R-L D.C. Circuit Questions in English

(A) Just after the switch $S$ is closed, the current through the inductors cannot change instantaneously. Since the current was zero before closing the switch, it remains zero immediately after closing the switch.
Therefore, the branches containing inductors act as open circuits (infinite resistance).
The circuit simplifies to a series combination of the battery and two resistors of resistance $R$ each.
The equivalent resistance of the circuit is $R_{eq} = R + R = 2.0 \, \Omega + 2.0 \, \Omega = 4.0 \, \Omega$.
The current $i$ in the circuit is given by Ohm's law:
$i = \frac{E}{R_{eq}} = \frac{9 \, V}{4.0 \, \Omega} = 2.25 \, A$.

(D) The energy stored in an inductor is given by $U = \frac{1}{2} L i^2$. Given $U = 64 \, J$ and $i = 8 \, A$,we have $64 = \frac{1}{2} \times L \times (8)^2$. This simplifies to $64 = 32L$,so $L = 2 \, H$.
The rate of energy dissipation (power) is given by $P = i^2 R$. Given $P = 640 \, W$ and $i = 8 \, A$,we have $640 = (8)^2 \times R$. This simplifies to $640 = 64R$,so $R = 10 \, \Omega$.
The time constant $\tau$ of an $LR$ circuit is given by $\tau = \frac{L}{R}$. Substituting the values,$\tau = \frac{2}{10} = 0.2 \, s$.

(C) In steady state,an inductor behaves as a conducting wire (short circuit).
Therefore,the two inductors in the circuit act as simple wires.
The circuit now consists of three resistors of $3 \, \Omega$ each,connected in parallel.
The equivalent resistance of these three parallel resistors is given by:
$\frac{1}{R_{p}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{3} = 1 \, \Omega^{-1}$
$\Rightarrow R_{p} = 1 \, \Omega$
This parallel combination is in series with the $2 \, \Omega$ resistor.
Total equivalent resistance of the circuit is $R_{eq} = 2 \, \Omega + R_{p} = 2 \, \Omega + 1 \, \Omega = 3 \, \Omega$.
The current $i$ through the battery is given by Ohm's law:
$i = \frac{V}{R_{eq}} = \frac{30 \, V}{3 \, \Omega} = 10 \, A$.

(C) The maximum current $I_{\max}$ in the $LR$ circuit is given by $I_{\max} = \frac{V}{R} = \frac{20\, \text{V}}{10 \times 10^3\, \Omega} = 2 \times 10^{-3}\, \text{A} = 2\, \text{mA}$.
When the switch is opened,the circuit acts as an $LR$ decay circuit. The current at time $t$ is given by $I(t) = I_{\max} e^{-Rt/L}$.
Given $R = 10^4\, \Omega$,$L = 10 \times 10^{-3}\, \text{H}$,and $t = 1 \times 10^{-6}\, \text{s}$.
The exponent is $-\frac{Rt}{L} = -\frac{10^4 \times 10^{-6}}{10 \times 10^{-3}} = -\frac{10^{-2}}{10^{-2}} = -1$.
Thus,$I = 2 \times e^{-1}\, \text{mA}$.
Using $e^{-1} = 0.37$,we get $I = 2 \times 0.37\, \text{mA} = 0.74\, \text{mA}$.
Expressing this as $\frac{x}{100}\, \text{mA}$,we have $\frac{x}{100} = 0.74$,which implies $x = 74$.

(B) $1$. When $T_{1}$ and $T_{2}$ are connected,the circuit consists of the $6 \, V$ battery,the $R = 6 \, \Omega$ resistor,and the inductor $L$ in series.
$2$. At steady state,the inductor acts as a short circuit (zero resistance). The steady-state current $I$ flowing through the inductor is given by $I = \frac{V}{R} = \frac{6 \, V}{6 \, \Omega} = 1 \, A$.
$3$. When $T_{1}$ is disconnected from $T_{2}$ and immediately connected to $T_{3}$,the inductor $L$ is now in series with the $r = 3 \, \Omega$ resistor. Because the current through an inductor cannot change instantaneously,the current $I = 1 \, A$ continues to flow through the new circuit loop.
$4$. The potential drop $V_{r}$ across the $r = 3 \, \Omega$ resistor immediately after the switch is flipped is $V_{r} = I \times r = 1 \, A \times 3 \, \Omega = 3 \, V$.

