Charging and Discharging of Capacitance and RC circuit (DC) Questions in English

(C) The dimension of $CR$ is given by the product of capacitance $(C)$ and resistance $(R)$.
$C = \frac{Q}{V}$,where $Q$ is charge and $V$ is potential difference.
$R = \frac{V}{I}$,where $I$ is current.
Therefore,$CR = \left( \frac{Q}{V} \right) \times \left( \frac{V}{I} \right) = \frac{Q}{I}$.
Since $I = \frac{Q}{t}$,where $t$ is time,we have $\frac{Q}{I} = t$.
Thus,the dimensions of $CR$ are the same as those of time,which corresponds to the time period.

(A) When the switch $S$ is closed,the capacitor begins to charge.
Initially,the capacitor acts as a short circuit (zero resistance),allowing current to flow through the circuit,which causes the bulb to light up for an instant.
As the capacitor charges,the potential difference across it increases,and the current in the circuit decreases.
Once the capacitor is fully charged,it acts as an open circuit,blocking the flow of direct current $(DC)$.
Therefore,the current becomes zero,and the bulb stops glowing.

(B) When a capacitor $C$ is charged by a battery of voltage $V$ through a resistance $R$,the total charge that flows from the battery is $Q = CV$.
The total work done by the battery is $W = QV = (CV)V = CV^2$.
The energy stored in the capacitor is $U = \frac{1}{2}CV^2$.
Since the energy supplied by the battery is $CV^2$ and the energy stored in the capacitor is $\frac{1}{2}CV^2$,the remaining energy $\frac{1}{2}CV^2$ is dissipated as heat in the resistance $R$.
Therefore,the energy supplied by the battery is $CV^2$,which is more than the energy stored in the capacitor,$\frac{1}{2}CV^2$.

(A) When a lamp is connected in series with a capacitor to a $D.C.$ source,the capacitor gets charged and eventually blocks the flow of current once it is fully charged.
Since $D.C.$ cannot pass through a capacitor,the circuit effectively becomes an open circuit.
Therefore,the lamp will not glow.

(A) In a $DC$ circuit,a capacitor acts as an open circuit in the steady state.
Therefore,no current flows through the branch containing the capacitor.
The entire potential difference of the battery,which is $2\,V$,appears across the capacitor.
Using the formula for charge on a capacitor,$Q = C \times V$,where $C = 10\,\mu F$ and $V = 2\,V$.
$Q = 10\,\mu F \times 2\,V = 20\,\mu C$.

(D) The potential difference across a discharging capacitor is given by $V = V_0 e^{-t/CR}$.
Given $V_0 = 50 \, V$ and $V = 40 \, V$ at $t = 1 \, s$,we have $40 = 50 e^{-1/CR}$,which implies $e^{-1/CR} = 4/5$.
For option $(b)$,the potential difference after $t = 2 \, s$ is $V' = V_0 e^{-2/CR} = 50 (e^{-1/CR})^2 = 50 (4/5)^2 = 50 \times (16/25) = 32 \, V$. Thus,$(b)$ is correct.
For option $(a)$,the energy stored in a capacitor is $U = \frac{1}{2} C V^2$. The fraction of energy remaining after $1 \, s$ is $U_f / U_i = (V_f / V_i)^2 = (40/50)^2 = (4/5)^2 = 16/25$. Thus,$(a)$ is correct.
Since both $(a)$ and $(b)$ are correct,the correct option is $(d)$.

(C) In the steady state,the capacitor acts as an open circuit,so no current flows through the branch containing the capacitor.
Therefore,the entire current $i$ from the battery flows through the resistor $R_2$.
The total resistance of the circuit is $R_{total} = R_2 + r$.
The current in the circuit is $i = \frac{E}{R_2 + r}$.
The potential difference across the resistor $R_2$ is $V = i \times R_2 = \frac{E R_2}{R_2 + r}$.
Since the capacitor is in parallel with the resistor $R_2$,the potential difference across the capacitor is the same as the potential difference across $R_2$.
Thus,the charge on the capacitor is $Q = C \times V = \frac{C E R_2}{R_2 + r}$.

(B) The potential difference across a discharging capacitor is given by $V(t) = V_0 e^{-\lambda t}$.
Given $V_0 = 320 \, V$.
After $t = 1 \, s$,$V_1 = 240 \, V$.
So,$240 = 320 e^{-\lambda} \implies e^{-\lambda} = \frac{240}{320} = \frac{3}{4}$.
After $t = 2 \, s$,the potential difference is $V_2 = V_0 (e^{-\lambda})^2 = 320 \times (\frac{3}{4})^2 = 320 \times \frac{9}{16} = 180 \, V$.
After $t = 3 \, s$,the potential difference is $V_3 = V_0 (e^{-\lambda})^3 = 320 \times (\frac{3}{4})^3 = 320 \times \frac{27}{64} = 135 \, V$.

