$12$ cells, each of emf $1 \mathrm{V}$ and internal resistances of each is $1\mathrm{\Omega }$, are connected in series such that $2$ cells are wrongly connected. The current sent by this battery in a resistance of $4\mathrm{\Omega }$

The current through $2\mathrm{\Omega }$ will be

The diagram shows a modified meter bridge, which is used for measuring two unknown resistances at the same time, when the first Galvanometer is used for obtaining the balanced point, it is found at point $C.$ Now the first galvanometer is removed and the $2\mathrm{nd}$ galvanometer is used which gives balance point at $D.$ Use the details given in diagram, $AB=L,AC=\frac{L}{4},AD=\frac{2L}{3}.$ The value of resistances $R_1$ and $R_2$ are respectively

In the circuit shown, switches $S_1$ and $S_2$ have been closed for a long time

After the switch $S$ shown in Figure A is closed, there is current $i$ through resistance $R.$ Figure B indicates current variation curves $a, b, c$ and $d$ for four sets of values of $R$ and capacitance $C:\left(1\right)R_0$ and $C_0,\left(2\right)$ $2R_0$ and $C_0$,$\left(3\right)$ $R_0$ and $2C_0$, $\left(4\right)2R_0$ and $2C_0$. Which set goes with which curve?

In the circuit shown, the charge on the $3\mu \mathrm{F}$ capacitor at steady state will be

Under steady state condition

In the circuit shown, the switch is shifted from position $1\to 2$ at $t=0.$ The switch was initially in position $1$ since long time. The graph between charge on capacitor $C$ and time$\text{'}f\text{'}$ is

In the circuit shown, $C_1=1\mu \mathrm{F},C_2=3\mu \mathrm{F}$ in steady state, the energy stored in these capacitors are respectively