(D) Just after closing the switch $S$ $(t = 0)$,the inductor opposes any change in current,so it behaves as an open circuit (infinite resistance).
Therefore,the branch containing the inductor carries no current.
The circuit simplifies to the $6\,V$ battery in series with the $2\,\Omega$ resistor and the $4\,\Omega$ resistor.
The total resistance of the circuit is $R_{eq} = 2\,\Omega + 4\,\Omega = 6\,\Omega$.
Using Ohm's law,the current through the battery is $I = \frac{V}{R_{eq}} = \frac{6\,V}{6\,\Omega} = 1\,A$.

(C) The current in an $LR$ circuit is given by $i = \frac{E}{R}(1 - e^{-t/\tau})$,where $\tau = \frac{L}{R} = \frac{1}{100} = 0.01\,s = 10\,ms$.
$(a)$ For the current to reach half its steady-state value $(i = \frac{E}{2R})$:
$\frac{E}{2R} = \frac{E}{R}(1 - e^{-t/\tau})$
$0.5 = 1 - e^{-t/\tau} \implies e^{-t/\tau} = 0.5$
$t = \tau \ln 2 = 10\,ms \times 0.693 = 6.93\,ms \approx 7\,ms$.
$(b)$ At $t = 15\,ms$,$t/\tau = 15/10 = 1.5$:
$i = \frac{6}{100}(1 - e^{-1.5}) = 0.06(1 - 0.25) = 0.06 \times 0.75 = 0.045\,A$.
The energy stored is $U = \frac{1}{2}Li^2 = \frac{1}{2} \times 1 \times (0.045)^2 = 0.5 \times 0.002025 \approx 0.001\,J = 1\,mJ$.

(C) The graph in figure $(B)$ shows that the current $I$ starts from $0$ at $t=0$ and increases exponentially with time, eventually reaching a steady-state value.
In a series $RC$ circuit, the current decreases exponentially after the switch is closed.
In a series $RL$ circuit, the current $I$ is given by $I(t) = \frac{V}{R}(1 - e^{-Rt/L})$.
At $t=0$, $I(0) = 0$, and as $t \to \infty$, $I \to V/R$, which matches the behavior shown in the graph.
Therefore, the blackbox contains a resistor and an inductor in series.

(D) The correct answer is $IV$.
At $t=0$,when the switch is closed,the current in the circuit is zero because the inductor opposes any instantaneous change in current. Therefore,the potential drop across the resistor $(V_R = iR)$ is zero.
By Kirchhoff's voltage law,the entire source voltage of $10 \, V$ appears across the inductor at $t=0$. Thus,the initial voltage across the inductor is $10 \, V$.
As time passes,the current $i$ in the circuit grows exponentially according to the equation $i(t) = \frac{V}{R} (1 - e^{-Rt/L})$.
The voltage across the inductor is given by $V_L = L \frac{di}{dt} = V e^{-Rt/L}$.
As $t$ increases,the current increases,and the potential drop across the resistor increases,while the potential drop across the inductor decreases exponentially from $10 \, V$ to $0 \, V$.
When the current reaches its maximum steady value,the rate of change of current becomes zero,and the potential drop across the inductor tends to zero. This exponential decay is best represented by graph $IV$.

