


Answers (1)

best_answer

For the capacitor to get charged up to 0.75 \mathrm{~V}, the charge on the plates should be as follows


\begin{aligned} & E=1.5 \mathrm{~V} \\ & \mathrm{q=C E\left[1-e^{-t / R C}\right] }\\ & \mathrm{0.75 \times 10^{-5}=10^{-5} \times 1.5\left[1-e \frac{-t}{10^5 \times 10^{-5}}\right] }\\ &\mathrm{ \frac{1}{2}=\left[1-e^{-t}\right] \Rightarrow e^{-t}=\frac{1}{2}} \end{aligned}

Taking log on both sides, we get

\mathrm{-t=-\ln 2 \Rightarrow t=0.693 \mathrm{~s}}

 

Posted by

Gautam harsolia

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