R-L D.C. Circuit Questions in English

(A) The time constant $\tau$ of an $L-R$ circuit is given by the formula $\tau = \frac{L}{R}$.
Here,$L$ is the inductance measured in Henry $(H)$ and $R$ is the resistance measured in Ohm $(\Omega)$.
Since $\tau$ represents time,its unit is second $(s)$.
Therefore,the unit $Henry/ohm$ is equivalent to second.

(A) The dimensions of inductance $L$ are $[M L^2 T^{-2} A^{-2}]$.
The dimensions of resistance $R$ are $[M L^2 T^{-3} A^{-2}]$.
Therefore,the dimensions of $L/R$ are:
$\frac{[M L^2 T^{-2} A^{-2}]}{[M L^2 T^{-3} A^{-2}]} = [T^1]$.
Since the dimension of $L/R$ is time,its unit is the second $(s)$.

(D) In an $LR$ circuit,the current $I$ at any time $t$ is given by $I = I_0(1 - e^{-Rt/L})$.
For the exponent to be dimensionless,the term $\frac{Rt}{L}$ must be dimensionless.
This implies that the dimensions of $\frac{Rt}{L}$ are equal to the dimensions of $1$ (a dimensionless constant).
Therefore,the dimensions of $\frac{L}{R}$ must be equal to the dimensions of $t$ (time).
Thus,the time constant $\tau = \frac{L}{R}$ has the dimensions of time.

(B) In a $DC$ circuit,at steady state,the inductor acts as a simple wire with zero resistance (ideal) or its internal resistance.
Given that the solenoid has a resistance $R = 10\,\Omega$ and is connected to a battery of $E = 10\,V$.
At steady state,the inductive reactance $X_L = 2\pi f L$ is zero because the frequency $f$ of a $DC$ source is $0$.
Therefore,the current $i$ is determined solely by Ohm's law:
$i = \frac{E}{R} = \frac{10\,V}{10\,\Omega} = 1\,A$.

(B) The current in an $LR$ circuit at time $t$ is given by the formula: $i = i_0(1 - e^{-Rt/L})$.
Here, the maximum current $i_0 = V/R = 5 \, V / 5 \, \Omega = 1 \, A$.
The time constant $\tau = L/R = 10 \, H / 5 \, \Omega = 2 \, s$.
Given $t = 2 \, s$, the current $i$ is:
$i = 1 \times (1 - e^{-(5 \times 2)/10})$
$i = 1 \times (1 - e^{-10/10})$
$i = (1 - e^{-1}) \, A$.

(C) The time constant $\tau$ of an $LR$ circuit is defined as the time taken for the current to decay to $1/e$ (approximately $37\%$) of its initial steady value.
The formula for the time constant is $\tau = \frac{L}{R}$.
Given:
Self-inductance $L = 50\,H$
Resistance $R = 10\,\Omega$
Substituting the values:
$\tau = \frac{50}{10} = 5\,s$.
Therefore,the current will decay to $1/e$ of its steady value in $5\,seconds$.

(D) The current in an $LR$ circuit at time $t$ is given by $I = I_0(1 - e^{-Rt/L})$,where $I_0 = V/R$ is the final steady-state current.
When the current rises to $0.63$ of its final value,$I = 0.63 I_0$.
Substituting this into the equation: $0.63 I_0 = I_0(1 - e^{-Rt/L})$.
$0.63 = 1 - e^{-Rt/L}$,which implies $e^{-Rt/L} = 1 - 0.63 = 0.37$.
Since $e^{-1} \approx 0.368 \approx 0.37$,we have $Rt/L = 1$.
Therefore,the time $t$ is equal to the time constant $\tau = L/R$.
Given $L = 2.5\,H$ and $R = 0.5\,\Omega$,we get $t = 2.5 / 0.5 = 5\,s$.

(B) In an $L-R$ circuit connected to a $D.C.$ source of $e.m.f.$ $E$,the current $I$ at any time $t$ is given by the expression: $I(t) = \frac{E}{R} (1 - e^{-Rt/L})$.
As time $t$ approaches infinity (a long time),the term $e^{-Rt/L}$ approaches $0$.
Therefore,the steady-state current becomes $I = \frac{E}{R} (1 - 0) = \frac{E}{R}$.
At this stage,the inductor acts as a simple wire with zero resistance,and the current is limited only by the resistor $R$.

(B) For an $RL$ circuit,the current at any time $t$ is given by $i = i_0(1 - e^{-Rt/L})$,where $i_0 = E/R$.
Taking the derivative with respect to time $t$,we get $\frac{di}{dt} = \frac{i_0 R}{L} e^{-Rt/L}$.
Substituting $i_0 = E/R$,we have $\frac{di}{dt} = \frac{E}{L} e^{-Rt/L}$.
At the instant of closing the switch $(t = 0)$,the expression becomes $\frac{di}{dt} = \frac{E}{L}$.
Given $\frac{di}{dt} = 4 \, A/s$,$L = 20 \, H$,and $R = 5 \, \Omega$.
Substituting these values: $4 = \frac{E}{20}$.
Therefore,$E = 4 \times 20 = 80 \, V$.