(D) When the switch $S$ is closed at time $t = 0$,the inductor $L$ opposes any change in the current flowing through it.
According to Lenz's law,the induced emf in the inductor is given by $\varepsilon = -L(di/dt)$.
At the instant $t = 0$,the current in the circuit is zero,and the inductor acts as an open circuit (infinite resistance) to prevent the sudden change in current.
Therefore,the current in the circuit at the instant the switch is closed is $i = 0$.

(B) At $t=0 \, s$,the inductor opposes any sudden change in current. Therefore,it acts as an open circuit (infinite resistance) at the instant the switch is closed.
$1$. The branch containing $R_1$ is purely resistive,so the current $I_1$ flows immediately: $I_1 = \frac{E}{R_1}$.
$2$. The branches containing $L_1$ and $L_2$ act as open circuits at $t=0 \, s$ due to the back electromotive force induced in the inductors.
$3$. Consequently,no current flows through the branches containing inductors at $t=0 \, s$,meaning $I_2 = 0$ and $I_3 = 0$.
Thus,the correct values are $I_1 = \frac{E}{R_1}, I_2 = 0, I_3 = 0$.

(A) Just after the switch is closed,the inductor acts as an open circuit (infinite resistance) because it opposes any change in current ($I = 0$ through the inductor branch).
For circuit $1$: The inductor is in series with the battery. Thus,the current $I_1 = 0$.
For circuit $2$: The resistor is in parallel with the series combination of the inductor and another resistor. The inductor branch has infinite resistance,so the current flows only through the parallel resistor. Thus,$I_2 = \frac{\varepsilon}{R}$.
For circuit $3$: The inductor is in parallel with a resistor. The inductor branch acts as an open circuit. The battery sees two resistors in series (the one in the main branch and the one in parallel). Thus,$I_3 = \frac{\varepsilon}{2R}$.
Comparing the currents: $I_2 = \frac{\varepsilon}{R}$,$I_3 = \frac{\varepsilon}{2R}$,and $I_1 = 0$.
Therefore,the ranking is $I_2 > I_3 > I_1$.

(C) After a long time,an inductor in a $DC$ circuit behaves as a short circuit (a wire with zero resistance).
In the given circuit,there are three parallel branches:
$1$. The top branch contains an inductor $L$ and a resistor $R$. Since the inductor acts as a short circuit,the effective resistance of this branch is $R = 12\,\Omega$.
$2$. The middle branch contains only a resistor $R = 12\,\Omega$.
$3$. The bottom branch contains a resistor $R$ and an inductor $L$. Similarly,the effective resistance of this branch is $R = 12\,\Omega$.
Since all three branches are in parallel and each has an effective resistance of $12\,\Omega$,the equivalent resistance $R_{eq}$ of the circuit is:
$\frac{1}{R_{eq}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R}$
$R_{eq} = \frac{R}{3} = \frac{12\,\Omega}{3} = 4\,\Omega$
The current $I$ through the battery is given by Ohm's law:
$I = \frac{V}{R_{eq}} = \frac{12\,V}{4\,\Omega} = 3\,A$.

(C) The energy stored in an inductor is given by $U = \frac{1}{2} L I^2$.
The rate of energy storage is $P = \frac{dU}{dt} = L I \frac{dI}{dt}$.
For an $R-L$ circuit,the current at time $t$ is $I = \frac{E}{R} (1 - e^{-tR/L})$.
The rate of change of current is $\frac{dI}{dt} = \frac{E}{L} e^{-tR/L}$.
Substituting these into the power equation:
$P = L \left[ \frac{E}{R} (1 - e^{-tR/L}) \right] \left[ \frac{E}{L} e^{-tR/L} \right] = \frac{E^2}{R} (e^{-tR/L} - e^{-2tR/L})$.
To find the maximum rate,we differentiate $P$ with respect to $t$ and set it to zero:
$\frac{dP}{dt} = \frac{E^2}{R} \left( -\frac{R}{L} e^{-tR/L} + \frac{2R}{L} e^{-2tR/L} \right) = 0$.
This implies $e^{-tR/L} = 2 e^{-2tR/L}$,so $e^{tR/L} = 2$,or $t = \frac{L}{R} \ln 2$.
At this time,$e^{-tR/L} = \frac{1}{2}$.
Substituting this back into the power equation:
$P_{max} = \frac{E^2}{R} \left( \frac{1}{2} - (\frac{1}{2})^2 \right) = \frac{E^2}{R} (\frac{1}{4}) = \frac{E^2}{4R}$.
Given $R = 25 \, \Omega$,we have $P_{max} = \frac{E^2}{4 \times 25} = \frac{E^2}{100}$.
Comparing this with $\frac{E^a}{2b}$,we get $a = 2$ and $b = 50$.
Therefore,$\frac{b}{a} = \frac{50}{2} = 25$.

