JEE MAIN - Physics (2021 - 1st September Evening Shift - No. 16)
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 $$\mu$$s. The capacitance of the capacitor is __________ $$\mu$$F. Given : $$\ln \left( {{{10} \over 9}} \right) = 0.105$$
9.52
0.95
0.105
1.85
Explanation
Given, the peak voltage of the battery, V0 = 20V
The voltage of the battery, V = 2V
Time, t = 1 $$\mu$$s = 1 $$\times$$ 10$$-$$6s
Resistance of the capacitor, R = 10 $$\Omega$$
As we know that,
V = V0(1 $$-$$ e$$-$$t/RC)
Substituting the values in the above equation, we get
2 = 20(1 $$-$$ e$$-$$t/RC)
$$ \Rightarrow {t \over {RC}} = \ln \left( {{{10} \over 9}} \right)$$
$$ \Rightarrow C = {t \over {R\ln (10/9)}} = {{{{10}^{ - 6}}} \over {10 \times \ln (10/9)}} = 0.95$$ $$\mu$$F
$$\therefore$$ The capacitance of the capacitor is 0.95 $$\mu$$F.
The voltage of the battery, V = 2V
Time, t = 1 $$\mu$$s = 1 $$\times$$ 10$$-$$6s
Resistance of the capacitor, R = 10 $$\Omega$$
As we know that,
V = V0(1 $$-$$ e$$-$$t/RC)
Substituting the values in the above equation, we get
2 = 20(1 $$-$$ e$$-$$t/RC)
$$ \Rightarrow {t \over {RC}} = \ln \left( {{{10} \over 9}} \right)$$
$$ \Rightarrow C = {t \over {R\ln (10/9)}} = {{{{10}^{ - 6}}} \over {10 \times \ln (10/9)}} = 0.95$$ $$\mu$$F
$$\therefore$$ The capacitance of the capacitor is 0.95 $$\mu$$F.
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