ExamBro
ExamBro
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, \(\log _{10} 2=0.3010\) )

  1. A \(2 \mathrm{~s}\)
  2. B \(0.693 \mathrm{~s}\)
  3. C \(0.5 \mathrm{~s}\)
  4. D \(1.0 \mathrm{~s}\)
Verified Solution

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…
Same subject
Explore more questions on app
From TS EAMCET
Explore more questions on app