(B) When a direct current source is connected to an $L-R$ series circuit,the current does not reach its maximum value instantaneously due to the self-induced electromotive force in the inductor.
The growth of current in an $L-R$ circuit is given by the differential equation $L(dI/dt) + IR = E$.
Solving this equation with the initial condition $I(0) = 0$,we get the expression for current as a function of time:
$I(t) = I_0 (1 - e^{-Rt/L})$
where $I_0 = E/R$ is the maximum steady-state current and $\tau = L/R$ is the time constant of the circuit.

(C) When the battery is disconnected,the current through the $RL$ circuit starts decreasing exponentially according to the equation $i = i_0 e^{-Rt/L}$.
Given that the current reduces to $37\%$ of its initial value,we have $i = 0.37 i_0$.
Substituting this into the equation: $0.37 i_0 = i_0 e^{-Rt/L}$.
Since $e^{-1} \approx 0.3678 \approx 0.37$,we can write $0.37 \approx e^{-1}$.
Therefore,$e^{-1} = e^{-Rt/L}$.
Comparing the exponents,we get $1 = \frac{Rt}{L}$.
Solving for $t$,we find $t = \frac{L}{R}$ seconds.

(A) The current $I$ at any instant $t$ after closing an $LR$ circuit is given by the formula: $I = {I_0}(1 - e^{-\frac{R}{L}t})$.
The time constant $\tau$ is defined as $\tau = \frac{L}{R}$.
Substituting $t = \tau = \frac{L}{R}$ into the current equation:
$I = {I_0}(1 - e^{-\frac{R}{L} \cdot \frac{L}{R}}) = {I_0}(1 - e^{-1})$.
Since $e^{-1} \approx \frac{1}{2.718} \approx 0.37$,we get:
$I = {I_0}(1 - 0.37) = 0.63{I_0}$.
Thus,the current grows to $63\%$ of its steady-state value in one time constant.

(D) The time constant $\tau$ of an $RL$ circuit is defined as the ratio of the inductance $L$ to the resistance $R$.
Formula: $\tau = \frac{L}{R}$
Given:
Inductance $L = 40 \, H$
Resistance $R = 8 \, \Omega$
Substituting the values:
$\tau = \frac{40}{8} = 5 \, s$
Therefore,the time constant of the circuit is $5 \, s$.

(B) The current in an $LR$ circuit at time $t$ is given by the formula $I(t) = I_0(1 - e^{-t/\tau})$,where $I_0 = V/R$ is the maximum steady-state current and $\tau = L/R$ is the time constant of the circuit.
We are asked to find the time $t$ when the current reaches $\frac{e-1}{e} \approx 63.2\%$ of its final value $I_0$.
This condition corresponds to $I(t) = I_0(1 - e^{-1}) = I_0(1 - 1/e)$.
Comparing this with the general formula $I(t) = I_0(1 - e^{-t/\tau})$,we see that $t = \tau$.
The time constant $\tau$ is calculated as $\tau = \frac{L}{R} = \frac{60 \, H}{30 \, \Omega} = 2 \, s$.
Therefore,it will take $2 \, s$ for the current to reach approximately $63.2\%$ of its final value.

(A) When the switch $S$ is closed, the current in the branch containing the inductor $L$ grows slowly due to the back $EMF$ induced by the inductor, which opposes the change in current $(e = -L \frac{di}{dt})$.
In the branch containing the resistor $R$, the current reaches its steady state value almost instantaneously.
Therefore, the bulb $B_2$ (in the resistor branch) lights up earlier than $B_1$ (in the inductor branch).
However, the problem states that the resistance of the coil $L$ is equal to the resistance $R$.
In the steady state, the inductor acts as a simple wire with resistance. Since the total resistance of both branches is the same (assuming the bulbs have identical resistance), the steady-state current in both branches will be equal.
Thus, both bulbs will eventually shine with equal brightness.
Wait, re-evaluating the standard problem: If the coil $L$ has internal resistance $r_L$ and we are given $r_L = R$, then the total resistance of the top branch is $R_{total} = r_L + R_{bulb1}$ and the bottom branch is $R_{total} = R + R_{bulb2}$. If $R_{bulb1} = R_{bulb2}$, then the final brightness is equal.
Looking at the provided options, option $(A)$ is the standard physical outcome for this circuit configuration.

(B) In an $LR$ circuit,the current $I$ at time $t$ is given by the equation $I(t) = I_0(1 - e^{-t/\tau})$,where $I_0$ is the maximum steady-state current and $\tau = L/R$ is the time constant.
When $t = \tau$,the current becomes $I(\tau) = I_0(1 - e^{-1})$.
Since $e^{-1} \approx 0.368$,we have $I(\tau) = I_0(1 - 0.368) = I_0(0.632)$.
This means the current reaches approximately $63\%$ of its final maximum value.

