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

Find the current in the sliding rod $AB$ (resistance $= R$) for the arrangement shown in the figure. $\vec{B}$ is constant and is directed out of the paper. The parallel wires have no resistance. $v$ is constant. The switch $S$ is closed at time $t = 0$.

$A$ coil of inductance $300\, mH$ and resistance $2\, \Omega$ is connected to a source of voltage $2\, V$. The current reaches half of its steady state value in $t$ seconds. Find the value of $t$. (in $, s$)

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

In an $L-R$ decay circuit,the initial current at $t = 0$ is $I$. The total charge that has flown through the resistor until the energy in the inductor has reduced to one-fourth of its initial value is:

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 $\text{ } \Omega$)