R-L D.C. Circuit Questions in English

(A) When $X$ is joined to $Y$ for a long time,the inductor $L$ acts as a short circuit in the steady state.
The steady-state current flowing through the inductor is $i_0 = \frac{\varepsilon}{R_1}$.
The energy stored in the magnetic field of the inductor is $U = \frac{1}{2}Li_0^2 = \frac{1}{2}L\left(\frac{\varepsilon}{R_1}\right)^2 = \frac{L\varepsilon^2}{2R_1^2}$.
When $X$ is joined to $Z$,the battery is disconnected and the inductor discharges through the resistor $R_2$.
Since the entire energy stored in the inductor is dissipated as heat in the circuit during the discharge process,the total heat produced in $R_2$ is equal to the initial stored energy $U = \frac{L\varepsilon^2}{2R_1^2}$.

(C) The current in bulb $B_1$ is $I_1 = V/R = 4\text{V} / 2\Omega = 2\text{A}$.
The current in the circuit containing bulb $B_2$ and inductor $L$ is given by $I_2(t) = I_{max}(1 - e^{-t/\tau})$,where $I_{max} = 4\text{V} / 2\Omega = 2\text{A}$ and $\tau = L/R = 20\text{H} / 2\Omega = 10\text{s}$.
For the same illumination,the current in both bulbs must be the same. Since $B_1$ is purely resistive,its current is constant at $2\text{A}$. However,the problem implies the bulbs have the same illumination when the transient current in $B_2$ reaches the steady-state value of $B_1$. Re-evaluating: $I_1 = 2\text{A}$. For $I_2(t) = I_1$,we need $2(1 - e^{-t/10}) = 2$,which is impossible. Looking at the circuit,$B_1$ is in a branch with $2\Omega$ resistance. $B_2$ is in a branch with $2\Omega$ resistance and $20\text{H}$ inductor. The steady state current in both is $2\text{A}$. The question likely implies the time constant $\tau = L/R = 20/2 = 10\text{s}$. If the illumination is same at $I_2 = 1\text{A}$,then $1 = 2(1 - e^{-t/10}) \Rightarrow 0.5 = 1 - e^{-t/10} \Rightarrow e^{-t/10} = 0.5 \Rightarrow t = 10 \ln 2 = 7\text{s}$.
Angle rotated by the wheel: $\theta = \omega_i t - \frac{1}{2} \alpha t^2 = (2.5\pi)(7) - \frac{1}{2}(2)(7)^2 = 17.5\pi - 49 \text{ rad}$.
$17.5\pi \approx 54.978$. $\theta = 54.978 - 49 = 5.978 \text{ rad}$.
$5.978 \text{ rad} \times (180/\pi) \approx 342.5^{\circ}$.
Since the wheel starts at the vertical pointer (between Violet and Indigo),$342.5^{\circ}$ clockwise rotation places the pointer in the Green sector.

(B) The circuit can be analyzed by splitting it into two independent loops connected to the battery.
Loop $1$ (Left loop): Contains the battery $(20 \, V)$,inductor $(L = 5 \, mH)$,and resistors ($4 \, \Omega$ and $6 \, \Omega$). The total resistance is $R_1 = 4 + 6 = 10 \, \Omega$. The time constant is $\tau_1 = L/R_1 = 5 \times 10^{-3} / 10 = 5 \times 10^{-4} \, s$. The current $I_1(t) = \frac{V}{R_1} (1 - e^{-t/\tau_1}) = \frac{20}{10} (1 - e^{-t / (5 \times 10^{-4})}) = 2 (1 - e^{-2000t})$.
Loop $2$ (Right loop): Contains the battery $(20 \, V)$,capacitor $(C = 0.1 \, mF = 10^{-4} \, F)$,and resistors ($5 \, \Omega$ and $5 \, \Omega$). The total resistance is $R_2 = 5 + 5 = 10 \, \Omega$. The time constant is $\tau_2 = R_2 C = 10 \times 10^{-4} = 10^{-3} \, s$. The current $I_2(t) = \frac{V}{R_2} e^{-t/\tau_2} = \frac{20}{10} e^{-t/10^{-3}} = 2 e^{-1000t}$.
At $t = 10^{-3} \ln 2 \, s$:
$I_1 = 2 (1 - e^{-2000 \times 10^{-3} \ln 2}) = 2 (1 - e^{-2 \ln 2}) = 2 (1 - e^{\ln(2^{-2})}) = 2 (1 - 1/4) = 2(3/4) = 1.5 \, A$.
$I_2 = 2 e^{-1000 \times 10^{-3} \ln 2} = 2 e^{-\ln 2} = 2 (1/2) = 1 \, A$.
The total current through the key $K$ is $I = I_1 + I_2 = 1.5 + 1 = 2.5 \, A$.

(A) For an $L-R$ circuit connected to a $DC$ source $\varepsilon$,the Kirchhoff's voltage law equation is $\varepsilon = L \frac{dI}{dt} + IR$.
The energy stored in the inductor is given by $U = \frac{1}{2} LI^2$.
For option $A$,the energy is $U = \frac{L\varepsilon^2}{8R^2}$.
Equating this to $\frac{1}{2} LI^2$,we get $\frac{1}{2} LI^2 = \frac{L\varepsilon^2}{8R^2}$,which simplifies to $I^2 = \frac{\varepsilon^2}{4R^2}$,so $I = \frac{\varepsilon}{2R}$.
Substituting $I$ into the circuit equation: $\varepsilon = L \frac{dI}{dt} + (\frac{\varepsilon}{2R})R = L \frac{dI}{dt} + \frac{\varepsilon}{2}$.
This gives $L \frac{dI}{dt} = \frac{\varepsilon}{2}$,or $\frac{dI}{dt} = \frac{\varepsilon}{2L}$.
Thus,option $A$ is correct.