(D) The current in an $LR$ circuit is given by $i = i_0(1 - e^{-Rt/L})$,where $i_0 = V/R$ is the steady state current.
We are given that the current reaches half of its steady state value,so $i = i_0/2$.
Substituting this into the equation: $i_0/2 = i_0(1 - e^{-Rt/L})$.
This simplifies to $1/2 = 1 - e^{-Rt/L}$,which means $e^{-Rt/L} = 1/2$.
Taking the natural logarithm on both sides: $-Rt/L = \ln(1/2) = -\ln(2)$.
Thus,$t = (L/R) \ln(2)$.
Given $L = 300\, mH = 0.3\, H$ and $R = 2\, \Omega$,we have $t = (0.3 / 2) \times 0.693$.
$t = 0.15 \times 0.693 = 0.10395\, s \approx 0.1\, s$.

(B) The current in an $R-L$ circuit at any time $t$ is given by $i(t) = i_0(1 - e^{-Rt/L})$,where $i_0 = E/R$ is the steady-state current.
Given: $E = 15 \ V$,$L = 5 \ H$,$R = 10 \ \Omega$.
Steady-state current $(t = \infty)$: $i_{\infty} = E/R = 15/10 = 1.5 \ A$.
At $t = 1 \ s$: $i_1 = i_0(1 - e^{-R(1)/L}) = 1.5(1 - e^{-10/5}) = 1.5(1 - e^{-2})$.
The ratio of currents is $\frac{i_{\infty}}{i_1} = \frac{1.5}{1.5(1 - e^{-2})} = \frac{1}{1 - e^{-2}} = \frac{1}{1 - 1/e^2} = \frac{e^2}{e^2 - 1}$.

(C) The current in an $LR$ series circuit is given by $i = i_0(1 - e^{-Rt/L})$,where $i_0 = E/R$.
Differentiating with respect to time $t$,we get the rate of change of current:
$\frac{di}{dt} = \frac{d}{dt} [i_0(1 - e^{-Rt/L})] = i_0 \cdot (\frac{R}{L}) e^{-Rt/L} = (\frac{E}{R}) \cdot (\frac{R}{L}) e^{-Rt/L} = \frac{E}{L} e^{-Rt/L}$.
At the initial moment,$t = 0$,the exponential term $e^0 = 1$.
Therefore,the initial rate of change of current is $\frac{di}{dt} = \frac{E}{L}$.
Given $E = 5 \, V$ and $L = 2 \, H$:
$\frac{di}{dt} = \frac{5}{2} = 2.5 \, A/s$.

(D) The maximum current in the circuit is given by $i_0 = \frac{V}{R} = \frac{12 \, V}{6 \, \Omega} = 2 \, A$.
The current in an $LR$ circuit at time $t$ is given by $i(t) = i_0(1 - e^{-Rt/L})$.
We want to find the time $t$ when $i(t) = 1.0 \, A$. Substituting the values:
$1.0 = 2(1 - e^{-6t / (8.4 \times 10^{-3})})$
$0.5 = 1 - e^{-6t / (8.4 \times 10^{-3})}$
$e^{-6t / (8.4 \times 10^{-3})} = 0.5$
Taking the natural logarithm on both sides:
$-\frac{6t}{8.4 \times 10^{-3}} = \ln(0.5) \approx -0.693$
$t = 0.693 \times \frac{8.4 \times 10^{-3}}{6} = 0.693 \times 1.4 \times 10^{-3} \approx 0.97 \times 10^{-3} \, s \approx 1 \, ms$.

(D) The current in an $LR$ circuit at time $t$ is given by $i = i_0(1 - e^{-Rt/L})$, where $i_0 = V/R$.
The rate of increase of current is $\frac{di}{dt} = \frac{d}{dt} [i_0(1 - e^{-Rt/L})] = i_0 \cdot \frac{R}{L} e^{-Rt/L} = \frac{V}{R} \cdot \frac{R}{L} e^{-Rt/L} = \frac{V}{L} e^{-Rt/L}$.
Given $V = 50 \, V$, $L = 5 \times 10^{-3} \, H$, $R = 180 \, \Omega$, and $t = 0.001 \, s$.
Calculate the exponent: $\frac{Rt}{L} = \frac{180 \times 0.001}{5 \times 10^{-3}} = \frac{0.18}{0.005} = 36$.
Now, $\frac{di}{dt} = \frac{50}{5 \times 10^{-3}} e^{-36} = 10^4 \times e^{-36} \, A/s$.
Since $e^{-36}$ is an extremely small value, the result is approximately $10^4 \times 1.7 \times 10^{-16} \approx 1.7 \times 10^{-12} \, A/s$.
None of the given options match this value.

(B) When a battery of electromotive force $E$ is connected in series with an inductor $L$ and a resistor $R$,the circuit is known as an $RL$ circuit.
According to Kirchhoff's voltage law,the equation for the circuit is $E - L \frac{di}{dt} = iR$.
Rearranging the terms,we get $\frac{di}{dt} = \frac{E - iR}{L}$.
Solving this differential equation with the initial condition that at $t = 0$,$i = 0$,we get the expression for current as a function of time:
$i(t) = \frac{E}{R} (1 - e^{-\frac{R}{L}t})$.
Here,$i_0 = \frac{E}{R}$ is the maximum steady-state current.
The function $i(t) = i_0 (1 - e^{-\frac{R}{L}t})$ represents an exponential growth curve that starts from the origin $(0,0)$ and asymptotically approaches the value $i_0$ as $t$ approaches infinity.
Among the given options,the graph in option $B$ correctly depicts this exponential growth behavior.

(C) The current in an $RL$ circuit is given by $i(t) = \frac{E}{R}(1 - e^{-Rt/L})$.
The rate of change of current is $\frac{di}{dt} = \frac{d}{dt} [\frac{E}{R}(1 - e^{-Rt/L})] = \frac{E}{R} \cdot (\frac{R}{L}) e^{-Rt/L} = \frac{E}{L} e^{-Rt/L}$.
Given: $E = 5 \, V$, $R = 20 \, \Omega$, $L = 5 \, H$, and $t = 0.25 \, s$.
Calculate the exponent: $\frac{Rt}{L} = \frac{20 \times 0.25}{5} = \frac{5}{5} = 1$.
Substitute the values: $\frac{di}{dt} = \frac{5}{5} e^{-1} = 1 \cdot e^{-1} = e^{-1} \, A/s$.

(B) In an $R-L$ circuit,when the switch $S$ is closed at $t = 0$,the current $i$ starts from zero and grows exponentially according to the equation $i(t) = \frac{E}{R}(1 - e^{-Rt/L})$. Thus,the graph of $i$ vs $t$ is an increasing curve.
The induced $emf$ across the inductor is given by $e = L \frac{di}{dt}$. Differentiating the current equation,we get $e(t) = E e^{-Rt/L}$. At $t = 0$,$e = E$ (maximum),and as $t$ increases,$e$ decays exponentially to zero. Thus,the graph of $e$ vs $t$ is a decreasing curve.

(B) The growth of current in an $L-R$ circuit is given by $i = i_0(1 - e^{-t/\tau})$,where $\tau = \frac{L}{R}$ is the time constant.
Given that at $t = 4 \ s$,$i = \frac{3}{4}i_0$.
Substituting these values: $\frac{3}{4}i_0 = i_0(1 - e^{-4/\tau})$.
$\frac{3}{4} = 1 - e^{-4/\tau} \implies e^{-4/\tau} = 1 - \frac{3}{4} = \frac{1}{4}$.
Taking the natural logarithm on both sides: $-\frac{4}{\tau} = \ln(\frac{1}{4}) = -\ln(4)$.
$\frac{4}{\tau} = \ln(2^2) = 2 \ln 2$.
$\tau = \frac{4}{2 \ln 2} = \frac{2}{\ln 2} \ s$.

(C) The current $i$ in an $LR$ circuit is given by $i = i_0(1 - e^{-Rt/L})$,where $i_0 = E/R$.
Taking the derivative with respect to time $t$,we get the rate of increase of current:
$\frac{di}{dt} = \frac{d}{dt} [\frac{E}{R}(1 - e^{-Rt/L})] = \frac{E}{R} \cdot (\frac{R}{L}) e^{-Rt/L} = \frac{E}{L} e^{-Rt/L}$.
Given values: $E = 5 \ V$,$R = 20 \ \Omega$,$L = 5 \ H$,and $t = 0.25 \ s$.
Calculate the exponent: $\frac{Rt}{L} = \frac{20 \times 0.25}{5} = \frac{5}{5} = 1$.
Substitute the values into the expression for $\frac{di}{dt}$:
$\frac{di}{dt} = \frac{5}{5} e^{-1} = 1 \cdot e^{-1} = e^{-1} \ A/s$.

(A) In an $L-R$ circuit,the energy stored in the magnetic field of the inductor is given by $U = \frac{1}{2} L I^2$.
The rate of dissipation of energy in the resistance is given by $P = I^2 R$.
The ratio of the energy stored to the rate of dissipation is $\frac{U}{P} = \frac{\frac{1}{2} L I^2}{I^2 R} = \frac{1}{2} \frac{L}{R}$.
Since the time constant $\tau$ is defined as $\tau = \frac{L}{R}$,we have $\frac{U}{P} = \frac{1}{2} \tau$.
Therefore,the time constant $\tau = 2 \times \left( \frac{U}{P} \right)$,which is twice the ratio of the energy stored in the magnetic field to the rate of dissipation of energy in the resistance.