(A) The potential difference across the capacitor is $V_C = V_0(1 - e^{-t/RC})$ and across the resistor is $V_R = V_0 e^{-t/RC}$.
Given that $V_C = 3 V_R$,we have $V_0(1 - e^{-t/RC}) = 3 V_0 e^{-t/RC}$.
Dividing by $V_0$,we get $1 - e^{-t/RC} = 3 e^{-t/RC}$,which simplifies to $1 = 4 e^{-t/RC}$.
Thus,$e^{t/RC} = 4$,or $t/RC = \ln(4) = 2 \ln(2)$.
Given $R = 2.5 \times 10^6 \, \Omega$ and $C = 4 \times 10^{-6} \, F$,the time constant $\tau = RC = (2.5 \times 10^6) \times (4 \times 10^{-6}) = 10 \, s$.
Substituting the values,$t = 10 \times 2 \times 0.693 = 13.86 \, s$.

(A) During the charging of a capacitor in an $RC$ circuit,the charge $q$ on the capacitor at any time $t$ is given by $q = q_0(1 - e^{-t/CR})$.
Since the potential difference $V$ across the capacitor is $V = q/C$,we have $V = V_0(1 - e^{-t/CR})$,where $V_0$ is the maximum potential.
This equation represents an exponential growth curve that starts from $V = 0$ at $t = 0$ and asymptotically approaches $V_0$ as $t \to \infty$.
Therefore,the correct graph is the one showing exponential growth.

(C) During the discharge of a capacitor through a resistance,the charge at any instant $t$ is given by $Q = Q_0 e^{-t/CR}$.
Rearranging the equation: $\frac{Q_0}{Q} = e^{t/CR} \implies t = CR \ln\left(\frac{Q_0}{Q}\right)$.
For a constant charge $Q$,the time $t$ is directly proportional to the resistance $R$ $(t \propto R)$.
Now,draw a line parallel to the time axis as shown in the graph. Suppose this line intersects the curves at points corresponding to times $t_1, t_2,$ and $t_3$ for resistances $R_1, R_2,$ and $R_3$ respectively.
From the graph,it is clear that $t_1 < t_2 < t_3$.
Since $t \propto R$,it follows that $R_1 < R_2 < R_3$.
Therefore,the smallest resistance is $R_1$.

(B) For an $RC$ circuit during charging,the current $i$ at time $t$ is given by $i = \frac{E}{R} e^{-t/RC}$.
Taking the natural logarithm on both sides,we get $\ln i = -\frac{t}{RC} + \ln(\frac{E}{R})$.
This is an equation of a straight line $y = mx + c$,where the slope $m = -\frac{1}{RC}$ and the intercept $c = \ln(\frac{E}{R})$.
When the resistance $R$ is doubled $(R' = 2R)$:
$1$. The intercept $c' = \ln(\frac{E}{2R}) = \ln(\frac{E}{R}) - \ln 2$. Since $\ln 2 > 0$,the new intercept $c'$ is lower than the original intercept $c$.
$2$. The magnitude of the slope $|m'| = \frac{1}{R'C} = \frac{1}{2RC} = \frac{1}{2} |m|$. The new slope is half the magnitude of the original slope,meaning the line becomes less steep.
Comparing these changes with the given graph,curve $Q$ starts at a lower intercept and has a smaller magnitude of slope compared to the dotted line. Therefore,curve $Q$ represents the variation.

(B) When a capacitor is connected to a cell,it charges until the potential difference across the capacitor plates becomes equal to the $emf$ of the cell.
Once the capacitor is fully charged,no current flows through the circuit $(i = 0)$.
The potential difference across the cell is given by $V = E - ir$.
Since $i = 0$,the potential difference across the cell is $V = E - 0 \times r = E$.
Therefore,the potential difference across the cell is $E$.

(D) At time $t = 0$, the capacitor acts as a short circuit (zero resistance). The total resistance in the circuit is $1000 \, \Omega$. Thus, the current $I = \frac{2 \,V}{1000 \, \Omega} = 2 \,mA$.
As time passes, the capacitor charges up. When it is fully charged, it acts as an open circuit. The current then flows through the series combination of the $1000 \, \Omega$ resistor and the $1000 \, \Omega$ resistor. The total resistance becomes $2000 \, \Omega$. Thus, the steady-state current $I = \frac{2 \,V}{2000 \, \Omega} = 1 \,mA$. Therefore, the current decreases from $2 \,mA$ to $1 \,mA$ over time.

(D) When a charged capacitor is discharged through a resistance,the total energy stored in the capacitor is dissipated as heat in the resistance.
Energy stored in the capacitor is given by the formula $U = \frac{1}{2}CV^2$.
Given: $C = 4\,\mu F = 4 \times 10^{-6}\,F$ and $V = 400\,V$.
Substituting the values:
$U = \frac{1}{2} \times (4 \times 10^{-6}) \times (400)^2$
$U = 2 \times 10^{-6} \times 160000$
$U = 2 \times 10^{-6} \times 1.6 \times 10^5$
$U = 3.2 \times 10^{-1} = 0.32\,J$.
Therefore,the heat produced in the resistance is $0.32\,J$.

