(B) The voltage across a charging capacitor in an $RC$ circuit is given by $V(t) = V_0(1 - e^{-t/RC})$.Here,$V_0 = 200 \ V$,$V(t) = 120 \ V$,$C = 2 \t

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(B) The voltage across a charging capacitor in an $RC$ circuit is given by $V(t) = V_0(1 - e^{-t/RC})$.Here,$V_0 = 200 \ V$,$V(t) = 120 \ V$,$C = 2 \times 10^{-6} \ F$,and $t = 5 \ s$.Substituting the values: $120 = 200(1 - e^{-5/(R \times 2 \times 10^{-6})})$.$0.6 = 1 - e^{-5/(R \times 2 \times 10^{-6})} \Rightarrow e^{-5/(R \times 2 \times 10^{-6})} = 0.4$.Taking the natural logarithm on both sides: $-5/(R \times 2 \times 10^{-6}) = \ln(0.4) = -\ln(2.5)$.$5/(R \times 2 \times 10^{-6}) = \ln(2.5) = 2.303 \times \log_{10}(2.5)$.Given $\log_{10}(2.5) = 0.4$,so $\ln(2.5) = 2.303 \times 0.4 = 0.9212$.$R = 5 / (2 \times 10^{-6} \times 0.9212) \approx 2.71 \times 10^6 \ \Omega$.

Class 12 Physics Electrostatic Potential and Capacitance Charging and Discharging of Capacitance and RC circuit (DC)

Medium

$A$ resistor $R$ and a $2 \ \mu F$ capacitor in series are connected through a switch to a $200 \ V$ direct supply. Across the capacitor is a neon bulb that lights up at $120 \ V$. Calculate the value of $R$ to make the bulb light up $5 \ s$ after the switch has been closed. (Given: $\log_{10} 2.5 = 0.4$)

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A
$1.7 \times 10^5 \ \Omega$
B
$2.7 \times 10^6 \ \Omega$
C
$3.3 \times 10^7 \ \Omega$
D
$1.3 \times 10^4 \ \Omega$

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Electrostatic Potential and Capacitance

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Charging and Discharging of Capacitance and RC circuit (DC)

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The capacitor $C$ is initially uncharged. Switch $S_1$ is closed for a long time while $S_2$ remains open. Now at $t = 0$, $S_2$ is closed while $S_1$ is opened. All the batteries are ideal and connecting wires are resistanceless. Find the $\text{INCORRECT}$ statement.

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As shown in the figure,if a capacitor $C$ is charged by connecting it with a battery of voltage $V$ through a resistance $R$,then the energy supplied by the battery will be:

Easy

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Identify the valid statements relevant to the given circuit at the instant when the key is closed.
$A.$ There will be no current through resistor $R.$
$B.$ There will be maximum current in the connecting wires.
$C.$ Potential difference between the capacitor plates $A$ and $B$ is minimum.
$D.$ Charge on the capacitor plates is minimum.
Choose the correct answer from the options given below $:$

MediumJEE MAIN 2025

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$A$ capacitor of capacity $0.1 \mu F$ connected in series to a resistor of $10 M \Omega$ is charged to a certain potential and then made to discharge through the resistor. The time in which the potential will take to fall to half its original value is (Given, $\log _{10} 2=0.3010$ ) (in $\,s$)

DifficultAP EAMCET 2010

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The electric field between the two parallel plates of a capacitor of $1.5 \mu F$ capacitance drops to one third of its initial value in $6.6 \mu s$ when the plates are connected by a thin wire. The resistance of this wire is . . . . . . . $\Omega$. (Given,$\log_{e} 3 = 1.1$)

DifficultJEE MAIN 2024

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The capacitor $C$ is initially uncharged. Switch $S_1$ is closed for a long time while $S_2$ remains open. Now at $t = 0$, $S_2$ is closed while $S_1$ is opened. All the batteries are ideal and connecting wires are resistanceless. Find the $\text{INCORRECT}$ statement.

As shown in the figure,if a capacitor $C$ is charged by connecting it with a battery of voltage $V$ through a resistance $R$,then the energy supplied by the battery will be:

$A$ capacitor of capacity $0.1 \mu F$ connected in series to a resistor of $10 M \Omega$ is charged to a certain potential and then made to discharge through the resistor. The time in which the potential will take to fall to half its original value is (Given, $\log _{10} 2=0.3010$ ) (in $\,s$)

The electric field between the two parallel plates of a capacitor of $1.5 \mu F$ capacitance drops to one third of its initial value in $6.6 \mu s$ when the plates are connected by a thin wire. The resistance of this wire is . . . . . . . $\Omega$. (Given,$\log_{e} 3 = 1.1$)

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