In the circuit shown,the final current through the $30\, \Omega$ resistance when the circuit is completed is ......... $A$.

The time constant of an inductance coil is $3 \text{ ms}$. When a $90 \Omega$ resistance is joined in series,the time constant becomes $0.5 \text{ ms}$. The inductance and the resistance of the coil are:

As shown in the figure,a battery of $emf \; \varepsilon$ is connected to an inductor $L$ and resistance $R$ in series. The switch $S$ is closed at $t=0$. The total charge that flows from the battery between $t=0$ and $t=t_{c}$ ($t_{c}$ is the time constant of the circuit) is

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

$A$ coil carrying a steady current is short-circuited. The current in it decreases $\alpha$ times in $t_0$. The time constant of the circuit is:-

An inductance $L$ and a resistance $R$ are first connected to a battery. After some time,the battery is disconnected,but $L$ and $R$ remain connected in a closed circuit. Then the current reduces to $37\%$ of its initial value in