(B) The capacitive reactance of a capacitor is given by the formula $X_C = \frac{1}{\omega C} = \frac{1}{2\pi \nu C}$.
For direct current $(dc)$,the frequency $\nu = 0$.
Substituting this into the formula,we get $X_C = \frac{1}{2\pi (0) C} = \infty$.
Since the resistance (reactance) offered to $dc$ is infinite,a capacitor acts as a perfect insulator for $dc$.

(B) The capacitive reactance $X_C$ is given by the formula $X_C = \frac{1}{2\pi \nu C}$,where $\nu$ is the frequency of the source and $C$ is the capacitance.
In a $DC$ circuit,the frequency $\nu = 0 \, Hz$.
Substituting this into the formula: $X_C = \frac{1}{2\pi \times 0 \times 1} = \frac{1}{0} = \infty$.
Therefore,the effective resistance (reactance) of a pure capacitor in a $DC$ circuit is infinite.

(C) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.
Since the charge $q$ decays as $q(t) = q_0 e^{-t/RC}$,the energy decays as $U(t) = \frac{q_0^2}{2C} e^{-2t/RC} = U_0 e^{-2t/RC}$,where $RC = \tau$ is the time constant.
For $t_1$,the energy reduces to half: $U(t_1) = U_0/2$.
$U_0/2 = U_0 e^{-2t_1/\tau} \implies 1/2 = e^{-2t_1/\tau} \implies \ln(1/2) = -2t_1/\tau \implies -\ln(2) = -2t_1/\tau \implies t_1 = \frac{\tau \ln(2)}{2}$.
For $t_2$,the charge reduces to one-fourth: $q(t_2) = q_0/4$.
$q_0/4 = q_0 e^{-t_2/\tau} \implies 1/4 = e^{-t_2/\tau} \implies \ln(1/4) = -t_2/\tau \implies -2\ln(2) = -t_2/\tau \implies t_2 = 2\tau \ln(2)$.
Taking the ratio $t_1/t_2 = \frac{\tau \ln(2) / 2}{2\tau \ln(2)} = \frac{1/2}{2} = 1/4 = 0.25$.

(B) The time constant of the circuit is given by $\tau = RC = (1 \times 10^3 \,\Omega) \times (2500 \times 10^{-6} \,F) = 2.5 \,s$.
For a charging capacitor,the voltage at time $t$ is given by $V(t) = V_0(1 - e^{-t/\tau})$.
Substituting the given values: $V(5) = 12(1 - e^{-5/2.5}) = 12(1 - e^{-2})$.
Using $e^{-2} \approx 0.135$,we get $V(5) = 12(1 - 0.135) = 12(0.865) = 10.38 \,V$.

(B) For an $R-C$ charging circuit,the current at time $t$ is given by $I = \frac{E}{R} e^{-\frac{t}{RC}}$.
Taking the natural logarithm on both sides,we get $\ln I = \ln(\frac{E}{R}) - \frac{t}{RC}$.
This is a linear equation of the form $y = mx + c$,where the slope $m = -\frac{1}{RC}$ and the intercept $c = \ln(\frac{E}{R})$.
When the resistance $R$ is doubled $(R' = 2R)$:
$1$. The new slope $m' = -\frac{1}{R'C} = -\frac{1}{2RC} = \frac{1}{2}m$. Since the slope becomes half,the line becomes less steep.
$2$. The new intercept $c' = \ln(\frac{E}{2R}) = \ln(\frac{E}{R}) - \ln 2$. Since $\ln 2 > 0$,the intercept decreases.
Comparing the options,line $B$ has a smaller (less negative) slope and a lower intercept than the dashed line,which matches the derived conditions.

(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 $\ln(2.5) = 2.303 \times \log_{10}(2.5) = 2.303 \times 0.4 = 0.9212$.
$5 / (2 \times 10^{-6} R) = 0.9212$.
$R = 5 / (2 \times 10^{-6} \times 0.9212) \approx 2.71 \times 10^6\ \Omega$.

(A) The potential difference across a discharging capacitor is given by $V(t) = V_0 e^{-t/\tau}$,where $\tau$ is the time constant.
At $t = \tau$,the potential difference is $V(\tau) = V_0 e^{-1} \approx 0.37 V_0$.
From the graph,the initial potential difference $V_0 = 25 \ V$.
Therefore,at $t = \tau$,$V(\tau) = 0.37 \times 25 \ V = 9.25 \ V$.
Looking at the graph,at $t = 100 \ s$,$V \approx 13 \ V$ and at $t = 150 \ s$,$V \approx 9 \ V$.
Since $9.25 \ V$ lies between $13 \ V$ and $9 \ V$,the time constant $\tau$ must lie between $100 \ s$ and $150 \ s$.

