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Question

A parallel plate capacitor of $$1\mu F$$ capacity is discharging through a resister. If its energy reduces to half in one second. The value of resistance will be?

A
$$\dfrac{4}{ln(2)}M\Omega$$
B
$$\dfrac{\theta}{ln(2)}M\Omega$$
C
$$\dfrac{2}{ln(2)}M\Omega$$
D
$$\dfrac{16}{ln(2)}M\Omega$$
Solution
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Correct option is A. $$\dfrac{2}{ln(2)}M\Omega$$
$$q=q_0e^{t/\tau}$$ when energy is $$50\%$$
then $$q=\dfrac{q_0}{\sqrt{2}}$$
$$\dfrac{q_0}{\sqrt{2}}=q_0e^{-t/\tau}$$
$$e^{t/\tau}=\sqrt{2}$$
$$\dfrac{t}{\tau}=ln(\sqrt{2})$$
$$\tau =\dfrac{t}{ln(\sqrt{2})}$$
$$R_c=\dfrac{1}{ln\sqrt{2})}$$
$$R=\dfrac{1}{Cln(\sqrt{2})}=\dfrac{1}{10^{-6}ln(\sqrt{2})}=\dfrac{10^6}{ln(\sqrt{2})}=\dfrac{2}{ln(2)}=M\Omega$$.

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