A capacitor is connected to a 20 V battery through a resistance of $10 \Omega$. It is found that the potential difference across the capacitor rises to 2 V in $1 \mu \mathrm{~s}$. The capacitance of the capacitor is
$\_\_\_\_$ $\mu \mathrm{F}$. Given :
$$
\ln \left(\frac{10}{9}\right)=0.105
$$
Select the correct option:
A
9.52
B
0.95
C
0.105
D
1.85
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{gathered}V=V_0\left(1-e^{-U R C}\right) \\ 2=20\left(1-e^{-U R C}\right) \\ \frac{1}{10}=1-e^{-U / R C} \\ e^{-U / R C}=\frac{10}{9} \\ \frac{t}{R C}=\ln \left(\frac{10}{9}\right) \Rightarrow C=\frac{t}{R \ln \left(\frac{10}{9}\right)} \\ C=\frac{10^{-6}}{10 \times 0.105}=0.95 \mu \mathrm{~F}\end{gathered}$
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