(A) At $t = 0$,the capacitor acts as a short circuit (resistance $R_C = 0$). The circuit consists of the $2 \ V$ battery and the $1000 \ \Omega$ resistor in series with the shorted capacitor branch. Thus,the current $i$ is given by $i = \frac{V}{R} = \frac{2 \ V}{1000 \ \Omega} = 2 \ mA$.
As $t \to \infty$,the capacitor becomes fully charged and acts as an open circuit (resistance $R_C = \infty$). The current now flows through both $1000 \ \Omega$ resistors in series. The total resistance is $R_{total} = 1000 \ \Omega + 1000 \ \Omega = 2000 \ \Omega$. Thus,the steady-state current is $i = \frac{V}{R_{total}} = \frac{2 \ V}{2000 \ \Omega} = 1 \ mA$.
Therefore,at $t = 0$,$i = 2 \ mA$,and as time passes,the current decreases exponentially to $1 \ mA$.

(D) The charging voltage of a capacitor in an $RC$ circuit is given by $V(t) = V_0(1 - e^{-t/RC_{eq}})$.
For parallel combination,equivalent capacitance $C_p = C + C = 2C$. The time constant is $\tau_p = R(2C) = 2RC$.
Given $V(t_p) = V_0/2$ at $t_p = 10 \ s$,we have $V_0/2 = V_0(1 - e^{-t_p/2RC})$,which implies $1/2 = e^{-t_p/2RC}$,so $t_p = 2RC \ln(2) = 10 \ s$.
For series combination,equivalent capacitance $C_s = C/2$. The time constant is $\tau_s = R(C/2) = RC/2$.
We need to find $t_s$ such that $V(t_s) = V_0/2$,so $1/2 = e^{-t_s/(RC/2)}$,which implies $t_s = (RC/2) \ln(2)$.
Comparing the two expressions,$t_s = t_p / 4 = 10 / 4 = 2.5 \ s$.

(D) The energy stored in a capacitor is given by $U = \frac{1}{2}CV^2$.
Initial energy $U_0 = \frac{1}{2}CV_0^2$,where $V_0 = 100\, V$.
Final energy $U = \frac{1}{2}CV^2$,where $V = 80\, V$.
The energy lost is $\Delta U = U_0 - U = \frac{1}{2}CV_0^2 - \frac{1}{2}CV^2$.
The fractional loss of energy is $\frac{\Delta U}{U_0} = \frac{\frac{1}{2}CV_0^2 - \frac{1}{2}CV^2}{\frac{1}{2}CV_0^2} = \frac{V_0^2 - V^2}{V_0^2}$.
Substituting the values: $\frac{\Delta U}{U_0} = \frac{100^2 - 80^2}{100^2} = \frac{(100 - 80)(100 + 80)}{10000} = \frac{20 \times 180}{10000} = \frac{3600}{10000} = \frac{36}{100} = \frac{9}{25}$.

(D) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.
For energy to reduce to half its initial value: $\frac{U_0}{2} = \frac{q^2}{2C} \Rightarrow \frac{q_0^2}{2} = q^2 \Rightarrow q = \frac{q_0}{\sqrt{2}}$.
Using the discharging equation $q = q_0 e^{-t/RC}$,we have $\frac{q_0}{\sqrt{2}} = q_0 e^{-t_1/RC}$.
Taking the natural logarithm: $-\frac{t_1}{RC} = \ln(1/\sqrt{2}) = -\frac{1}{2} \ln 2$,so $t_1 = \frac{RC \ln 2}{2}$.
For charge to reduce to one-fourth its initial value: $\frac{q_0}{4} = q_0 e^{-t_2/RC}$.
Taking the natural logarithm: $-\frac{t_2}{RC} = \ln(1/4) = -2 \ln 2$,so $t_2 = 2RC \ln 2$.
The ratio $\frac{t_1}{t_2} = \frac{(RC \ln 2) / 2}{2RC \ln 2} = \frac{1}{4} = 0.25$.

(B) $(i)$ At $t \rightarrow 0$,the capacitor acts as a short circuit (zero resistance wire). In this state,the total resistance of the circuit is $R_1$. Therefore,the current $I = \frac{E}{R_1}$.
$(ii)$ At $t \rightarrow \infty$,the capacitor is fully charged and acts as an open circuit (no current flows through it). In this state,the resistors $R_1$ and $R_2$ are in series. The total resistance is $R_1 + R_2$. Therefore,the current $I = \frac{E}{R_1 + R_2}$.