(B) When the switch is closed (charging),the inductor $L$ is in series with the resistor $2R$. The resistor $R$ is in parallel with the series combination of $2R$ and $L$. The time constant for the charging circuit is determined by the equivalent resistance $R_{eq}$ seen by the inductor. When the battery is shorted (or considered ideal),the equivalent resistance is $R_{eq} = 2R + R = 3R$. Thus,the charging time constant is $\tau_{1} = \frac{L}{3R}$.
When the switch is opened (discharging),the inductor $L$ forms a closed loop with the resistors $2R$ and $R$ in series. The total resistance in this loop is $R_{eq} = 2R + R = 3R$. Thus,the discharging time constant is $\tau_{2} = \frac{L}{3R}$.
Wait,re-evaluating the circuit: When the switch is closed,the inductor is in series with $2R$. The resistor $R$ is in parallel with the branch containing $2R$ and $L$. The effective resistance seen by the inductor is $R_{eq} = 2R + 0 = 2R$ (since the battery is ideal,the parallel branch $R$ does not affect the time constant). So $\tau_{1} = \frac{L}{2R}$.
When the switch is opened,the inductor discharges through the series combination of $2R$ and $R$. So $R_{eq} = 2R + R = 3R$. Thus,$\tau_{2} = \frac{L}{3R}$.
The ratio $\frac{\tau_{1}}{\tau_{2}} = \frac{L/2R}{L/3R} = \frac{3}{2}$.

(A) Given: Inductance $L = 5\,H$,Resistance $R = 10\,\Omega$,$emf$ $V = 6\,V$.
The current in an $RL$ circuit at time $t$ is given by $i(t) = i_0(1 - e^{-Rt/L})$,where $i_0 = V/R = 6/10 = 0.6\,A$.
The time constant $\tau = L/R = 5/10 = 0.5\,s$. Thus,$i(t) = 0.6(1 - e^{-t/0.5}) = 0.6(1 - e^{-2t})$.
The $emf$ across the coil is given by $\varepsilon_L = L \frac{di}{dt}$.
Calculating $\frac{di}{dt}$: $\frac{di}{dt} = 0.6 \times (-2) \times (-e^{-2t}) = 1.2 e^{-2t}$.
Therefore,$\varepsilon_L = 5 \times 1.2 e^{-2t} = 6 e^{-2t}$.
At $t = \ln \sqrt{2} = \frac{1}{2} \ln 2$,we have $e^{-2t} = e^{-2(\frac{1}{2} \ln 2)} = e^{-\ln 2} = \frac{1}{2}$.
Substituting this value: $\varepsilon_L = 6 \times \frac{1}{2} = 3\,V$.

(A) At $t = 0$,when the key is just closed,the inductor acts as an open circuit because it opposes any change in current. Therefore,the $10\, \Omega$ resistor and $20\, \Omega$ resistor are in series.
Total resistance $R_{eq} = 10\, \Omega + 20\, \Omega = 30\, \Omega$.
The initial current $I_0 = \frac{V}{R_{eq}} = \frac{2\, V}{30\, \Omega} = \frac{1}{15}\, A$.
As $t \rightarrow \infty$,the inductor acts as a short circuit (ideal inductor has zero resistance). The $10\, \Omega$ resistor is now in parallel with the short-circuited inductor,so all current flows through the inductor,bypassing the $10\, \Omega$ resistor.
Now,only the $20\, \Omega$ resistor is in the circuit.
The final current $I_{\infty} = \frac{V}{20\, \Omega} = \frac{2\, V}{20\, \Omega} = \frac{1}{10}\, A$.

(D) Given: Electromotive force $E = 2V$,Inductance $L = 4H$,Fuse current $I = 5A$.
For an ideal inductor in a $DC$ circuit,the voltage across the inductor is given by $E = L \frac{di}{dt}$.
Rearranging the equation,we get $di = \frac{E}{L} dt$.
Integrating both sides from $t=0$ to $t$ and $i=0$ to $I$:
$\int_{0}^{I} di = \int_{0}^{t} \frac{E}{L} dt$
$I = \frac{E}{L} t$
Substituting the given values:
$5 = \frac{2}{4} t$
$5 = 0.5 t$
$t = \frac{5}{0.5} = 10s$
Thus,the fuse will blow after $10s$.

(A) When $X$ is connected to $Y$ for a long time,the inductor $L$ acts as a short circuit (steady state). The current flowing through the inductor is $I_0 = \frac{E}{R_1}$.
The energy stored in the inductor is $U = \frac{1}{2} L I_0^2 = \frac{1}{2} L \left(\frac{E}{R_1}\right)^2 = \frac{L E^2}{2 R_1^2}$.
When $X$ is switched to $Z$,the inductor $L$ and resistor $R_2$ form a closed series circuit. The energy stored in the inductor is dissipated as heat in the resistor $R_2$ as the current decays to zero.
Since all the energy stored in the inductor is dissipated as heat in the circuit,the total heat produced in $R_2$ is equal to the initial energy stored in the inductor.
Therefore,the total heat $H = U = \frac{L E^2}{2 R_1^2}$.

