In the figure magnetic energy stored in the coil is:

A solenoid of inductance L and resistance R is connected to a battery of e.m.f. E. Maximum value of magnetic energy stored in the inductor is:

$1.$ $\frac{E^2}{2R}$
$2.$ $\frac{E^{2}L}{2R^2}$
$3.$ $\frac{E^{2}L}{R}$ $$
$4.$ $\frac{E^{2}L}{2R}$

The figure shows three circuits with identical batteries, inductors, and resistors. Rank the circuits according to the current, in descending order, through the battery (i) just after the switch is closed and (ii) a long time later:

The key K is inserted at time t= 0. The initial (t=0) and final $\left(t\to \infty \right)$ currents through the battery are:
                               

1. \(\frac{1}{15}~A,~\frac{1}{10}~A\)
2. \(\frac{1}{10}~A,~\frac{1}{15}~A\)
3. \(\frac{2}{15}~A,~\frac{1}{10}~A\)
4. \(\frac{1}{15}~A,~\frac{2}{25}~A\)

In the circuit diagram shown in figure, R = 10 \(\Omega\), L = 5 H, E = 20 V and i = 2 A. This current is decreasing at a rate of 1.0 A/s. $V_{ab}$ at this instant will be:

The resistance in the following circuit is increased at a particular instant. At this instant the value of resistance is 10 Ω. The current in the circuit will be:

1. i = 0.5 A

2. i > 0.5 A

3. i < 0.5 A

4. i = 0