(C) At $t = 0$,the capacitor acts as a short circuit (uncharged). The total resistance in the circuit is $1000\,\Omega$. Thus,the current $I = \frac{V}{R} = \frac{2\,V}{1000\,\Omega} = 2\,mA$.
As $t \to \infty$,the capacitor becomes fully charged and acts as an open circuit. The current now flows through the series combination of the two $1000\,\Omega$ resistors. The total resistance is $1000\,\Omega + 1000\,\Omega = 2000\,\Omega$. Thus,the current $I = \frac{V}{R_{total}} = \frac{2\,V}{2000\,\Omega} = 1\,mA$.

(B) The voltage across a discharging capacitor is given by $V(t) = V_0 e^{-t/RC}$.
Let $k = e^{-1/RC}$. Given $V_0 = 320\ V$ and $V(1) = 240\ V$.
$240 = 320 \cdot k \implies k = \frac{240}{320} = \frac{3}{4}$.
For $t = 2\ s$,$V(2) = V_0 \cdot k^2 = 320 \cdot (\frac{3}{4})^2 = 320 \cdot \frac{9}{16} = 180\ V$.
For $t = 3\ s$,$V(3) = V_0 \cdot k^3 = 320 \cdot (\frac{3}{4})^3 = 320 \cdot \frac{27}{64} = 135\ V$.

(A) The current in a discharging $RC$ circuit is given by $I(t) = I_0 e^{-t/\tau}$,where $\tau = R_{eq}C$ is the time constant.
Here,$R_{eq} = R + r$,where $R = 2 \, \Omega$ is the external resistance and $r$ is the resistance of the ammeter.
The capacitance is $C = 0.5 \, \mu F = 0.5 \times 10^{-6} \, F$.
Given that at $t = \ln 2 \, \mu s = \ln 2 \times 10^{-6} \, s$,the current $I(t) = I_0 / 2$.
Substituting these values into the equation: $I_0 / 2 = I_0 e^{-t/\tau} \Rightarrow 1/2 = e^{-t/\tau}$.
Taking the natural logarithm on both sides: $\ln(1/2) = -t/\tau \Rightarrow -\ln 2 = -t/\tau \Rightarrow \tau = t / \ln 2$.
Substituting $t = \ln 2 \, \mu s$: $\tau = (\ln 2 \, \mu s) / \ln 2 = 1 \, \mu s = 10^{-6} \, s$.
Since $\tau = (R + r)C$,we have $10^{-6} = (2 + r) \times 0.5 \times 10^{-6}$.
Dividing both sides by $10^{-6}$: $1 = (2 + r) \times 0.5$.
$2 = 2 + r \Rightarrow r = 0 \, \Omega$.

(D) When the switch $S$ is opened,the battery and the resistor in series with it are removed from the circuit. The capacitor $C$ is left in a loop with two resistors of $1 \, k\Omega$ each,which are connected in parallel to each other.
The equivalent resistance $R_{eq}$ of these two resistors in parallel is given by:
$R_{eq} = \frac{1 \, k\Omega \times 1 \, k\Omega}{1 \, k\Omega + 1 \, k\Omega} = 0.5 \, k\Omega = 500 \, \Omega$.
The time constant $\tau$ for the discharging circuit is given by:
$\tau = R_{eq} C$
$\tau = (500 \, \Omega) \times (100 \times 10^{-6} \, F)$
$\tau = 50000 \times 10^{-6} \, s = 0.05 \, s$
Converting the time constant to milliseconds $(ms)$:
$\tau = 0.05 \times 1000 \, ms = 50 \, ms$.

(C) To find the time constant $\tau = R_{eq}C$,we need to determine the equivalent resistance $R_{eq}$ as seen by the capacitor $C$ when the voltage sources are short-circuited.
$1$. Short-circuit the voltage sources $V_1$ and $V_2$. The resistor $2R$ and $R$ (in the left branch) are in series,giving $3R$.
$2$. This $3R$ is in parallel with the middle resistor $R$. The equivalent resistance of this part is $R_{th} = \frac{(3R \times R)}{(3R + R)} = \frac{3R^2}{4R} = \frac{3}{4}R$.
$3$. The capacitor $C$ is in series with the resistor $R$ on the right side. Thus,the total equivalent resistance $R_{eq}$ seen by the capacitor is $R_{eq} = R_{th} + R = \frac{3}{4}R + R = \frac{7}{4}R$.
$4$. Therefore,the time constant is $\tau = R_{eq}C = \frac{7}{4}RC$.