(B) When the key $K$ is closed,the inductor opposes the change in current. However,after a long time,the current reaches a steady state.
In the steady state,the inductor acts as a simple wire with zero resistance (ideal inductor).
The circuit behaves as a simple $DC$ circuit with only the battery and the resistor.
Using Ohm's Law: $I = \frac{V}{R}$.
Given $V = 10 \ V$ and $R = 2 \ \Omega$.
Therefore,$I = \frac{10}{2} = 5 \ A$.

(A) In an inductor, the magnetic flux $\phi = LI$ cannot change instantaneously.
Initially, the inductor has an iron core, so its inductance is $L_1 = L_{core}$. The steady-state current is $I_1 = V/R$.
The magnetic flux linked with the inductor is $\phi_1 = L_1 I_1$.
When the iron core is removed, the inductance changes to $L_2$ (where $L_2 < L_1$).
Since the flux cannot change instantaneously, the current $I_2$ immediately after removal must satisfy $\phi_1 = \phi_2$, which means $L_1 I_1 = L_2 I_2$.
Since $L_2 < L_1$, it follows that $I_2 = (L_1/L_2) I_1 > I_1$.
Therefore, the current flowing through the inductor increases immediately after the removal of the iron core.

(A) The current in an $LR$ circuit at time $t$ is given by $I(t) = \frac{E}{R}(1 - e^{-Rt/L})$.
As $t \to \infty$,the steady-state current is $I_{max} = \frac{E}{R}$.
From the graph,both curves approach the same steady-state current value,which implies that the resistance $R$ remains unchanged.
The time constant of the circuit is $\tau = \frac{L}{R}$.
Curve $2$ reaches the steady-state value more slowly than curve $1$,which means the time constant for curve $2$ is larger than that for curve $1$ $(\tau_2 > \tau_1)$.
Since $\tau = \frac{L}{R}$ and $R$ is constant,an increase in the time constant $\tau$ implies that the inductance $L$ must have been increased.

(D) The time constant $\tau$ of an $LR$ circuit is given by $\tau = \frac{L}{R}$.
Initially,$\frac{L}{R} = 10$,which implies $L = 10R$ --- $(1)$.
When a resistance of $10 \, \Omega$ is added in series,the new resistance becomes $(R + 10) \, \Omega$. The new time constant is $\frac{L}{R + 10} = 2$,which implies $L = 2(R + 10) = 2R + 20$ --- $(2)$.
Equating $(1)$ and $(2)$:
$10R = 2R + 20$
$8R = 20$
$R = 2.5 \, \Omega$.
Substituting $R$ into equation $(1)$:
$L = 10 \times 2.5 = 25 \, H$.

(B) The growth of current in an $LR$ circuit is given by the formula: $I = I_0(1 - e^{-t/\tau})$,where $\tau$ is the time constant.
Given that at $t = 4 \, s$,$I = \frac{3}{4} I_0$.
Substituting these values into the equation:
$\frac{3}{4} I_0 = I_0(1 - e^{-4/\tau})$
$\frac{3}{4} = 1 - e^{-4/\tau}$
$e^{-4/\tau} = 1 - \frac{3}{4} = \frac{1}{4}$
Taking the natural logarithm on both sides:
$-\frac{4}{\tau} = \ln(\frac{1}{4})$
$-\frac{4}{\tau} = -\ln(4) = -\ln(2^2) = -2 \ln 2$
$\tau = \frac{4}{2 \ln 2} = \frac{2}{\ln 2} \, s$.

(C) To find the time constant $\tau$ of an $LR$ circuit,we need to determine the equivalent resistance $R_{eq}$ as seen from the terminals of the inductor $L$ by short-circuiting the voltage source $E$.
$1$. Short-circuit the voltage source $E$. The circuit now consists of two resistors $R$ in parallel with each other,and this combination is in series with the third resistor $R$.
$2$. The two resistors $R$ connected in parallel have an equivalent resistance of $R_p = \frac{R \times R}{R + R} = \frac{R}{2}$.
$3$. This parallel combination is in series with the remaining resistor $R$. Therefore,the total equivalent resistance $R_{eq}$ seen by the inductor is $R_{eq} = R + \frac{R}{2} = \frac{3}{2}R$.
$4$. The time constant $\tau$ for an $LR$ circuit is given by $\tau = \frac{L}{R_{eq}}$.
$5$. Substituting the value of $R_{eq}$,we get $\tau = \frac{L}{\frac{3}{2}R} = \frac{2L}{3R}$.

(D) At $t=0$,the inductor $L$ opposes any change in current,so it acts as an open circuit (current through it is zero).
For circuit $(i)$: The entire circuit is open,so the current $i_1 = 0$.
For circuit $(ii)$: The inductor $L$ is in series with resistor $R$. At $t=0$,the branch containing $L$ is open. The current flows through the parallel resistor $R$ connected directly to the source $E$. Thus,$i_2 = E/R$.
For circuit $(iii)$: The inductor $L$ is in parallel with resistor $R$. At $t=0$,the branch containing $L$ is open. The current flows through the resistor $R$ in parallel. Thus,$i_3 = E/R$.
Comparing the currents,$i_2 = i_3 = E/R$,which is greater than $i_1 = 0$. Therefore,the current is maximum in both circuits $(ii)$ and $(iii)$.