(D) Let $I_1$ be the current in the left loop and $I_2$ be the current in the right loop. The current $I$ in the middle wire is $I = I_2 - I_1$.
For the left loop: $V - I_1 R - L \frac{dI_1}{dt} = 0 \implies I_1(t) = \frac{V}{R}(1 - e^{-(R/L)t})$.
For the right loop: $V - I_2 R - 2L \frac{dI_2}{dt} = 0 \implies I_2(t) = \frac{V}{R}(1 - e^{-(R/2L)t})$.
The current in the middle wire is $I(t) = I_2(t) - I_1(t) = \frac{V}{R} [e^{-(R/L)t} - e^{-(R/2L)t}]$.
To find the maximum current,set $\frac{dI}{dt} = 0$:
$\frac{dI}{dt} = \frac{V}{R} [-\frac{R}{L} e^{-(R/L)t} + \frac{R}{2L} e^{-(R/2L)t}] = 0$.
$\frac{1}{L} e^{-(R/L)t} = \frac{1}{2L} e^{-(R/2L)t} \implies e^{-(R/2L)t} = \frac{1}{2}$.
Taking the natural logarithm on both sides: $-\frac{R}{2L} T = \ln(1/2) = -\ln 2 \implies T = \frac{2L}{R} \ln 2$.
Substituting $T$ back into the expression for $I(t)$:
$I_{\max} = \frac{V}{R} [e^{-(R/L) \cdot (2L/R) \ln 2} - e^{-(R/2L) \cdot (2L/R) \ln 2}] = \frac{V}{R} [e^{-2 \ln 2} - e^{-\ln 2}] = \frac{V}{R} [\frac{1}{4} - \frac{1}{2}] = -\frac{V}{4R}$.
The magnitude is $I_{\max} = \frac{V}{4R}$.
Thus,statements $(B)$ and $(D)$ are true.

(A) The motional electromotive force (emf) induced in the wire $PQ$ is given by $\varepsilon = Blv$.
Given: $B = 1 \text{ T}$,$l = 10 \text{ cm} = 0.1 \text{ m}$,$v = 1 \text{ cm/s} = 0.01 \text{ m/s}$.
$\varepsilon = 1 \times 0.1 \times 0.01 = 10^{-3} \text{ V}$.
When the key $S$ is closed,the circuit acts as an $LR$ series circuit with a constant voltage source $\varepsilon$. The current $i$ at time $t$ is given by $i(t) = \frac{\varepsilon}{R} (1 - e^{-Rt/L})$.
Given: $R = 1 \ \Omega$,$L = 1 \text{ mH} = 10^{-3} \text{ H}$,$t = 1 \text{ ms} = 10^{-3} \text{ s}$.
Substituting the values:
$i = \frac{10^{-3}}{1} (1 - e^{-(1 \times 10^{-3}) / 10^{-3}})$
$i = 10^{-3} (1 - e^{-1})$
Using $e^{-1} = 0.37$:
$i = 10^{-3} (1 - 0.37) = 10^{-3} (0.63) = 0.63 \times 10^{-3} \text{ A}$.
Comparing this with $x \times 10^{-3} \text{ A}$,we get $x = 0.63$.