(B) The circuit consists of two $RC$ branches connected in parallel to a voltage source $V$. When the switch $S$ is closed at $t = 0$,each branch charges independently.
For the first branch with capacitor $C_1$ and resistor $R$,the current is given by $i_1(t) = \frac{V}{R} e^{-t / (R C_1)}$.
For the second branch with capacitor $C_2$ and resistor $R$,the current is given by $i_2(t) = \frac{V}{R} e^{-t / (R C_2)}$.
The ratio of the currents is $\frac{i_1}{i_2} = \frac{\frac{V}{R} e^{-t / (R C_1)}}{\frac{V}{R} e^{-t / (R C_2)}} = e^{-t / (R C_1) + t / (R C_2)} = e^{\frac{t}{R} \left( \frac{1}{C_2} - \frac{1}{C_1} \right)}$.
Given $C_1 = 2C_2$,we have $\frac{1}{C_2} = \frac{2}{C_1}$.
Substituting this into the ratio: $\frac{i_1}{i_2} = e^{\frac{t}{R} \left( \frac{2}{C_1} - \frac{1}{C_1} \right)} = e^{\frac{t}{R C_1}}$.
Since $t$ increases,the exponent $\frac{t}{R C_1}$ increases,and therefore the ratio $i_1 / i_2 = e^{\frac{t}{R C_1}}$ increases with time $t$.

(D) At $t = 0$,the capacitor acts as a short circuit (it is uncharged). The total resistance in the circuit is $1000\,\Omega$. Thus,the current $I$ through the resistor $AB$ is $I = \frac{2\,V}{1000\,\Omega} = 2\,mA$.
As time passes,the capacitor charges. As it charges,its potential difference increases,which opposes the flow of current.
After a long time $(t \to \infty)$,the capacitor is fully charged and acts as an open circuit. The current then flows through both resistors in series. The total resistance becomes $1000\,\Omega + 1000\,\Omega = 2000\,\Omega$. The current $I$ through the resistor $AB$ becomes $I = \frac{2\,V}{2000\,\Omega} = 1\,mA$.
Therefore,at $t = 0$,$I = 2\,mA$,and as $t$ increases,$I$ decreases to $1\,mA$.

(B) When the switch $S$ is closed, the energy stored in the capacitor is completely dissipated as heat through the resistor.
The energy stored in a capacitor with capacitance $C$ and initial charge $Q$ is given by $U = \frac{Q^2}{2C}$.
Given, $U = 3.6 \times 10^{-3} \ J$ and $C = 2 \ \mu F = 2 \times 10^{-6} \ F$.
Equating the energy, we have:
$\frac{Q^2}{2 \times 2 \times 10^{-6}} = 3.6 \times 10^{-3}$
$Q^2 = 3.6 \times 10^{-3} \times 4 \times 10^{-6}$
$Q^2 = 14.4 \times 10^{-9} = 144 \times 10^{-10}$
Taking the square root on both sides:
$Q = \sqrt{144 \times 10^{-10}} = 12 \times 10^{-5} \ C$
Converting to microcoulombs $(\mu C)$:
$Q = 120 \times 10^{-6} \ C = 120 \ \mu C$.

(C) During discharging,the current in the circuit is given by $i(t) = i_0 e^{-t / RC}$.
Given that at $t = (\ln 4) \, \mu s$,the current $i = i_0 / 2$.
Substituting these values into the equation: $i_0 / 2 = i_0 e^{-(\ln 4 \, \mu s) / RC}$.
This simplifies to $1/2 = e^{-(\ln 4 \, \mu s) / RC}$,or $2 = e^{(\ln 4 \, \mu s) / RC}$.
Taking the natural logarithm on both sides: $\ln 2 = (\ln 4 \, \mu s) / RC$.
Since $\ln 4 = 2 \ln 2$,we have $\ln 2 = (2 \ln 2 \, \mu s) / RC$.
Thus,$RC = 2 \, \mu s$.
Given $C = 0.5 \, \mu F$,we find $R = (2 \, \mu s) / (0.5 \, \mu F) = 4 \, \Omega$.
The total resistance $R$ is the sum of the resistor and the ammeter resistance $r$: $R = 2 \, \Omega + r$.
Therefore,$2 \, \Omega + r = 4 \, \Omega$,which gives $r = 2 \, \Omega$.

(C) Just after the switch $S$ is closed,the capacitors act as short circuits (wires) because they are initially uncharged. However,looking at the circuit,the capacitors are in series with the $4\,\Omega$ resistor and in parallel with the $8\,\Omega$ resistor. At $t=0$,the capacitor acts as a wire,so the $4\,\Omega$ branch is effectively shorted. The circuit simplifies to the $6\,V$ battery in series with the $2\,\Omega$ resistor and the $8\,\Omega$ resistor. Thus,$I_1 = I_3 = 6\,V / (2\,\Omega + 8\,\Omega) = 0.6\,A$ and $I_2 = 0$.
Long time after the switch $S$ is closed,the capacitors are fully charged and act as open circuits. The branch containing the $4\,\Omega$ resistor and the capacitor will have no current $(I_2 = 0)$. The circuit simplifies to the $6\,V$ battery in series with the $2\,\Omega$ and $8\,\Omega$ resistors. Thus,$I_1 = I_3 = 6\,V / (10\,\Omega) = 0.6\,A$ and $I_2 = 0$.