(A) Let the inductance of the coil be $L$ and its resistance be $R$.
The magnetic energy stored in the inductor is given by $U = \frac{1}{2} L I^2$.
Given $U = 32 \ J$ and $I = 4 \ A$,we have:
$32 = \frac{1}{2} \times L \times (4)^2$
$32 = \frac{1}{2} \times L \times 16$
$32 = 8L \Rightarrow L = 4 \ H$.
The power dissipated as heat is given by $P = I^2 R$.
Given $P = 320 \ W$ and $I = 4 \ A$,we have:
$320 = (4)^2 \times R$
$320 = 16 \times R$
$R = \frac{320}{16} = 20 \ \Omega$.
The time constant $\tau$ of an $RL$ circuit is defined as $\tau = \frac{L}{R}$.
Substituting the values:
$\tau = \frac{4}{20} = 0.2 \ s$.

(B) The energy stored in an inductor is given by $U = \frac{1}{2} L i^2$.
Initially,$U_0 = \frac{1}{2} L I^2$.
When the energy reduces to one-fourth of its initial value,$U = \frac{1}{4} U_0 = \frac{1}{4} (\frac{1}{2} L I^2)$.
Thus,$\frac{1}{2} L i^2 = \frac{1}{4} (\frac{1}{2} L I^2)$,which implies $i^2 = \frac{1}{4} I^2$,so $i = \frac{I}{2}$.
In an $L-R$ decay circuit,the current at time $t$ is $i = I e^{-t/\tau}$,where $\tau = L/R$.
Setting $i = I/2$,we get $I/2 = I e^{-t/\tau}$,so $e^{-t/\tau} = 1/2$,which means $t/\tau = \ln 2$,or $t = \tau \ln 2$.
The total charge $q$ flown through the resistor is $q = \int_{0}^{t} i dt = \int_{0}^{\tau \ln 2} I e^{-t/\tau} dt$.
$q = I [-\tau e^{-t/\tau}]_{0}^{\tau \ln 2} = -I \tau (e^{-\ln 2} - e^0) = -I \tau (1/2 - 1) = -I \tau (-1/2) = \frac{I \tau}{2}$.
Substituting $\tau = L/R$,we get $q = \frac{IL}{2R}$.

(D) The expression for current growth in an $L-R$ circuit is given by $i = \frac{E}{R}(1 - e^{-t / \tau}) = I_0(1 - e^{-t / \tau})$,where $\tau = \frac{L}{R}$ is the time constant.
From the graph,both curves reach the same steady-state current $I_0 = \frac{E}{R}$. Since the battery voltage $E$ is the same for both circuits,we have $I_{0,1} = I_{0,2} \Rightarrow \frac{E}{R_1} = \frac{E}{R_2}$,which implies $R_1 = R_2$.
The time constant $\tau$ determines the rate of growth of current. $A$ larger time constant means the current takes longer to reach its steady state. From the graph,curve $(c)$ grows more slowly than curve $(b)$,so $\tau_c > \tau_b$.
Since $\tau = \frac{L}{R}$ and $R_1 = R_2$,we have $\frac{L_2}{R_2} > \frac{L_1}{R_1} \Rightarrow L_2 > L_1$,or $L_1 < L_2$.
Thus,both statements $(B)$ and $(C)$ are correct.

(D) At $t = 0$,the inductor opposes any change in current,acting as an open circuit.
$1$. Current in the circuit $(i)$: Since the inductor acts as an open circuit,$i = 0$.
$2$. Magnetic field energy $(U_B = \frac{1}{2} L i^2)$: Since $i = 0$,$U_B = 0$.
$3$. Power delivered by the battery $(P = V i)$: Since $i = 0$,$P = 0$.
$4$. Induced $emf$ in the inductor $(\varepsilon = L \frac{di}{dt})$: At $t = 0$,the rate of change of current $\frac{di}{dt}$ is maximum,making the induced $emf$ equal to the battery voltage $V$. Thus,it is non-zero.

(A) In an $L-R$ circuit,the energy stored in the magnetic field of the inductor is given by $U = \frac{1}{2} L I^2$.
The rate of energy dissipation in the resistor is given by $P = I^2 R$.
The ratio of the energy stored to the rate of dissipation is $\frac{U}{P} = \frac{\frac{1}{2} L I^2}{I^2 R} = \frac{1}{2} \frac{L}{R}$.
Since the time constant $\tau$ is defined as $\tau = \frac{L}{R}$,we have $\frac{U}{P} = \frac{1}{2} \tau$.
Therefore,$\tau = 2 \times \frac{U}{P}$,which means the time constant is twice the ratio of the energy stored in the magnetic field to the rate of dissipation of energy in the resistance.

