JEE MAIN - Physics (2021 - 27th July Morning Shift - No. 6)
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A capacitor of capacitance C = 1 $$\mu$$F is suddenly connected to a battery of 100 volt through a resistance R = 100 $$\Omega$$. The time taken for the capacitor to be charged to get 50 V is :
[Take ln 2 = 0.69]
1.44 $$\times$$ 10$$-$$4 s
3.33 $$\times$$ 10$$-$$4 s
0.69 $$\times$$ 10$$-$$4 s
0.30 $$\times$$ 10$$-$$4 s
Explanation
$$V = {V_0}\left( {1 - {e^{ - {t \over {RC}}}}} \right)$$
$$50 = 100\left( {1 - {e^{ - {t \over {RC}}}}} \right)$$
$$ = t = 0.69 \times {10^{ - 4}}$$ sec.
$$50 = 100\left( {1 - {e^{ - {t \over {RC}}}}} \right)$$
$$ = t = 0.69 \times {10^{ - 4}}$$ sec.
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