(D) At $t = 0$,the capacitor acts as a short circuit. The equivalent resistance of the circuit is $R_{eq} = R_3 + (R_1 || R_2) = 1 + \frac{2 \times 2}{2 + 2} = 1 + 1 = 2 \, \Omega$. The total current from the battery is $I = \frac{\varepsilon}{R_{eq}} = \frac{10}{2} = 5 \, A$. This current flows through $R_3$,so option $A$ is correct.
As $t \to \infty$,the capacitor is fully charged and acts as an open circuit. The current flows through the series combination of $R_1 || R_2$ and $R_3$. The total resistance is $R_{eq} = 2 \, \Omega$. The current through $R_3$ is $I = \frac{10}{2} = 5 \, A$. Note: The original option $B$ was incorrect; the current is $5 \, A$ at $t \to \infty$ as well.
Since $R_1$ and $R_2$ are in parallel,the voltage across them is the same. Thus,$I_{R_1} R_1 = I_{R_2} R_2$,which implies $\frac{I_{R_1}}{I_{R_2}} = \frac{R_2}{R_1} = \frac{2}{2} = 1$. This ratio is constant. Thus,option $C$ is correct.
Given the options,$D$ is the most appropriate choice as $A$ and $C$ are correct.

(B) The circuit represents the discharging of a capacitor through a resistor $R$. The current $I$ at time $t$ is given by $I = I_0 e^{-t / RC}$,where $I_0 = V/R$.
Taking the natural logarithm on both sides: $\ln I = \ln(V/R) - \frac{t}{RC}$.
Converting to base $10$ logarithm: $\log_{10} I = \log_{10}(V/R) - \frac{t}{2.303 RC}$.
Comparing this with the equation of a straight line $y = mx + c$,the intercept is $\log_{10}(V/R)$ and the slope is $m = -\frac{1}{2.303 RC}$.
From the graph,the intercept for both lines $(1)$ and $(2)$ is the same,which implies that $V$ and $R$ remain constant.
The magnitude of the slope of line $(2)$ is less than that of line $(1)$.
Since $|m| = \frac{1}{2.303 RC}$,a decrease in the magnitude of the slope implies that the product $RC$ must increase.
Since $R$ is constant,$C$ must be increased.

(D) From the graph,both circuits reach the same maximum charge $q_{\max}$.
Since $q_{\max} = C_1 V_1 = C_2 V_2$,it is clear that both capacitors are charged to the same magnitude of charge. Thus,statement $A$ is correct.
The time constant $\tau = RC$ determines the rate of charging. From the graph,circuit $2$ takes more time to reach $q_{\max}$ compared to circuit $1$,implying $\tau_2 > \tau_1$,or $R_2 C_2 > R_1 C_1$.
Since $q_{\max} = CV$,we have $V = q_{\max} / C$. Because we do not know the individual values of $C_1$ and $C_2$,we cannot determine the relationship between $V_1$ and $V_2$. Therefore,the $emf's$ ($V_1$ and $V_2$) may be different. Thus,statement $C$ is also correct.
Since both $A$ and $C$ are correct,the correct option is $D$.

(D) The charge on a capacitor during charging is given by $q(t) = CE(1 - e^{-t/RC})$.
The time constant is $\tau = RC$.
From the graph,circuit $2$ takes more time to reach the maximum charge compared to circuit $1$,which means the time constant of circuit $2$ is greater than that of circuit $1$.
Thus,$\tau_2 > \tau_1$,which implies $R_2C_2 > R_1C_1$.
Rearranging this,we get $\frac{R_1}{R_2} < \frac{C_2}{C_1}$.
Therefore,option $D$ is correct.

(D) $1$. When $X$ is connected to $Y$,the capacitor charges to a potential difference $V = E$. The charge on the capacitor is $q_1 = CE$. The energy stored is $U_1 = \frac{1}{2}CE^2$.
$2$. When $X$ is connected to $Z$,the capacitor is connected to cell $B$ with reversed polarity. The potential difference across the capacitor changes from $+E$ to $-E$. The final charge on the capacitor is $q_2 = -CE$.
$3$. The total charge that flows through the circuit is $\Delta q = q_1 - q_2 = CE - (-CE) = 2CE$.
$4$. The work done by the cell $B$ is $W = \Delta q \cdot E = (2CE) \cdot E = 2CE^2$.
$5$. The change in energy stored in the capacitor is $\Delta U = U_f - U_i = \frac{1}{2}CE^2 - \frac{1}{2}CE^2 = 0$.
$6$. By the work-energy theorem,$W = \Delta U + H$,where $H$ is the heat produced. Thus,$H = W = 2CE^2$.
$7$. Since the final energy stored is $U_f = \frac{1}{2}CE^2$,we have $H = 4 \cdot (\frac{1}{2}CE^2) = 4U_f$. Therefore,the heat produced is $4$ times the energy finally stored in the capacitor.

