Growth and Decay of Current in LR Circuits

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-

What is the time constant in LR DC circuit?

How do you find the half life of a circuit?

How do DC circuits work?

What is basic DC circuit?

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

In the inductive circuit given in the figure, the current rises after the switch is put on. At instant when the current is $15 \mathrm{mA},$ then potential difference across the inductor will be :-

Identify the correct statement out of the following options regarding the given circuit.

For a certain inductive coil, it is known that its time constant is $2.0\times {10}^{-3} \mathrm{s}$. If a $90 \mathrm{\Omega }$ resistance is joined in series with the coil, the time constant becomes $0.5\times {10}^{-3} \mathrm{s}$. Hence, find the inductance and resistance of the coil.

With respect to the given L-R circuit, $i_1$ is the initial current and $i_2$ is the steady state current through the battery. Then, calculate the ratio $\frac{i_1}{i_2}$.

In a $\mathrm{LR}$ circuit, an inductance of $300 \mathrm{mH}$ and a resistance of $2 \mathrm{\Omega }$ are connected in series to a source of voltage $2 \mathrm{V}$. What is the time when the current in the circuit reaches half of its steady state value?

Calculate the time constant for the adjoining circuit that contains an inductor with a few resistors.

How much time does a $10 \mathrm{H}$ solenoid of $2 \mathrm{\Omega }$ resistance require, to attain the magnetic energy equal to ${\left(\frac{1}{4}\right)}^{\mathrm{th}}$ of its maximum magnetic energy, when connected to a $10 \mathrm{V}$ battery?

At $t=\mathrm{0,}$ the switch is closed. Find the time after which the rate of dissipation of energy in the resistor is equal to the rate at which energy is being stored in the inductor.

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 ' $t$ ' is given by:

In the given $L-R$ circuit, the switch $S_1$ is initially closed and $S_2$ is open. At any time $t=t_0$ , the switch $S_1$ is opened and $S_2$ is closed. The current $\left(i\right)$ vs time $\left(t\right)$ graph for the entire time during which it passes through the inductor is shown in figure. Then the best suitable graph is.

For given $\mathrm{L}-\mathrm{R}$ circuit, growth of current as function of time $\mathrm{t}$ is shown in graph. Which of the following option represents value of time constant most closely for the circuit?

An inductor $\left(L=20\mathrm{ m}H\right),$ a resistor $\left(R =100\mathrm{\Omega }\right)$ and a battery $\left(E=10V\right)$ are connected in series. After a long time, the circuit is short-circuited and then the battery is disconnected. Find the current in the circuit at 1ms after short circuiting.

The current in $\mathrm{LR}$ circuit builds up to ${\left(\frac{3}{4}\right)}^{\mathrm{th}}$ of its maximum value in $4 \mathrm{s}$. The time constant of the circuit is

In a circuit inductance L and capacitance C are connected as shown in figure. $A_1$ and $A_2$ are ammeters. When key K is pressed to complete the circuit, then just after closing key (K), the readings of $A_1$ and $A_2$ will be