(B) When the switch $S$ is closed,the current in the circuit starts to increase from zero.
As the current changes,a changing magnetic flux is linked with the inductor $L$.
According to Faraday's law of electromagnetic induction,this changing flux induces an $emf$ in the inductor.
According to Lenz's law,this induced $emf$ opposes the growth of the current in the circuit.
Therefore,the current takes some time to reach its steady-state value,causing the lamp $P$ to reach its full brightness slowly.

(A) Initially,when steady state is achieved,the current in the circuit is $i_0 = \frac{E}{R} = \frac{100 \ V}{100 \ \Omega} = 1 \ A$.
When the battery is disconnected and points $A$ and $B$ are short-circuited,the circuit becomes an $LR$ decay circuit.
The equation for the decay of current is given by $i(t) = i_0 e^{-\frac{R}{L}t}$.
Given $R = 100 \ \Omega$,$L = 100 \ mH = 0.1 \ H$,and $t = 1 \ ms = 10^{-3} \ s$.
The time constant of the circuit is $\tau = \frac{L}{R} = \frac{0.1 \ H}{100 \ \Omega} = 10^{-3} \ s$.
Substituting the values into the decay equation:
$i(t) = 1 \times e^{-\frac{10^{-3}}{10^{-3}}} = 1 \times e^{-1} = \frac{1}{e} \ A$.

(B) The growth of current in an $LR$ circuit is given by the formula: $I = I_0(1 - e^{-\frac{R}{L}t})$.
Here,$I_0 = \frac{E}{R}$ is the maximum steady-state current.
Given: $L = 10 \ H$,$R = 5 \ \Omega$,$E = 5 \ V$,and $t = 2 \ s$.
First,calculate the maximum current: $I_0 = \frac{5 \ V}{5 \ \Omega} = 1 \ A$.
Now,substitute the values into the growth equation:
$I = 1 \times (1 - e^{-\frac{5}{10} \times 2})$
$I = 1 - e^{-1} \ A$.

(D) When the switch $S$ is closed at $t = 0$,the branch containing $R_1$ is in parallel with the branch containing $L$ and $R_2$. The potential difference across the $L-R_2$ branch is equal to the emf of the battery,$E = 12 \ V$.
The current $i$ in the $L-R_2$ branch grows according to the equation: $i = \frac{E}{R_2}(1 - e^{-R_2 t / L})$.
The potential drop across the inductor $L$ is given by $V_L = L \frac{di}{dt}$.
Differentiating the current expression with respect to time $t$:
$\frac{di}{dt} = \frac{E}{R_2} \cdot \frac{R_2}{L} e^{-R_2 t / L} = \frac{E}{L} e^{-R_2 t / L}$.
Substituting this into the expression for $V_L$:
$V_L = L \left( \frac{E}{L} e^{-R_2 t / L} \right) = E e^{-R_2 t / L}$.
Given $E = 12 \ V$,$R_2 = 2 \ \Omega$,and $L = 400 \ mH = 0.4 \ H$,the time constant $\tau = \frac{L}{R_2} = \frac{0.4}{2} = 0.2 \ s$.
Thus,$V_L = 12 e^{-t / 0.2} = 12 e^{-5t} \ V$.

(A) At $t = 0$,the inductor $L$ acts as an open circuit because it opposes any change in current. Therefore,no current flows through the branch containing $L$ and $R_1$.
Thus,the current through the battery is $I = \frac{V}{R_2}$.
At $t = \infty$,the inductor $L$ acts as a short circuit (ideal inductor with zero resistance). The resistors $R_1$ and $R_2$ are now in parallel.
The effective resistance of the circuit is $R_{eff} = \frac{R_1 R_2}{R_1 + R_2}$.
Therefore,the current through the battery is $I = \frac{V}{R_{eff}} = \frac{V(R_1 + R_2)}{R_1 R_2}$.

(B) $1$. When point '$C$' is connected to '$A$', the current in the circuit becomes constant, $I_0 = E/R$. The inductor acts as a short circuit.
$2$. When '$C$' is connected to '$B$' at $t = 0$, the battery is removed, and the circuit becomes a decaying $LR$ circuit.
$3$. The current in the circuit at time $t$ is given by $I(t) = I_0 e^{-Rt/L}$.
$4$. At $t = L/R$, the current is $I = I_0 e^{-R(L/R)/L} = I_0 e^{-1} = I_0/e$.
$5$. The voltage across the resistance is $V_R = I R = (I_0/e) R = (E/R \cdot 1/e) R = E/e$.
$6$. Applying Kirchhoff's voltage law in the closed loop (without battery): $V_R + V_L = 0$, which implies $V_L = -V_R$.
$7$. Therefore, the ratio of the voltage across the resistance $(V_R)$ to the voltage across the inductor $(V_L)$ is $V_R / V_L = V_R / (-V_R) = -1$.

