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JEE Mains · Physics · STD 12 - Semicondutor electronics
- JEE Mains 2020
- Medium
- MCQ
Two identical capacitors \(\mathrm{A}\) and \(\mathrm{B}\), charged to the same potential \(5\; \mathrm{V}\) are connected in two different circuits as shown below at time \(\mathrm{t}=0\) If the charge on capacitors \(A\) and \(B\) at time \(\mathrm{t}=\mathrm{CR}\) is \(\mathrm{Q}_{\mathrm{A}}\) and \(\mathrm{Q}_{\mathrm{B}}\) respectively, then (Here \(e\) is the base of natural logarithm)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(\mathrm{Q}_{\mathrm{A}}=\mathrm{VC}, \mathrm{Q}_{\mathrm{B}}=\frac{\mathrm{VC}}{\mathrm{e}}\)
Step-by-step Solution
Detailed explanation
For (A)
No current flows
Hence \(Q_{A}=C V\)
For \((\mathrm{B})\)
\(\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}} \mathrm{e}^{-\frac{\mathrm{t}}{\mathrm{RC}}}\)
\(\mathrm{q}=\mathrm{CV} \mathrm{e}^{\frac{\mathrm{t}}{\mathrm{RC}}}\)
at \(\mathrm{t}=\mathrm{CR}\)
\(\mathrm{Q}_{\mathrm{B}}=\mathrm{CV} \mathrm{e}^{-1}=\frac{\mathrm{CV}}{\mathrm{e}}\)
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