In the following figure,what is the final value of current in the $10 \, \Omega$ resistor when the plug of key $K$ is inserted?

An inductor $(L = 100 \, mH)$,a resistor $(R = 100 \, \Omega)$ and a battery $(E = 100 \, V)$ are initially connected in series as shown in the figure. After a long time,the battery is disconnected by short-circuiting the points $A$ and $B$. The current in the circuit $1 \, ms$ after the short circuit is

$A$ series $L-R$ circuit is connected to a battery of emf $V$. If the circuit is switched on at $t = 0$,then the time at which the energy stored in the inductor reaches $\left(\frac{1}{n}\right)$ times of its maximum value,is

$A$ bulb of $100\, W$ is connected in parallel with an ideal inductance of $1\, H$. This arrangement is connected to a $90\, V$ battery through a switch. On pressing the switch,the

Given $L_1 = 1\, mH, R_1 = 1\,\Omega, L_2 = 2\,mH, R_2 = 2\,\Omega$. Neglecting mutual inductance,find the time constant (in $ms$) for the circuit shown.

Regarding the given circuit,the correct statement is: