A capacitor is connected to a 20, V battery t | vedclass

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(B) The potential difference across a charging capacitor is given by $V = V_0(1 - e^{-t/RC})$.Given $V = 2\, V$,$V_0 = 20\, V$,$t = 1\, \mu s = 10^{-6

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$0.95$

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(B) The potential difference across a charging capacitor is given by $V = V_0(1 - e^{-t/RC})$.Given $V = 2\, V$,$V_0 = 20\, V$,$t = 1\, \mu s = 10^{-6}\, s$,and $R = 10\, \Omega$.Substituting the values: $2 = 20(1 - e^{-t/RC})$.$1/10 = 1 - e^{-t/RC} \Rightarrow e^{-t/RC} = 9/10$.Taking the natural logarithm on both sides: $-t/RC = \ln(9/10) = -\ln(10/9)$.$t/RC = \ln(10/9)$.Rearranging for capacitance $C$: $C = t / (R \cdot \ln(10/9))$.$C = 10^{-6} / (10 	imes 0.105) = 10^{-6} / 1.05 \approx 0.952\, \mu F$.Rounding to the nearest option,$C = 0.95\, \mu F$.

$A$ 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: $\ln(10/9) = 0.105$.

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