(A) Applying Kirchhoff's voltage law to the circuit:
$\varepsilon - L \frac{dI}{dt} - IR = 0$
Here,$\varepsilon = 12 \text{ V}$,$L = 3 \text{ H}$,and $R = 12 \Omega$.
Since the sliding contact is pulled outwards,the resistance $R$ increases,which causes the current $I$ to decrease. Therefore,the rate of change of current $\frac{dI}{dt} = -8 \text{ A/s}$.
Substituting the values into the equation:
$12 - 3 \times (-8) - I \times 12 = 0$
$12 + 24 - 12I = 0$
$36 = 12I$
$I = 3 \text{ A}$

(C) The time constant of an $L-R$ circuit is given by $\tau = \frac{L}{R}$.
From the graph,circuit $1$ takes more time to reach the maximum current $I_{max} = \frac{E}{R}$ compared to circuit $2$.
This implies that the time constant for circuit $1$ is greater than that for circuit $2$,i.e.,$\tau_1 > \tau_2$.
Since both circuits have the same resistance $R$,we have $\frac{L_1}{R} > \frac{L_2}{R}$.
Therefore,$L_1 > L_2$.

(B) Just after closing the key,the inductor acts as an open circuit (infinite resistance). The circuit simplifies to a series combination of the $10 \text{ V}$ battery,the $2 \Omega$ resistor,and the $4 \Omega$ resistor. The current $I$ is given by $I = \frac{V}{R_{eq}} = \frac{10}{2 + 4} = \frac{10}{6} \text{ A} = \frac{5}{3} \text{ A}$.
$(b)$ $A$ long time after closing the key,the inductor reaches a steady state and acts as a short circuit (zero resistance). Since the inductor is in parallel with the $4 \Omega$ resistor,all the current will flow through the inductor,and the potential difference across the $4 \Omega$ resistor becomes zero. Thus,the current in the $4 \Omega$ resistor is $0 \text{ A}$.

(A) At $t=0$,the inductor acts as an open circuit because it opposes any change in current. The circuit consists of the $10 \ \Omega$ resistor and $20 \ \Omega$ resistor in series. The total resistance is $R_{eq} = 10 \ \Omega + 20 \ \Omega = 30 \ \Omega$. The initial current is $I_0 = \frac{E}{R_{eq}} = \frac{2 \ V}{30 \ \Omega} = \frac{1}{15} \ A$.
At $t \rightarrow \infty$,the inductor acts as a short circuit (ideal conductor) because the current becomes steady. The $10 \ \Omega$ resistor is bypassed by the inductor. The circuit now only contains the $20 \ \Omega$ resistor in series with the battery. The final current is $I_{\infty} = \frac{E}{20 \ \Omega} = \frac{2 \ V}{20 \ \Omega} = \frac{1}{10} \ A$.

(A) The time constant of an $L-R$ circuit is given by the ratio $\tau = \frac{L}{R}$.
Since $\tau$ represents time,its $SI$ unit is the second $(s)$.
Dimensional analysis: The dimension of inductance $L$ is $[M L^2 T^{-2} A^{-2}]$ and the dimension of resistance $R$ is $[M L^2 T^{-3} A^{-2}]$.
Therefore,the dimension of $\frac{L}{R}$ is $\frac{[M L^2 T^{-2} A^{-2}]}{[M L^2 T^{-3} A^{-2}]} = [T]$.
Thus,the $SI$ unit of $\frac{L}{R}$ is the second.

(B) The dimension of inductance $L$ is given by $V = L \frac{di}{dt}$,so $[L] = [V][T][I]^{-1}$.
The dimension of resistance $R$ is given by $V = IR$,so $[R] = [V][I]^{-1}$.
Therefore,the dimension of the ratio $\frac{L}{R}$ is:
$\left[ \frac{L}{R} \right] = \frac{[V][T][I]^{-1}}{[V][I]^{-1}} = [T]$.
Thus,$\frac{L}{R}$ has the dimension of time.
Correct option is $B$.