(C) $1$. When key $K_1$ is closed for a long time,the inductor acts as a short circuit. The steady-state current $i_0$ in the circuit is given by $i_0 = \frac{V}{R} = \frac{15}{0.15 \times 10^3} = 0.1 \; A$.
$2$. At $t = 0$,$K_1$ is opened and $K_2$ is closed. The battery is removed from the circuit,and the inductor discharges through the resistor. This is a decaying $LR$ circuit.
$3$. The current at time $t$ is given by $i = i_0 e^{-\frac{Rt}{L}}$.
$4$. Substituting the values: $i = 0.1 \times e^{-\frac{0.15 \times 10^3 \times 10^{-3}}{0.03}} = 0.1 \times e^{-\frac{0.15}{0.03}} = 0.1 \times e^{-5}$.
$5$. Given $e^5 \cong 150$,then $e^{-5} = \frac{1}{150}$.
$6$. Therefore,$i = 0.1 \times \frac{1}{150} = \frac{0.1}{150} \; A = \frac{0.1}{150} \times 1000 \; mA = \frac{100}{150} \; mA = \frac{2}{3} \; mA \approx 0.67 \; mA$.

(C) At $t = 0$,the inductor $L$ acts as an open circuit. The current flows only through resistor $R_2 = 10 \ \Omega$. Therefore,the current $I_0 = V / R_2 = 10 \ V / 10 \ \Omega = 1 \ A$.
At $t = \infty$,the inductor $L$ acts as a short circuit. The circuit consists of two resistors $R_1 = 5 \ \Omega$ and $R_2 = 10 \ \Omega$ in parallel. The equivalent resistance $R_{eq} = (R_1 \times R_2) / (R_1 + R_2) = (5 \times 10) / (5 + 10) = 50 / 15 = 10 / 3 \ \Omega$.
The current $I_{\infty} = V / R_{eq} = 10 / (10 / 3) = 3 \ A$.

(A) At $t = 0$,the switch $S$ is open. The inductor acts as an open circuit,but the current flows through the $20 \, V$ battery,the $5 \, \Omega$ resistor,the $5 \, \Omega$ resistor,and the $10 \, V$ battery. The current $I_0$ through the inductor branch is $0 \, A$ because the switch is open.
At steady state $(t \to \infty)$,the inductor acts as a short circuit (zero resistance). The capacitor acts as an open circuit. The circuit simplifies to a loop with the $20 \, V$ battery,the $5 \, \Omega$ resistor,and the inductor branch (with $5 \, \Omega$ resistance). The $10 \, V$ battery branch is in parallel with the inductor branch. The current $I$ through the inductor is determined by the $20 \, V$ source and the $5 \, \Omega$ resistor in series with the inductor branch: $I = \frac{20 \, V}{5 \, \Omega} = 4 \, A$.
The change in flux is $\Delta \phi = L \Delta I = L(I_{final} - I_{initial})$.
Given $L = 500 \, mH = 0.5 \, H$,$I_{initial} = 0 \, A$,and $I_{final} = 4 \, A$.
$\Delta \phi = 0.5 \times (4 - 0) = 2 \, Wb$.

(A) When the switch is closed,the circuit is connected to a $90\, V$ $DC$ battery.
An ideal inductor has zero resistance,so it acts as a short circuit for $DC$ current.
Since the bulb and the inductor are connected in parallel,the entire current from the battery will flow through the path of least resistance,which is the ideal inductor.
Consequently,no current flows through the bulb,and it does not glow.

(C) At the instant the key $K$ is closed $(t=0)$:
$(i)$ $A$ capacitor acts as a short circuit (offers zero resistance) because it is initially uncharged,allowing maximum current to flow through the branch containing $A_1$.
$(ii)$ An inductor opposes any change in current. Since the current was zero before closing the key,the inductor acts as an open circuit (offers infinite resistance) at $t=0$,resulting in zero current through the branch containing $A_2$.
Therefore,the reading of $A_1$ is maximum and the reading of $A_2$ is zero.

(C) The rate at which magnetic energy is stored in the coil is given by $P_m = \frac{d}{dt}(\frac{1}{2}Li^2) = Li \frac{di}{dt}$.
The rate at which energy is supplied by the battery is $P_b = Ei$.
The ratio of these rates is $\frac{P_m}{P_b} = \frac{Li \frac{di}{dt}}{Ei} = \frac{L}{E} \frac{di}{dt}$.
For an $RL$ circuit,the current at time $t$ is $i = \frac{E}{R}(1 - e^{-Rt/L})$.
The rate of change of current is $\frac{di}{dt} = \frac{E}{R} \cdot \frac{R}{L} e^{-Rt/L} = \frac{E}{L} e^{-Rt/L}$.
Substituting this into the ratio expression:
$\frac{P_m}{P_b} = \frac{L}{E} \left( \frac{E}{L} e^{-Rt/L} \right) = e^{-Rt/L}$.
Given $R = 10\, \Omega$,$L = 1\, H$,and $t = 0.1\, s$:
$\frac{P_m}{P_b} = e^{-(10 \times 0.1) / 1} = e^{-1} \approx \frac{1}{2.718} \approx 0.368 \approx 0.36$.