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A capacitor is being charged through a resistance of 3 mega ohm. If it reaches 75 % of its final potential in 0.5 sec, find its capacitance (log₁₀4=0.6021).

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0.12 µF

1.2 µF

2.4 µF

0.24 µF

answer is A.

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Detailed Solution

$v = v_{0} (1 - e^{- t / C R})$

$\frac{v}{v_{0}} = 0.75, t = 0.5, R = 3 \times 10^{6} Ω$

$0.75 = 1 - e^{\frac{- 0.5}{(3 \times 10^{6}) c}}$

$e^{\frac{- 0.5}{c (3 \times 10^{6})}} = 1 - 0.75 = 0.25$

$e^{\frac{0.5}{c (3 \times 10^{6})}} = 4$

$\frac{0.5}{c (3 \times 10^{6})} = \ln 4$

$c = 0.12 μ F$

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