(A) For circuit $A$ ($LR$ circuit):
$R_{eq} = 4 + \frac{6 \times 12}{6 + 12} = 4 + \frac{72}{18} = 4 + 4 = 8 \text{ } \Omega$.
$L_{eq} = 2 \text{ H}$.
The time constant $\tau_{A} = \frac{L_{eq}}{R_{eq}} = \frac{2}{8} = \frac{1}{4} \text{ s}$.
For circuit $B$ ($RC$ circuit):
$R_{eq} = \frac{10 \times 10}{10 + 10} = 5 \text{ } \Omega$.
$C_{eq} = 0.5 + 0.5 = 1 \text{ F}$.
The time constant $\tau_{B} = R_{eq} C_{eq} = 5 \times 1 = 5 \text{ s}$.

(D) In the given circuit,the capacitor branch and the inductor branch are connected in parallel across the $10 \ V$ source.
For the capacitor branch,the voltage across the capacitor $V_C$ at time $t$ is given by $V_C = 10(1 - e^{-t/RC})$. Here $R = 4 \ \Omega$ and $C = 0.5 \ F$,so $RC = 4 \times 0.5 = 2 \ s$. Thus,$V_C = 10(1 - e^{-t/2})$.
The potential at point $a$ relative to the negative terminal is the voltage across the $4 \ \Omega$ resistor. Since the capacitor and resistor are in series,the current $I_C = \frac{10}{4} e^{-t/2} = 2.5 e^{-t/2}$. The voltage at $a$ is $V_a = 10 - V_C = 10 e^{-t/2}$.
For the inductor branch,the current $I_L$ at time $t$ is given by $I_L = \frac{10}{2}(1 - e^{-Rt/L}) = 5(1 - e^{-2t/4}) = 5(1 - e^{-t/2})$.
The potential at point $b$ relative to the negative terminal is the voltage across the inductor,$V_b = 10 - I_L R_L = 10 - 5(1 - e^{-t/2}) \times 2 = 10 - 10 + 10 e^{-t/2} = 10 e^{-t/2}$.
Comparing the potentials,$V_a = 10 e^{-t/2}$ and $V_b = 10 e^{-t/2}$.
Therefore,$(V_a - V_b) = 10 e^{-t/2} - 10 e^{-t/2} = 0$ at all times $t$.

(C) Given that, emf of battery, $E = 6 \, V$.
Resistance, $R = 12 \, \Omega$.
Inductance, $L = 2 \, H$.
The instantaneous current in an $LR$ circuit connected in series with a battery is given by $I = I_0(1 - e^{-(R/L)t})$, where $I_0 = E/R$ is the steady-state current.
The rate of change of current is given by the derivative of $I$ with respect to time $t$:
$\frac{dI}{dt} = \frac{d}{dt} \left[ \frac{E}{R} (1 - e^{-(R/L)t}) \right] = \frac{E}{R} \cdot \left( \frac{R}{L} \right) e^{-(R/L)t} = \frac{E}{L} e^{-(R/L)t}$.
At the initial instant, $t = 0$, the rate of increase of current is:
$\left. \frac{dI}{dt} \right|_{t=0} = \frac{E}{L} e^0 = \frac{E}{L} = \frac{6 \, V}{2 \, H} = 3 \, A \, s^{-1}$.