(A) When the circuit is connected for a long time,the inductor acts as a short circuit,and the steady-state current $I_0$ is given by:
$I_0 = \frac{E}{R} = \frac{100 \, V}{100 \, \Omega} = 1 \, A$
When the battery is disconnected and points $A$ and $B$ are short-circuited,the circuit becomes a decaying $LR$ circuit. The current at time $t$ is given by:
$I(t) = I_0 e^{-t/\tau}$
Here,the time constant $\tau$ is:
$\tau = \frac{L}{R} = \frac{100 \times 10^{-3} \, H}{100 \, \Omega} = 1 \times 10^{-3} \, s = 1 \, ms$
Given $t = 1 \, ms$,the current is:
$I = 1 \times e^{-1 \, ms / 1 \, ms} = 1 \times e^{-1} = \frac{1}{e} \, A$

(C) At $t = 0$, the inductor acts as an open circuit because it opposes any sudden change in current. The circuit consists of the $10\, V$ battery, $2\, \Omega$ resistor, and $4\, \Omega$ resistor in series.
Total resistance $R_{eq} = 2\, \Omega + 4\, \Omega = 6\, \Omega$.
The current $I = \frac{V}{R_{eq}} = \frac{10\, V}{6\, \Omega} = \frac{5}{3}\, A$.
At $t = \infty$, the inductor acts as a short circuit (a wire with zero resistance) because the current has reached a steady state. The $4\, \Omega$ resistor is bypassed by the short-circuited inductor.
The circuit now consists of only the $10\, V$ battery and the $2\, \Omega$ resistor.
The current $I = \frac{10\, V}{2\, \Omega} = 5\, A$.
Thus, the currents are $\frac{5}{3}\, A$ and $5\, A$ respectively.

(B) For the growth of current,the switch is closed. 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 equivalent resistance $R_{eq}$ seen by the inductor during growth is $2R$ (since the battery acts as a short circuit for the ideal case or is considered part of the loop). Thus,the time constant for growth is $\tau_{growth} = \frac{L}{2R}$.
For the decay of current,the switch is opened. The inductor $L$ now forms a closed loop with the resistors $R$ and $2R$ in series. The equivalent resistance $R_{eq}$ seen by the inductor during decay is $R + 2R = 3R$. Thus,the time constant for decay is $\tau_{decay} = \frac{L}{3R}$.
The ratio of time constants during growth and decay is $\frac{\tau_{growth}}{\tau_{decay}} = \frac{L/2R}{L/3R} = \frac{3R}{2R} = \frac{3}{2}$ or $3 : 2$.

(C) In steady state,the inductor acts as a short circuit (zero resistance).
The upper branch has a $6 \, \Omega$ resistor in series with a parallel combination of a $4 \, H$ inductor and a $2 \, \Omega$ resistor. Since the inductor is a short circuit,the $2 \, \Omega$ resistor is bypassed. Thus,the resistance of the upper branch is $6 \, \Omega$.
The lower branch has a $2 \, H$ inductor in series with a $2 \, \Omega$ resistor and a $4 \, \Omega$ resistor. The inductor acts as a short circuit,so the resistance of the lower branch is $2 \, \Omega + 4 \, \Omega = 6 \, \Omega$.
The total current entering the parallel combination is $2 \, A$ in the upper branch. Since both branches have equal resistance $(6 \, \Omega)$,the current in the lower branch is also $2 \, A$.
The total current flowing through the $3 \, \Omega$ resistor is $I_{total} = 2 \, A + 2 \, A = 4 \, A$.
The potential difference $V_A - V_B$ is the sum of the potential drop across the upper branch and the $3 \, \Omega$ resistor: $V_A - V_B = (I_{upper} \times R_{upper}) + (I_{total} \times R_{series}) = (2 \, A \times 6 \, \Omega) + (4 \, A \times 3 \, \Omega) = 12 \, V + 12 \, V = 24 \, V$.

(A) The current in an $LR$ circuit at time $t$ is given by $I(t) = \frac{E}{R}(1 - e^{-t/T})$,where $T = L/R$ is the time constant.
The charge $q$ passing through the battery in time $T$ is the integral of current with respect to time:
$q = \int_{0}^{T} I(t) dt = \int_{0}^{T} \frac{E}{R}(1 - e^{-t/T}) dt$
Integrating the expression:
$q = \frac{E}{R} \left[ t - (-T)e^{-t/T} \right]_{0}^{T}$
$q = \frac{E}{R} \left[ t + Te^{-t/T} \right]_{0}^{T}$
Substituting the limits:
$q = \frac{E}{R} \left[ (T + Te^{-T/T}) - (0 + Te^{0}) \right]$
$q = \frac{E}{R} \left[ T + \frac{T}{e} - T \right]$
$q = \frac{ET}{eR}$

(A) In a steady-state condition,an inductor acts as a short circuit (ideal wire) because the current is constant $(di/dt = 0)$.
Given: Steady-state current $I = 2.0 \ A$,$emf$ $E = 4.0 \ V$,and inductance $L = 1.0 \ H$.
Using Ohm's law for the steady state: $R = E / I = 4.0 / 2.0 = 2.0 \ \Omega$.
The time constant $\tau$ of an $LR$ circuit is given by the formula: $\tau = L / R$.
Substituting the values: $\tau = 1.0 \ H / 2.0 \ \Omega = 0.5 \ s$.
Therefore,the time constant of the circuit is $0.5 \ s$.

(D) In circuit $(a)$,we have an $RC$ series circuit. When the switch is closed at $t=0$,the capacitor starts charging. The current in the circuit is given by $i(t) = \frac{E}{R} e^{-t/RC}$. This shows that the current starts at a maximum value of $\frac{E}{R}$ and decays exponentially to zero as $t \to \infty$.
In circuit $(b)$,we have an $RL$ series circuit. When the switch is closed at $t=0$,the inductor opposes the change in current. The current in the circuit is given by $i(t) = \frac{E}{R} (1 - e^{-Rt/L})$. This shows that the current starts at $0$ and increases exponentially to a maximum steady-state value of $\frac{E}{R}$ as $t \to \infty$.
Comparing these behaviors with the given options,figure $(d)$ shows curve $(a)$ starting at $\frac{E}{R}$ and decaying,and curve $(b)$ starting at $0$ and increasing to $\frac{E}{R}$. Thus,figure $(d)$ is the correct representation.

