Capacitors

JEE Main level, single correct, medium

16 of 180

Question 16 of 180

A capacitor is connected to a 20V20V battery through a resistance of 10Ω.10\Omega . It is found that the potential difference across the capacitor rises to 2V2V in 1μs.1\mu s. The capacitance of the capacitor is ________\mu F.
Given In (109)=0.105\left(\dfrac{10}{9}\right)=0.105

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Answer: (A) 0.950.95

V=V0(1−e−t/RC)V={V}_{0}\left(1-{e}^{-t/RC}\right)
2=20(1−e−1μs10C)2=20\left(1-{e}^{-\dfrac{1\mu s}{10C}}\right)
110=(1−e−1×10−610C)\dfrac{1}{10}=\left(1-{e}^{-\dfrac{1\times {10}^{-6}}{10C}}\right)
e−1×10−610C=910{e}^{-\dfrac{1\times {10}^{-6}}{10C}}=\dfrac{9}{10}
1×10−610C=ln⁡(109)\dfrac{1\times {10}^{-6}}{10C}=\ln \left(\dfrac{10}{9}\right)
C=1×10−610×ln⁡(109)=10−71.05=11.05μF=100105μF=0.95μFC=\dfrac{1\times {10}^{-6}}{10\times \ln \left(\dfrac{10}{9}\right)}=\dfrac{{10}^{-7}}{1.05}=\dfrac{1}{1.05}\mu F=\dfrac{100}{105}\mu F=0.95\mu F

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