(D) From the graph,the potential difference $V$ as a function of time $t$ is a linear equation: $V(t) = mt + c$.
At $t=0$,$V=2\, V$,so the intercept $c=2$.
At $t=3\, s$,$V=8\, V$,so the slope $m = \frac{8-2}{3-0} = \frac{6}{3} = 2\, V/s$.
Thus,$V(t) = 2t + 2$.
For a capacitor,$V = \frac{q}{C} = \frac{1}{C} \int i \, dt$.
Given $i = 1\, A$,we have $V = \frac{1}{C} \int 1 \, dt = \frac{t}{C} + V_0$,where $V_0$ is the initial voltage at $t=0$.
Comparing $V(t) = \frac{1}{C}t + 2$ with $V(t) = 2t + 2$,we get $\frac{1}{C} = 2$,which implies $C = 0.5\, F$.

(C) In the steady state,the capacitor acts as an open circuit,so no current flows through the branch containing the capacitor $C$ and resistor $r_1$.
The circuit reduces to a simple series circuit consisting of the battery $E$ and resistors $r$ and $r_2$.
The current $i$ flowing through the circuit is given by:
$i = \frac{E}{r + r_2}$
The potential difference across the capacitor $V_c$ is equal to the potential difference across the resistor $r_2$ because they are connected in parallel.
$V_c = i \cdot r_2 = \left( \frac{E}{r + r_2} \right) r_2$
The charge $Q$ on the capacitor is given by $Q = C V_c$.
Substituting the value of $V_c$:
$Q = C \left( \frac{E r_2}{r + r_2} \right) = CE \frac{r_2}{r + r_2}$

(D) The charge on a discharging capacitor is given by $q(t) = q_0 e^{-t/RC}$.
The energy stored in the capacitor is $U(t) = \frac{q^2}{2C} = \frac{q_0^2}{2C} e^{-2t/RC} = U_0 e^{-2t/RC}$.
For time $t_1$,the energy reduces to half: $U_0/2 = U_0 e^{-2t_1/RC} \implies 1/2 = e^{-2t_1/RC} \implies \ln(2) = 2t_1/RC \implies t_1 = \frac{RC \ln(2)}{2}$.
For time $t_2$,the charge reduces to one-fourth: $q_0/4 = q_0 e^{-t_2/RC} \implies 1/4 = e^{-t_2/RC} \implies \ln(4) = t_2/RC \implies t_2 = RC \ln(4) = 2RC \ln(2)$.
The ratio $t_1 / t_2 = \frac{RC \ln(2) / 2}{2RC \ln(2)} = 1/4 = 0.25$.

(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 $\ln(2.5) = 2.303 \times 0.4 = 0.9212$.
$R = 5 / (2 \times 10^{-6} \times 0.9212) \approx 2.71 \times 10^6 \ \Omega$.

(A) The equation for the discharging of a capacitor is given by $V = V_{0} e^{-t/\tau}$,where $\tau = RC$ is the time constant.
At $t = \tau$,the potential difference becomes $V = V_{0} / e \approx 0.37 V_{0}$.
From the graph,the initial potential difference $V_{0} = 25\; V$.
Therefore,at $t = \tau$,$V = 0.37 \times 25\; V = 9.25\; V$.
Looking at the graph,at $t = 100\; sec$,the potential is slightly above $10\; V$,and at $t = 150\; sec$,the potential is slightly below $10\; V$ (specifically around $8\; V$ to $9\; V$).
Since $9.25\; V$ corresponds to the time constant $\tau$,and this value lies between $100\; sec$ and $150\; sec$ on the time axis,the correct range is $100\; sec$ and $150\; sec$.

(D) When switch $S_1$ is closed and $S_2$ is open,the circuit behaves as a simple series $RC$ circuit consisting of the battery $V$,resistor $R$,and capacitor $C$.
The charge $q$ on the capacitor at any time $t$ is given by the charging formula: $q(t) = CV(1 - e^{-t/\tau})$,where $\tau = RC$.
Checking option $D$: At $t = 2\tau$,substituting the value into the formula gives $q = CV(1 - e^{-2\tau/\tau}) = CV(1 - e^{-2})$.
Thus,option $D$ is the correct statement.

(A) $Q_0$ is the maximum charge on the capacitor at steady state,i.e.,at $t = \infty$.
At $t = \infty$,the capacitor acts as an open circuit,meaning no current flows through it.
The circuit simplifies to a series combination of $R_1$ and $R_2$ connected to the voltage source $V$.
The potential difference across the capacitor is equal to the potential difference across the resistor $R_2$.
Using the voltage divider rule,the voltage across $R_2$ is $V_C = V \left( \frac{R_2}{R_1 + R_2} \right)$.
The charge on the capacitor is $Q_0 = C V_C$.
Substituting the value of $V_C$,we get $Q_0 = C \left( \frac{V R_2}{R_1 + R_2} \right) = \frac{CVR_2}{R_1 + R_2}$.