(A) For $0 \le t < t_0$,the circuit consists of a battery,resistor $R$,and inductor $L$ in series. The current grows exponentially as $I(t) = \frac{\epsilon}{R}(1 - e^{-Rt/L})$.
At $t = t_0$,the current reaches a value $I_0 = \frac{\epsilon}{R}(1 - e^{-Rt_0/L})$.
For $t \ge t_0$,the switch $S_1$ is opened and $S_2$ is closed. The battery is removed,and the inductor $L$ discharges through the resistor $R$. The current decays exponentially as $I(t) = I_0 e^{-R(t-t_0)/L}$.
The correct graph shows exponential growth followed by exponential decay to zero,which matches the provided solution image.

(D) After a long time,the inductor acts as a short circuit (a wire with zero resistance) because the current becomes steady.
The circuit now consists of two resistors of $5\,\Omega$ each,connected in parallel across the $15\, V$ battery.
The equivalent resistance $R_{eq}$ of the two parallel resistors is given by:
$R_{eq} = \frac{R \times R}{R + R} = \frac{5 \times 5}{5 + 5} = \frac{25}{10} = 2.5\,\Omega$
The current $i$ through the battery is given by Ohm's law:
$i = \frac{V}{R_{eq}} = \frac{15}{2.5} = 6\, A$

(D) The current in an $LR$ circuit at time $t$ is given by $I = I_0(1 - e^{-t/\tau})$,where $I_0 = E/R$ and $\tau = L/R$.
Given $L = 20 \, H$ and $R = 10 \, \Omega$,the time constant $\tau = 20/10 = 2 \, s$.
So,$I = \frac{E}{10}(1 - e^{-t/2})$.
The rate of dissipation of energy (Joule's heat) across the resistance is $P_R = I^2 R = I^2 \times 10$.
The rate at which magnetic energy is stored in the inductor is $P_L = \frac{d}{dt}(\frac{1}{2} L I^2) = L I \frac{dI}{dt}$.
Equating $P_R = P_L$,we get $I^2 R = L I \frac{dI}{dt}$,which simplifies to $I R = L \frac{dI}{dt}$.
Substituting $I = \frac{E}{10}(1 - e^{-t/2})$ and $\frac{dI}{dt} = \frac{E}{10} \times \frac{1}{2} e^{-t/2}$:
$\frac{E}{10}(1 - e^{-t/2}) \times 10 = 20 \times \frac{E}{20} e^{-t/2}$.
$1 - e^{-t/2} = e^{-t/2}$.
$1 = 2 e^{-t/2} \implies e^{-t/2} = 1/2$.
Taking the natural logarithm on both sides: $-t/2 = \ln(1/2) = -\ln 2$.
Therefore,$t = 2 \ln 2 \, s$.

(A) Given: Self-inductance $L = 10\,mH = 10 \times 10^{-3}\,H$,coil resistance $r_1 = 0.1\,\Omega$,internal resistance $r_2 = 0.9\,\Omega$.
The total resistance of the circuit is $R = r_1 + r_2 = 0.1 + 0.9 = 1.0\,\Omega$.
The time constant of the $LR$ circuit is $\tau = \frac{L}{R} = \frac{10 \times 10^{-3}}{1.0} = 10^{-2}\,s$.
The current at any time $t$ is given by $i = i_0(1 - e^{-t/\tau})$,where $i_0$ is the saturation current.
We want to find $t$ when $i = 0.8 i_0$.
$0.8 i_0 = i_0(1 - e^{-t/\tau})$
$0.8 = 1 - e^{-t/\tau}$
$e^{-t/\tau} = 0.2 = \frac{1}{5}$
$e^{t/\tau} = 5$
Taking natural logarithm on both sides: $\frac{t}{\tau} = \ln 5$.
Given $\ln 5 = 1.6$,so $t = 1.6 \times \tau = 1.6 \times 10^{-2}\,s = 0.016\,s$.

(B) The current in an $LR$ circuit as a function of time $t$ after closing the switch is given by
$i(t) = \frac{E}{R}\left(1 - e^{-\frac{Rt}{L}}\right)$.
The charge $q$ that passes through the battery in time interval $t = 0$ to $t = \frac{L}{R}$ is given by the integral of current with respect to time:
$q = \int_{0}^{\frac{L}{R}} i(t)\, dt = \int_{0}^{\frac{L}{R}} \frac{E}{R}\left(1 - e^{-\frac{Rt}{L}}\right) dt$
$q = \frac{E}{R} \int_{0}^{\frac{L}{R}} \left(1 - e^{-\frac{Rt}{L}}\right) dt$
$q = \frac{E}{R} \left[ t + \frac{L}{R} e^{-\frac{Rt}{L}} \right]_{0}^{\frac{L}{R}}$
$q = \frac{E}{R} \left[ \left(\frac{L}{R} + \frac{L}{R} e^{-1}\right) - \left(0 + \frac{L}{R} e^{0}\right) \right]$
$q = \frac{E}{R} \left[ \frac{L}{R} + \frac{L}{Re} - \frac{L}{R} \right]$
$q = \frac{E}{R} \cdot \frac{L}{Re} = \frac{EL}{eR^2}$
Since $e \approx 2.718$, we have:
$q = \frac{EL}{2.7R^2}$

(A) The time constant of an $LR$ circuit is given by $\tau = \frac{L}{R}$.
Initially,$\tau_1 = \frac{L}{R} = 4 \, sec$.
When the coil is cut into two equal parts,the resistance and inductance of each part become half of the original values,i.e.,$R' = \frac{R}{2}$ and $L' = \frac{L}{2}$.
When these two parts are connected in parallel,the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R'} + \frac{1}{R'} = \frac{2}{R'} = \frac{2}{R/2} = \frac{4}{R}$,so $R_{eq} = \frac{R}{4}$.
The equivalent inductance $L_{eq}$ for two inductors in parallel is $\frac{1}{L_{eq}} = \frac{1}{L'} + \frac{1}{L'} = \frac{2}{L'} = \frac{2}{L/2} = \frac{4}{L}$,so $L_{eq} = \frac{L}{4}$.
The new time constant is $\tau_2 = \frac{L_{eq}}{R_{eq}} = \frac{L/4}{R/4} = \frac{L}{R} = \tau_1$.
Therefore,the new time constant is $4 \, sec$.

(A) In a $DC$ circuit,after a long time $(t \to \infty)$,the inductor acts as a short circuit (a wire with zero resistance) because the current becomes steady and the induced $EMF$ across the inductor becomes zero.
Since the $50\,\Omega$ resistor is connected in parallel with the pure inductor,the entire current will flow through the path of least resistance,which is the inductor.
Therefore,the potential difference across the $50\,\Omega$ resistor becomes zero.
Consequently,the final current flowing through the $50\,\Omega$ resistor is $0\,A$.

(A) The time constant $\tau$ for an $RL$ circuit is given by $\tau = \frac{L}{R_{eq}}$.
When the switch is in position $a$,the inductor $L = 4\,mH$ is in series with the parallel combination of resistors $R$ and $500\,\Omega$.
The equivalent resistance $R_{eq}$ is the resistance seen by the inductor,which is $R_{eq} = \frac{R \times 500}{R + 500}$.
Given $\tau = 10\,\mu s = 10 \times 10^{-6}\,s$ and $L = 4\,mH = 4 \times 10^{-3}\,H$.
Substituting these values into the formula:
$10 \times 10^{-6} = \frac{4 \times 10^{-3}}{\left(\frac{R \times 500}{R + 500}\right)}$
$10 \times 10^{-6} = \frac{4 \times 10^{-3} \times (R + 500)}{500R}$
$10^{-5} = \frac{4 \times 10^{-3} \times (R + 500)}{500R}$
$500R \times 10^{-5} = 4 \times 10^{-3} \times (R + 500)$
$5 \times 10^{-3} R = 4 \times 10^{-3} R + 2000 \times 10^{-3}$
$10^{-3} R = 2$
$R = 2000\,\Omega = 2\,K\Omega$.

(B) The current in an $LR$ circuit at time $t$ is given by the formula: $I(t) = I_0(1 - e^{-\frac{R}{L}t})$.
We are given that the current reaches $\frac{e-1}{e}$ of its final value $I_0$.
So,$I(t) = \left(\frac{e-1}{e}\right)I_0 = I_0(1 - e^{-1})$.
Comparing this with the formula,we get $e^{-\frac{R}{L}t} = e^{-1}$.
This implies $\frac{R}{L}t = 1$,or $t = \frac{L}{R}$.
Given $L = 60\, H$ and $R = 30\, \Omega$,we have $t = \frac{60}{30} = 2\, s$.

(C) The energy stored in an inductor is given by $U = \frac{1}{2} L I^2$,where $L$ is the inductance and $I$ is the current.
From this,we can express the inductance as $L = \frac{2U}{I^2}$.
The rate of energy dissipation (power loss) in the inductor due to its internal resistance $R$ is given by $P = I^2 R$.
From this,we can express the resistance as $R = \frac{P}{I^2}$.
The time constant $\tau$ of an $LR$ circuit is defined as $\tau = \frac{L}{R}$.
Substituting the expressions for $L$ and $R$:
$\tau = \frac{2U / I^2}{P / I^2} = \frac{2U}{P}$.
Therefore,the time constant is $\frac{2U}{P}$.

(B) The circuit consists of a battery of $5\,V$,an inductor of $10\,mH$,and a variable resistor $R_H$.
According to Kirchhoff's voltage law,the equation for the circuit is $V = iR + L\frac{di}{dt}$.
If the resistance $R$ were constant at $10\,\Omega$,the steady-state current would be $i = \frac{V}{R} = \frac{5}{10} = 0.5\,A$.
However,the resistance is increasing,which means $\frac{dR}{dt} > 0$. As the resistance increases,the current $i$ tends to decrease.
According to Lenz's law,the inductor opposes this change by inducing an electromotive force $(EMF)$ that tries to maintain the current.
The induced $EMF$ is given by $\varepsilon = -L\frac{di}{dt}$. Since the current is decreasing,$\frac{di}{dt}$ is negative,making the induced $EMF$ positive in the direction of the main current.
This induced $EMF$ acts as a source that adds to the current,causing the actual current to be higher than the value it would have if the resistance were constant.
Therefore,$i > 0.5\,A$.

(A) The current in a decaying $L-R$ circuit is given by $i = i_0 e^{-t/\tau}$,where $\tau = L/R$ is the time constant.
The energy stored in the inductor is $U = \frac{1}{2} L i^2$.
Let $U_0$ be the initial energy,$U_0 = \frac{1}{2} L i_0^2$.
We are given that the energy reduces to one-fourth of its initial value,so $U = \frac{1}{4} U_0$.
Substituting the expressions for energy: $\frac{1}{2} L i^2 = \frac{1}{4} (\frac{1}{2} L i_0^2)$.
This simplifies to $i^2 = \frac{1}{4} i_0^2$,or $i = \frac{1}{2} i_0$.
Substituting $i = i_0 e^{-t/\tau}$,we get $i_0 e^{-t/\tau} = \frac{1}{2} i_0$.
$e^{-t/\tau} = \frac{1}{2} \Rightarrow e^{t/\tau} = 2$.
Taking the natural logarithm on both sides: $t/\tau = \ln 2$.
Therefore,$t = \tau \ln 2 = \frac{L}{R} \ln 2$.

(B) When the switch $S$ is closed at $t=0$,the current $i$ in the $RL$ circuit grows according to the equation:
$i = \frac{E}{R} (1 - e^{-tR/L})$
This is an increasing exponential function starting from $i=0$ at $t=0$ and approaching $i=E/R$ as $t \to \infty$. This behavior is represented by graph $D$.
The induced $emf$ $e$ across the inductor $L$ is given by:
$e = L \frac{di}{dt} = L \frac{d}{dt} [\frac{E}{R} (1 - e^{-tR/L})] = L \cdot \frac{E}{R} \cdot \frac{R}{L} e^{-tR/L} = E e^{-tR/L}$
At $t=0$,$e=E$,and as $t \to \infty$,$e \to 0$. This is a decreasing exponential function,which is represented by graph $C$.
Therefore,graph $C$ represents $e$ and graph $D$ represents $i$.

(B) For a short-circuited $RL$ circuit,the current at time $t$ is given by $I = I_0 e^{-t/\tau}$,where $\tau$ is the time constant.
Given that at $t = t_0$,the current becomes $I = \frac{I_0}{\alpha}$.
Substituting these values into the equation:
$\frac{I_0}{\alpha} = I_0 e^{-t_0/\tau}$
$\frac{1}{\alpha} = e^{-t_0/\tau}$
Taking the natural logarithm on both sides:
$\ln(1/\alpha) = -t_0/\tau$
$-\ln(\alpha) = -t_0/\tau$
$\ln(\alpha) = t_0/\tau$
Therefore,the time constant is $\tau = \frac{t_0}{\ln(\alpha)}$ or $\tau = \frac{t_0}{\log_e \alpha}$.

(B) When the switch $K$ is closed,the circuit reaches a steady state after a long time.
In the steady state,the inductor acts as a short circuit (ideal inductor with zero resistance).
Since the inductor is connected in parallel with the $20\, \Omega$ resistor,the entire current will flow through the inductor,effectively short-circuiting the $20\, \Omega$ resistor.
Therefore,the equivalent resistance of the circuit is only the $30\, \Omega$ resistor.
The current $I$ flowing through the $30\, \Omega$ resistor is given by Ohm's law:
$I = \frac{V}{R} = \frac{3\, V}{30\, \Omega} = 0.1\, A$.

(D) The equation for an $LR$ circuit is given by $V = IR + L \frac{dI}{dt}$.
At the instant the switch is closed $(t = 0)$,the current $I$ in the circuit is $0$.
Substituting $I = 0$ into the equation,we get $V = L \frac{dI}{dt}$.
Therefore,the initial rate of growth of the current is $\frac{dI}{dt} = \frac{V}{L}$.
Given $V = 1.6 \, V$ and $L = 0.20 \, H$,we have $\frac{dI}{dt} = \frac{1.6}{0.20} = 8 \, A \, s^{-1}$.

(B) The power dissipated in the resistor is given by $P = I^2 R$.
Since the current through the resistor varies with time,we must integrate the power over time to find the total heat energy $W$.
$W = \int_{0}^{\infty} I^2 R \, dt$
The current in an $RL$ circuit during decay is given by $I = I_0 e^{-(R/L)t}$.
Substituting this into the integral:
$W = \int_{0}^{\infty} (I_0 e^{-(R/L)t})^2 R \, dt = \int_{0}^{\infty} I_0^2 R e^{-(2R/L)t} \, dt$
$W = I_0^2 R \left[ \frac{e^{-(2R/L)t}}{-(2R/L)} \right]_{0}^{\infty} = I_0^2 R \left( 0 - \left( -\frac{L}{2R} \right) \right) = \frac{1}{2} L I_0^2$.
Alternatively,by the law of conservation of energy,the total heat produced in the resistor must equal the total energy initially stored in the magnetic field of the inductor,which is $U = \frac{1}{2} L I_0^2$.

(A) The circuit consists of two inductors $L_1$ and $L_2$ in series and two resistors $R_1$ and $R_2$ in series.
For inductors in series,the equivalent inductance is $L = L_1 + L_2 = 1\, mH + 2\, mH = 3\, mH$.
For resistors in series,the equivalent resistance is $R = R_1 + R_2 = 1\, \Omega + 2\, \Omega = 3\, \Omega$.
The time constant $\tau$ for an $RL$ circuit is given by $\tau = \frac{L}{R}$.
Substituting the values,we get $\tau = \frac{3\, mH}{3\, \Omega} = 1\, ms$.

(A) Just after pressing the key $K$ (at $t = 0$), the inductor acts as an open circuit because it opposes any instantaneous change in current.
Therefore, the branch containing the inductor and the $8 \, \Omega$ resistor is effectively disconnected.
The circuit simplifies to a $10 \, V$ battery in series with a $6 \, \Omega$ resistor and an $8 \, \Omega$ resistor connected in parallel to the source.
However, since the inductor branch is open, the current $I_1$ flows through the $6 \, \Omega$ resistor and then entirely through the $8 \, \Omega$ resistor connected in parallel.
The equivalent resistance of the circuit is $R_{eq} = 6 \, \Omega + 8 \, \Omega = 14 \, \Omega$.
The current $I_1$ is given by $I_1 = \frac{V}{R_{eq}} = \frac{10 \, V}{14 \, \Omega} = \frac{5}{7} \, A$.

(C) The growth of current in an $LR$ circuit is given by $I = I_0 (1 - e^{-t/\tau})$,where $\tau$ is the time constant.
Given that at $t = 4 \, sec$,$I = \frac{I_0}{4}$.
Substituting these values: $\frac{I_0}{4} = I_0 (1 - e^{-4/\tau})$.
$\frac{1}{4} = 1 - e^{-4/\tau} \implies e^{-4/\tau} = 1 - \frac{1}{4} = \frac{3}{4}$.
Taking the natural logarithm on both sides: $-\frac{4}{\tau} = \log_e (3/4)$.
$-\frac{4}{\tau} = \log_e 3 - \log_e 4$.
$\frac{4}{\tau} = \log_e 4 - \log_e 3 = \log_e (4/3)$.
Therefore,$\tau = \frac{4}{\log_e (4/3)} \, sec$.

(C) The circuit consists of two parallel branches connected to a $10 \, V$ battery.
Branch $1$ contains an inductor $(0.1 \, H)$ and a resistor $(10 \, \Omega)$ in series.
Branch $2$ contains a resistor $(10 \, \Omega)$.
At $t = 0$ (just after closing the switch),the inductor acts as an open circuit due to the back $EMF$. Thus,no current flows through the first branch. The total current in the circuit is only through the second branch: $i_{min} = \frac{10 \, V}{10 \, \Omega} = 1 \, A$.
At $t = \infty$ (long after closing the switch),the inductor acts as a short circuit (zero resistance). The total current is the sum of currents in both branches: $i_{branch1} = \frac{10 \, V}{10 \, \Omega} = 1 \, A$ and $i_{branch2} = \frac{10 \, V}{10 \, \Omega} = 1 \, A$.
Thus,$i_{max} = 1 \, A + 1 \, A = 2 \, A$.
The difference between the maximum and minimum current is $i_{max} - i_{min} = 2 \, A - 1 \, A = 1 \, A$.

(A) When the switch $S$ is closed,the current in the circuit containing the inductor $L$ grows gradually due to the back $EMF$ induced in 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 the bulb $B_1$ (in the inductor branch).
As time passes,the current in the inductor branch increases and eventually reaches the same steady-state value as the resistor branch,because the resistance of the inductor coil is equal to $R$.
Thus,finally,both bulbs shine with equal brightness.

(D) When the key $K$ is closed,the circuit reaches a steady state.
In a steady state,an ideal inductor acts as a short circuit (zero resistance).
Since the inductor is connected in parallel with the $10 \, \Omega$ resistor,the entire current will flow through the inductor path,bypassing the $10 \, \Omega$ resistor.
Therefore,the final current flowing through the $10 \, \Omega$ resistor is $0 \, A$.

(D) The time constant $\tau$ of an $LR$ circuit is given by $\tau = \frac{L}{R}$.
Initially,$\frac{L}{R} = 10$,which implies $L = 10R$ .......... $(1)$
When a resistance of $10 \, \Omega$ is added in series,the new resistance becomes $(R + 10) \, \Omega$.
The new time constant is $\frac{L}{R + 10} = 2$,which implies $L = 2(R + 10) = 2R + 20$ .......... $(2)$
Equating $(1)$ and $(2)$:
$10R = 2R + 20$
$8R = 20$
$R = 2.5 \, \Omega$
Substituting $R$ into equation $(1)$:
$L = 10 \times 2.5 = 25 \, H$.

(D) An inductor in a $d.c.$ circuit behaves as an open circuit at $t = 0$ (infinite resistance) and as a short circuit at steady state ($t \to \infty$,zero resistance).
Since the resistance changes from infinity to zero,the assertion that it is 'always constant' is incorrect.
In a steady state,an ideal inductor has zero resistance,so the statement that it is 'non-zero' is also incorrect.
Therefore,both the Assertion and the Reason are incorrect.

(B) The current in an $RL$ circuit at time $t$ is given by $i(t) = i_0(1 - e^{-Rt/L})$,where $i_0 = V/R$ is the steady-state current at $t = \infty$.
Given $V = 20\; V$,$R = 5\; \Omega$,and $L = 10\; mH = 10 \times 10^{-3}\; H = 0.01\; H$.
The time constant $\tau = L/R = 0.01 / 5 = 0.002\; s = 2\; ms$.
At $t = 40\; ms$,the exponent is $-Rt/L = -t/\tau = -40\; ms / 2\; ms = -20$.
Wait,checking the time provided: if $t = 40\; ms$,then $t/\tau = 20$. If $t = 40\; s$,then $t/\tau = 20000$. Assuming the question implies $t = 4\; ms$ to match the $e^2$ hint: $t/\tau = 4\; ms / 2\; ms = 2$.
Then $i(4\; ms) = i_0(1 - e^{-2})$.
The ratio $i(\infty) / i(4\; ms) = i_0 / [i_0(1 - e^{-2})] = 1 / (1 - 1/e^2) = e^2 / (e^2 - 1)$.
Using $e^2 = 7.389$,the ratio is $7.389 / (7.389 - 1) = 7.389 / 6.389 \approx 1.156$.

(B) The current in an $LR$ circuit at time $t$ is given by $i(t) = \frac{\varepsilon}{R} (1 - e^{-Rt/L})$.
Let $T_c = \frac{L}{R}$ be the time constant. Then $i(t) = \frac{\varepsilon}{R} (1 - e^{-t/T_c})$.
The total charge $q$ flowing from $t=0$ to $t=T_c$ is given by the integral of current with respect to time:
$q = \int_{0}^{T_c} i(t) dt = \int_{0}^{T_c} \frac{\varepsilon}{R} (1 - e^{-t/T_c}) dt$
$q = \frac{\varepsilon}{R} \left[ t - \frac{e^{-t/T_c}}{-1/T_c} \right]_{0}^{T_c} = \frac{\varepsilon}{R} \left[ t + T_c e^{-t/T_c} \right]_{0}^{T_c}$
$q = \frac{\varepsilon}{R} \left[ (T_c + T_c e^{-1}) - (0 + T_c e^{0}) \right]$
$q = \frac{\varepsilon}{R} [T_c + T_c e^{-1} - T_c] = \frac{\varepsilon}{R} T_c e^{-1}$
Substituting $T_c = \frac{L}{R}$:
$q = \frac{\varepsilon}{R} \cdot \frac{L}{R} \cdot \frac{1}{e} = \frac{\varepsilon L}{e R^2}$.

(N/A) rod of length $d$ moves with velocity $v$ perpendicular to a magnetic field $B$,inducing an electromotive force (emf) $\varepsilon = Bvd$ between its ends.
When the switch $S$ is closed at $t = 0$,the capacitor begins to charge. Let $Q(t)$ be the charge on the capacitor at time $t$.
Applying Kirchhoff's voltage law to the circuit:
$\varepsilon = V_R + V_C$
$Bvd = IR + \frac{Q}{C}$
Since $I = \frac{dQ}{dt}$,we have:
$R \frac{dQ}{dt} + \frac{Q}{C} = Bvd$
$\frac{dQ}{dt} + \frac{Q}{RC} = \frac{Bvd}{R}$
This is a first-order linear differential equation. The general solution for $Q(t)$ is:
$Q(t) = BvdC + A e^{-t/RC}$
At $t = 0$,the charge $Q(0) = 0$,so $0 = BvdC + A$,which gives $A = -BvdC$.
Thus,$Q(t) = BvdC(1 - e^{-t/RC})$.
The current $I(t)$ is the derivative of charge with respect to time:
$I(t) = \frac{dQ}{dt} = \frac{d}{dt} [BvdC(1 - e^{-t/RC})]$
$I(t) = BvdC \cdot \left( \frac{1}{RC} \right) e^{-t/RC}$
$I(t) = \frac{Bvd}{R} e^{-t/RC}$

(N/A) When a rod of length $d$ moves with velocity $v$ perpendicular to a magnetic field,an electromotive force (emf) is induced between the two ends of the rod.
The induced emf is $\varepsilon = Bvd$.
When the switch $S$ is closed at $t = 0$,the current in the inductor starts to increase.
Let the current in the circuit be $I$ at time $t$. According to Kirchhoff's voltage law:
$\varepsilon = V_R + V_L$
$Bvd = IR + L \frac{dI}{dt}$
Rearranging the terms,we get the linear differential equation:
$\frac{dI}{dt} + \frac{R}{L}I = \frac{Bvd}{L}$
The solution to this linear differential equation is of the form:
$I(t) = \frac{Bvd}{R} + A e^{-\frac{R}{L}t}$
Using the initial condition that at $t = 0$,$I = 0$:
$0 = \frac{Bvd}{R} + A \implies A = -\frac{Bvd}{R}$
Substituting $A$ back into the equation:
$I(t) = \frac{Bvd}{R} \left(1 - e^{-\frac{R}{L}t}\right)$
This is the required expression for the current in the rod as a function of time.

(A) The maximum energy stored in the inductor is $U_{\max} = \frac{1}{2} L I_{\max}^2$.
The current in the $L-R$ circuit at time $t$ is given by $i = I_{\max}(1 - e^{-Rt/L})$.
We want the energy $U$ to be $\frac{U_{\max}}{n}$. Since $U = \frac{1}{2} L i^2$,this implies $\frac{1}{2} L i^2 = \frac{1}{n} \left(\frac{1}{2} L I_{\max}^2\right)$,which simplifies to $i = \frac{I_{\max}}{\sqrt{n}}$.
Substituting this into the current equation: $\frac{I_{\max}}{\sqrt{n}} = I_{\max}(1 - e^{-Rt/L})$.
Dividing by $I_{\max}$,we get $\frac{1}{\sqrt{n}} = 1 - e^{-Rt/L}$,or $e^{-Rt/L} = 1 - \frac{1}{\sqrt{n}} = \frac{\sqrt{n}-1}{\sqrt{n}}$.
Taking the natural logarithm on both sides: $-\frac{Rt}{L} = \ln \left(\frac{\sqrt{n}-1}{\sqrt{n}}\right)$.
Multiplying by $-1$ and rearranging: $\frac{Rt}{L} = -\ln \left(\frac{\sqrt{n}-1}{\sqrt{n}}\right) = \ln \left(\frac{\sqrt{n}}{\sqrt{n}-1}\right)$.
Therefore,$t = \frac{L}{R} \ln \left(\frac{\sqrt{n}}{\sqrt{n}-1}\right)$.

(C) The magnetic energy stored in an inductor is given by $U = \frac{1}{2} L i^2$.
Let $U_{max}$ be the maximum energy,which occurs when the current reaches its maximum value $i_0$. Thus,$U_{max} = \frac{1}{2} L i_0^2$.
We are given that the energy reaches $25\%$ of its maximum value:
$U = 0.25 U_{max} = \frac{1}{4} (\frac{1}{2} L i_0^2) = \frac{1}{2} L (\frac{i_0}{2})^2$.
Comparing this with $U = \frac{1}{2} L i^2$,we find that the current $i$ must be $\frac{i_0}{2}$.
The expression for current in an $RL$ circuit during charging is $i = i_0 (1 - e^{-Rt/L})$.
Setting $i = \frac{i_0}{2}$,we get: $\frac{i_0}{2} = i_0 (1 - e^{-Rt/L})$.
$\frac{1}{2} = 1 - e^{-Rt/L} \Rightarrow e^{-Rt/L} = \frac{1}{2}$.
Taking the natural logarithm on both sides: $-\frac{Rt}{L} = \ln(\frac{1}{2}) = -\ln 2$.
Therefore,$t = \frac{L}{R} \ln 2$.

(D) At $t = 0$,the inductor acts as an open circuit (infinite resistance). The circuit simplifies to two parallel branches: one with $(5 \Omega + 1 \Omega) = 6 \Omega$ and the other with $(5 \Omega + 4 \Omega) = 9 \Omega$.
These two branches are in parallel,so the equivalent resistance is $R_{eq} = (6 \times 9) / (6 + 9) = 54 / 15 = 18 / 5 \ \Omega$.
The current $i$ at $t = 0$ is $i_1 = E / R_{eq} = E / (18 / 5) = 5E / 18$.
At $t = \infty$,the inductor acts as a short circuit (zero resistance). The circuit becomes a bridge-like structure where the $5 \Omega$ and $5 \Omega$ resistors are in parallel,and the $1 \Omega$ and $4 \Omega$ resistors are in parallel.
The equivalent resistance is $R_{eq} = (5 \parallel 5) + (1 \parallel 4) = (2.5) + (4 / 5) = 2.5 + 0.8 = 3.3 \ \Omega = 33 / 10 \ \Omega$.
The current $i$ at $t = \infty$ is $i_2 = E / R_{eq} = E / (33 / 10) = 10E / 33$.