TS EAMCET · Physics · Capacitance
A capacitor of capacity \(0.1 \mu \mathrm{F}\) connected in series to a resistor of \(10 \mathrm{M} \Omega\) is charged to a certain potential and then made to discharge through the resistor. The time in which the potential will take to fall to half its original value is (Given, \(\log _{10} 2=0.3010\) )
- A \(2 \mathrm{~s}\)
- B \(0.693 \mathrm{~s}\)
- C \(0.5 \mathrm{~s}\)
- D \(1.0 \mathrm{~s}\)
Answer & Solution
Correct Answer
(B) \(0.693 \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
By equation of charging \(q=q_0\left(1-e^{-t / C R}\right)\) According to question \(\frac{q}{q_0}=\frac{1}{2}=0.50\) \(\therefore \quad 0.50=1-e^{-t / C R}\) \(e^{-t / C R}=1-0.50=0.50\) \(e^{t / C R}=2\) or \(\quad \frac{t}{C R}=\log _e 2\) or…
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