In $R-L$ circuit $R=4 \mathrm{\Omega }, L=0.5 \mathrm{H},$ emf of cell, $\epsilon =6 \mathrm{V}$. Work done in changing current from $0.80 \mathrm{A}$ to $0.81 \mathrm{A}$ is:- 

When a certain circuit, consisting of a constant emf $E$, an inductance $L$ and a resistance $R$, is closed, the current in it increases with time according to curve $1$. After one parameter $E, L or R$ is changed, the increase in current follows curve $2$, when the circuit is closed second time. Which parameter was changed and in what direction?

In the branch, $AB$ of a circuit, as shown in the figure, a current, $I=(t+2) \mathrm{A}$ is flowing, where, $t$ is the time in second. At $t=0$, the value of $(V_A–V_B)$ will be:- 

The circuit in the figure consisting of three identical lamps and two coils, is connected to a dc source. The ohmic resistance of the coils is negligible. After some time switch S is opened. Select correct alternative(s) just afterward. 

As shown in the figure, a magnet is brought towards a fixed coil. Due to this the induced emf, current, and the charge are $E$, $I$ and $Q$ respectively. If the speed of the magnet is double then the following statement is wrong:-