(A) The voltage across the capacitor in a charging $RC$ circuit is given by $V(t) = V_0(1 - e^{-t/RC})$.Here,$V_0 = 200\ V$,$V(t) = 120\ V$,$t = 5\ s$

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

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

Easy

$A$ $2\ \mu F$ capacitor and an $R$ resistor are connected in series to a $200\ V$ $DC$ supply. $A$ neon bulb is connected across the capacitor,which glows at $120\ V$. Calculate the value of $R$ so that the bulb glows for $5\ s$ after the switch is closed. (Given: $\log_{10} 2.5 = 0.4$)

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

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Electric Potential and Capacitance

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Charging and Discharging of Capacitance and RC circuit (DC)

$A$ $4 \mu F$ capacitor and a resistance of $2.5 \, M\Omega$ are in series with a $12 \, V$ battery. Find the time after which the potential difference across the capacitor is $3$ times the potential difference across the resistor. (Given $\ln(2) = 0.693$)

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$A$ circuit element is placed in a closed box. At time $t=0$,a constant current generator supplying a current of $1\, A$ is connected across the box. The potential difference across the box varies according to the graph shown in the figure. The element in the box is:

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In the circuit shown,the cells are ideal and of equal emfs $E$. The capacitance of the capacitor is $C$ and the resistance of the resistor is $R$. $X$ is first joined to $Y$ and then to $Z$. After a long time,the total heat produced in the resistor will be:

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After the switch shown in Figure $A$ is closed,there is a current $i$ through the resistance $R$. Figure $B$ indicates current variation curves $a, b, c,$ and $d$ for four sets of values of $R$ and capacitance $C$:
$(i)$ $R_0$ and $C_0$
$(ii)$ $2R_0$ and $C_0$
$(iii)$ $R_0$ and $2C_0$
$(iv)$ $2R_0$ and $2C_0$
Which set goes with which curve?

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Given,
${R_1} = 1\,\Omega, R_2 = 2\,\Omega$
${C_1} = 2\,\mu F, C_2 = 4\,\mu F$
The time constants (in $\mu s$) for the circuits $I, II, III$ are respectively:

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$A$ $2\ \mu F$ capacitor and an $R$ resistor are connected in series to a $200\ V$ $DC$ supply. $A$ neon bulb is connected across the capacitor,which glows at $120\ V$. Calculate the value of $R$ so that the bulb glows for $5\ s$ after the switch is closed. (Given: $\log_{10} 2.5 = 0.4$)

Similar Questions

$A$ $4 \mu F$ capacitor and a resistance of $2.5 \, M\Omega$ are in series with a $12 \, V$ battery. Find the time after which the potential difference across the capacitor is $3$ times the potential difference across the resistor. (Given $\ln(2) = 0.693$)

$A$ circuit element is placed in a closed box. At time $t=0$,a constant current generator supplying a current of $1\, A$ is connected across the box. The potential difference across the box varies according to the graph shown in the figure. The element in the box is:

In the circuit shown,the cells are ideal and of equal emfs $E$. The capacitance of the capacitor is $C$ and the resistance of the resistor is $R$. $X$ is first joined to $Y$ and then to $Z$. After a long time,the total heat produced in the resistor will be:

Given,${R_1} = 1\,\Omega, R_2 = 2\,\Omega$${C_1} = 2\,\mu F, C_2 = 4\,\mu F$The time constants (in $\mu s$) for the circuits $I, II, III$ are respectively:

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Given,
${R_1} = 1\,\Omega, R_2 = 2\,\Omega$
${C_1} = 2\,\mu F, C_2 = 4\,\mu F$
The time constants (in $\mu s$) for the circuits $I, II, III$ are respectively: