LR Circuit with DC Source

A coil having inductance $L$ and resistance $R$ is connected to a battery of emf $\epsilon$ at $t=0$. If $t_1$ and $t_2$ are the time for $90\%$and $99\%$completion of the current growth in the circuit, then $\frac{t_2}{t_1}$ will be-

The time constant of the given circuit, containing an inductor and few resistors, is 

In the circuit shown here, the point '$C$' $A$is kept connected to point '' till the current flowing through the circuit becomes constant. Afterwards, suddenly, point '$C$' $A$is disconnected from point '' and connected to point '$B$' at time $t=0$. Ratio of the voltage across resistance and the inductor at $t=\frac{L}{R}$ will be equal to:

          

In the circuit shown below, the key $K$ is closed at $t=0$. The current through the battery is:

An inductor of inductance, $L=400 \mathrm{mH}$ and resistors of resistances, $R_1=2 \mathrm{\Omega }$ and $R_2=2 \mathrm{\Omega }$ are connected to a battery of emf $12 \mathrm{V}$ as shown in the figure. The internal resistance of the battery is negligible. The switch $S$ is closed at $t=0$. The potential drop across $L$ as a function of time is,

Figure shows a circuit that contains three identical resistors with resistance $R=9.0 \mathrm{\Omega }$ each two identical inductors, with inductance $L=2.0 \mathrm{mH}$ each and an ideal battery with emf $\epsilon =18 \mathrm{V}$. The current battery just after the switch closed is-

An inductor $\left(L=0.03 \mathrm{H}\right)$ and a resistor $\left(R=0.15 \mathrm{k\Omega }\right)$ are connected in series to a battery of $15 \mathrm{V} \mathrm{EMF}$ in a circuit shown below. The key $K_1$ has been kept closed for a long time. Then at $t=0, K_1$ is opened and key $K_2$ is closed simultaneously. At $t=1 \mathrm{ms}$, the current in the circuit will be $:\left(e^5\cong 150\right)$

In the circuit shown here, the point $C$ is kept connected to point $A$ till the current flowing through the circuit becomes constant. Afterward, suddenly point $C$ is disconnected from point $A$ and connected to point $B$ at time $t=0$. The ratio of the voltage across resistance and the inductor at $t=\frac{L}{R}$ will be equal to,

The battery shown in the figure is ideal: The values are $\epsilon =10 \mathrm{V},R=5\mathrm{\Omega },L=2\mathrm{H}.$Initially the current in the inductor is zero. The current through the battery at $t=2s$ is

An inductor $\left(L=100 \mathrm{mH}\right)$, a resistor $\left(R=100 \mathrm{\Omega }\right)$ and a battery $\left(E=100 \mathrm{V}\right)$ are initially connected in series as shown in the figure. After a long time, the battery is disconnected after short-circuiting the points $A$ and $B$. The current in the circuit, $1$ $\mathrm{ms}$ after the short circuit is:

In the given circuit find the ratio of $i_1$ to $i_2,$ where $i_1$ is the initial (at $t=0$) current, and $i_2$ is steady state (at $t=\infty$) current through the battery.

In the figure magnetic energy stored in the coil is (in steady state)

In the circuit shown the switch $S_1$ is closed at time $t=0$ and the switch $S_2$ is kept open. At some later time $\left(t_0\right)$, the switch $S_1$ is opened and $S_2$ is closed. The behavior of the current $I$ as a function of time '$t$' given by

In the given circuit $S_1$and $S_2$ are switches. $S_2$ is closed for a long time and $S_1$ remains open. Now $S_1$ is also closed. Here $R=10 \Omega ,L=1 \mathrm{mH}$ and $\epsilon =3 \mathrm{V}$. Let $\frac{\mathrm{di}}{\mathrm{dt}}$ is the magnitude of rate of change of current (in $\mathrm{A} {\mathrm{s}}^{-1}$), just after $S_1$ is closed. The value for the inductor $L$ is $\frac{\mathrm{d}i}{\mathrm{d}t}=x\times {10}^3 \mathrm{A} {\mathrm{s}}^{-1}$. Then find $x$.

In the figure shown the battery is ideal. The values are $\epsilon =10 \mathrm{V}, R=5 \mathrm{\Omega },\mathit{ }L=2 \mathrm{H}$. The current through the battery at $t=2 \mathrm{s}$ after closing the switch is:

When a $LR$ circuit with $L=3 \mathrm{mH}$ is connected in series to an $AC$ source of $\mathrm{rms}$ voltage $30 \mathrm{V},$ a maximum rms current of $6 \mathrm{A}$ is observed at frequency $\left(\frac{500}{\pi }\right)\mathrm{Hz}.$ If this $LR$ circuit is now connected to a battery of emf $21 \mathrm{V}$ and internal resistance of $3 \mathrm{\Omega },$ the current will be

The value of time constant for given circuit is $\dots ....s$

An e.m.f. of $15$ $\mathrm{V}$ is applied in a circuit containing $5 \mathrm{H}$ inductance and $10 \mathrm{\Omega }$ resistance. The ratio of the currents at time $t\to \infty$ and at $t=1 \sec$  is:

In an $L-R$ circuit, time constant is that time in which current grows from zero to the value... (where, $I_0$ is steady-state current).

In the given branch $\mathrm{AB}$ of a circuit a current $\mathrm{I}=(10 \mathrm{t}+5) \mathrm{A}$ is flowing, where $\mathrm{t}$ is time in second. At $\mathrm{t}=0,$ the potential difference between points $\mathrm{A}$ and $\mathrm{B}\left({\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}\right)$ is :