$A$ constant voltage of $25 \, V$ is applied to a series $L-R$ circuit at $t=0$ by closing a switch. What is the potential difference across the resistor and the inductor at time $t=0$?

If $L$ is the inductance and $R$ is the resistance,then the $SI$ unit of $\frac{L}{R}$ is

Three identical resistors with resistance $R = 12\,\Omega$ and two identical inductors with self-inductance $L = 5\,mH$ are connected to an ideal battery with an emf of $12\,V$ as shown in the figure. The current through the battery long after the switch has been closed will be $......A$.

In the circuit shown below,the switch is closed at time $t=0$. Which of the graphs shown below best represents the voltage across the inductor,as seen on an oscilloscope?

In the circuit shown, when the current $i$ is $3 \text{ A}$ and increasing at the rate of $1 \text{ A/s}$, the potential difference between $A$ and $B$ is $12 \text{ V}$. When the same current $3 \text{ A}$ is decreasing at the rate of $1 \text{ A/s}$, the potential difference between $A$ and $B$ is $6 \text{ V}$. The value of $R$ in the circuit is: (in $\text{ } \Omega$)

In the given circuit,the switch is in position $a$. For what value of $R$ will the circuit have a time constant of $10\,\mu s$ in $K\Omega$?