LR DC Circuit

In the given figure, an inductor and resistor are connected in series with a battery of $\mathrm{emf}$ $E$ volt. $\frac{E^a}{2b} \mathrm{J} {\mathrm{s}}^{-1}$ represents the maximum rate at which the energy is stored in the magnetic field (inductor). The numerical value of $\frac{b}{a}$ will be _____.

In the circuit shown, the power consumed by the bulb$B_1$ long after the key $K$ is closed is (inductor is ideal)

Two resistors of equal resistance $R$, an inductor of inductance $8 \mathrm{mH}$ and a capacitor of capacitance $20 \mathrm{\mu F}$ are connected across an ideal battery of emf $10 \mathrm{V}$. The switch $S$ is closed at time $t=0$. After closing the key, the current through battery is found to be constant. The value of $\frac{R}{10}$(in ohms)

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In the given circuit key is closed at $t=0$. At what time the potential difference across inductor is one fourth of emf of the cell.

In the circuit shown the reading of ammeter long time after the key is closed is

In a particular R-L series circuit a voltage of $10 \mathrm{V}$ at $50 \mathrm{Hz}$ produces a current of $700 \mathrm{mA}$ while the same voltage at $75 \mathrm{Hz}$ produces $500 \mathrm{mA}$. What are the values of $R$ and $L$ in the circuit ?

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-

In the given circuit, $R_1=4 \mathrm{\Omega }, R_2=2 \mathrm{\Omega } \text{and }{R}_3=10 \mathrm{\Omega }$ the battery has emf $E=10 \mathrm{V}$ and negligible internal resistance, the inductance of the inductor $L=400 \mathrm{mH}$. Find the potential drop across the inductor as a function of time, if at $t=0$ switch $S$ is closed.

In the given circuit, what is the current(in $\mathrm{A}$) drawn from battery at time $t=0$, when switch is closed.

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 $\dots ....s$

Consider a part of network as shown below:

If at a certain instant, the current $i$ is $5 \mathrm{A}$ and it is decreasing at the rate of ${10}^3 \mathrm{A} {\mathrm{s}}^{-1}$, then $V_{\mathrm{B}}-V_{\mathrm{A}}$ is,

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 :

In the circuit shown in figure switch S is closed at time $\mathrm{t}=0$.The charge which passes through the battery in one time constant is :

In which of the following circuit is the current maximum just after the switch S is closed:

A solenoid has an inductance of $60 \mathrm{henry}$ and a resistance of $30 \mathrm{ohm}$. If it connected to a $100 \mathrm{volt}$ battery, how long will it take for the current to reach $\frac{e-1}{e}\approx 63.2\%$ of its final value?

Two resistance $20\mathrm{\Omega }$ and $50\mathrm{\Omega }$ and a pure inductance of $50 \mathrm{H}$ are connected to a $10 \mathrm{V}$ battery through a key as shown in the figure. The key is tapped at $t=0$. Find the final value of current in the $50\mathrm{\Omega }$ resistor :

An $LR$ circuit having $L=4 \mathrm{H}, R=1 \mathrm{\Omega }$ and $E=6 \mathrm{V}$ is switched on at $t=0.$ The power dissipated in joule heating at $t=4 \mathrm{s}$ is