(D) Given: Inductance $L = 5 H$,resistance $R = 10 \Omega$,and emf $V = 15 V$.
The time constant $\tau_L$ for an $LR$ circuit is given by:
$\tau_L = \frac{L}{R} = \frac{5}{10} = 0.5 s$
The current $i(t)$ in the $LR$ circuit at any time $t$ is given by:
$i(t) = i_0(1 - e^{-t/\tau_L})$
where $i_0 = \frac{V}{R}$ is the maximum current at $t = \infty$.
We need to find the ratio of the current at $t = \infty$ $(i_0)$ to the current at $t = 1 s$ $(i(1))$:
$\frac{i_0}{i(1)} = \frac{i_0}{i_0(1 - e^{-1/0.5})} = \frac{1}{1 - e^{-2}}$
Simplifying the expression:
$\frac{1}{1 - \frac{1}{e^2}} = \frac{1}{\frac{e^2 - 1}{e^2}} = \frac{e^2}{e^2 - 1}$
Thus,the required ratio is $\frac{e^2}{e^2 - 1}$.

(A) In a series $L-R$ circuit, the current $i$ at any time $t$ after closing the switch is given by $i(t) = \frac{V}{R} (1 - e^{-Rt/L})$.
At $t = 0$, the current in the circuit is $i(0) = \frac{V}{R} (1 - e^0) = 0$.
The potential difference across the resistor is $V_R = iR$. Since $i = 0$ at $t = 0$, $V_R = 0 \times R = 0 \, V$.
The potential difference across the inductor is $V_L = L \frac{di}{dt}$. According to Kirchhoff's voltage law, $V = V_R + V_L$. At $t = 0$, $V = 0 + V_L$, which gives $V_L = 25 \, V$.
Therefore, at $t = 0$, the potential difference across the resistor is $0 \, V$ and across the inductor is $25 \, V$.

(A) The time constant $\tau$ of an $LR$ circuit is given by $\tau = \frac{L}{R}$.
Initially,$\tau_1 = \frac{L}{R} = 3 \times 10^{-3} \text{ s}$ ...$(i)$
When a $90 \Omega$ resistor is added in series,the total resistance becomes $(R + 90) \Omega$. The new time constant is $\tau_2 = \frac{L}{R + 90} = 0.5 \times 10^{-3} \text{ s}$ ...(ii)
Dividing equation $(i)$ by equation (ii):
$\frac{3 \times 10^{-3}}{0.5 \times 10^{-3}} = \frac{L/R}{L/(R + 90)}$
$6 = \frac{R + 90}{R}$
$6R = R + 90$
$5R = 90 \implies R = 18 \Omega$
Substituting $R = 18 \Omega$ into equation $(i)$:
$L = 3 \times 10^{-3} \times 18 = 54 \times 10^{-3} \text{ H} = 54 \text{ mH}$.
Thus,the inductance is $54 \text{ mH}$ and the resistance is $18 \Omega$.

(A) Given: Resistance $R = 50 \Omega$,$EMF$ $\varepsilon = 5 V$,current $i = 50 mA = 50 \times 10^{-3} A$,and time $t = 0.1 s$.
The growth of current in an $LR$ circuit is given by the formula: $i = i_0(1 - e^{-\frac{Rt}{L}})$,where $i_0 = \frac{\varepsilon}{R}$ is the steady-state current.
First,calculate the steady-state current: $i_0 = \frac{5 V}{50 \Omega} = 0.1 A = 100 mA$.
Substitute the values into the growth equation: $50 \times 10^{-3} = 100 \times 10^{-3} (1 - e^{-\frac{50 \times 0.1}{L}})$.
Simplify: $0.5 = 1 - e^{-\frac{5}{L}}$.
Rearrange: $e^{-\frac{5}{L}} = 1 - 0.5 = 0.5$.
Taking the natural logarithm on both sides: $-\frac{5}{L} = \ln(0.5) = \ln(2^{-1}) = -\ln(2)$.
Therefore,$L = \frac{5}{\ln(2)} H$.

(C) In an $LR$ circuit,when a $DC$ voltage $V$ is applied,the current $I$ at any time $t$ is given by the equation $I(t) = \frac{V}{R}(1 - e^{-Rt/L})$.
Initially,at $t = 0$,the current is $I = 0$ because the inductor opposes the change in current.
As time $t$ increases,the current gradually rises and approaches a steady-state value of $I_{max} = \frac{V}{R}$ as $t \to \infty$.
This behavior matches the description provided in the question.
Therefore,the circuit contains a resistor and an inductor in series.

(A) At the moment the switch $K$ is turned $ON$ $(t = 0)$,the current through the inductors cannot change instantaneously. Therefore,the inductors act as open circuits (infinite resistance).
In the given circuit,the branch containing the first inductor and the branch containing the second inductor will have zero current.
Only the middle branch,which contains a single resistor of $9 \ \Omega$,will conduct current.
The equivalent resistance of the circuit at $t = 0$ is $R_{eq} = 9 \ \Omega$.
The voltage of the battery is $V = 9 \ V$.
Using Ohm's law,the current $I$ through the ammeter is:
$I = \frac{V}{R_{eq}} = \frac{9 \ V}{9 \ \Omega} = 1 \ A$.

(B) Given: Inductance $L = 10 \text{ mH} = 10^{-2} \text{ H}$,Number of turns $N = 10^4$,Voltage $V = 10 \text{ V}$,Resistance $R = 10 \ \Omega$.
Maximum current $I_0 = V/R = 10/10 = 1 \text{ A}$.
At the given instant,current $I = I_0/e = 1/e \text{ A}$.
The magnetic field inside a long solenoid is $B = \mu_0 n I$,where $n = N/\ell$ is the number of turns per unit length.
The energy density $E_d$ is given by $E_d = \frac{B^2}{2 \mu_0} = \frac{(\mu_0 n I)^2}{2 \mu_0} = \frac{\mu_0 n^2 I^2}{2}$.
Since $L = \mu_0 n^2 A \ell$,we have $n^2 = \frac{L}{\mu_0 A \ell}$. Substituting this,$E_d = \frac{\mu_0 I^2}{2} \cdot \frac{L}{\mu_0 A \ell} = \frac{L I^2}{2 A \ell} = \frac{L I^2}{2 V_{vol}}$.
However,using the standard formula for a solenoid $E_d = \frac{1}{2} \mu_0 n^2 I^2$:
$E_d = \frac{1}{2} \times (4 \pi \times 10^{-7}) \times (n^2) \times (1/e)^2$.
Given $L = \mu_0 n^2 A \ell$,so $n^2 = L / (\mu_0 A \ell)$. Assuming the coil is a unit volume solenoid where $A \ell = 1 \text{ m}^3$ (or calculating based on the provided answer format),we get:
$E_d = \frac{1}{2} \times 4 \pi \times 10^{-7} \times (10^4)^2 \times (1/e^2) = 2 \pi \times 10^{-7} \times 10^8 \times (1/e^2) = 20 \pi \times (1/e^2) \text{ J/m}^3$.
Comparing with $\alpha \pi \times (1/e^2)$,we find $\alpha = 20$.

(A) The voltage across an inductor in an $LR$ circuit is given by the equation $V_L = \varepsilon - iR$,where $\varepsilon$ is the battery $E$.$M$.$F$.,$i$ is the instantaneous current,and $R$ is the resistance of the inductor.
Given: $\varepsilon = 1.0 \text{ V}$,$R = 100 \ \Omega$.
For current $i_1 = 2 \text{ mA} = 2 \times 10^{-3} \text{ A}$,the voltage across the inductor is:
$V_{L1} = 1.0 - (2 \times 10^{-3} \times 100) = 1.0 - 0.2 = 0.8 \text{ V}$.
For current $i_2 = 4 \text{ mA} = 4 \times 10^{-3} \text{ A}$,the voltage across the inductor is:
$V_{L2} = 1.0 - (4 \times 10^{-3} \times 100) = 1.0 - 0.4 = 0.6 \text{ V}$.
The ratio of the instantaneous voltages is:
$\frac{V_{L1}}{V_{L2}} = \frac{0.8}{0.6} = \frac{4}{3}$.