In the circuit diagram each resistor of resistance $5 \mathrm{\Omega }$. The points $A$ and $B$ are connected to the terminals of a cell of electromotive force $9 \mathrm{V}$ and internal resistance $\frac{2}{3} \mathrm{\Omega }$.

If the reading of ammeter is $2 \mathrm{A}$ then the reading of voltmeter

Write steady state current in the $2 \Omega$ resistor shown in the circuit (see figure). The internal resistance of the battery is negligible and the capacitance of the condenser $C$ is $0.2 \mathrm{mF}$ is terms of $N\times {10}^{-1}\mathrm{amp}$.

In the circuit shown in the figure initially switch $S$ is open and capacitor is uncharged. Internal resistances of the cells are $r_1=9\mathrm{\Omega }$ and $r_2=2\mathrm{\Omega }$ and their emf's are equal to $\epsilon$. The potential difference across the cell of internal resistance $r_1$ becomes zero long time after closing the switch. Write the value of $R$ in $\mathrm{\Omega }$.

Find the product of reading of voltmeter ammeter as shown in figure.

In the circuit shown what is the sum of potential at $B,C$ and $D$ in volt ?

A network of nine conductors connects six points $A, B, C, D, E$ and $F$ as shown in figure. The figure denotes resistances in ohms. Find the equivalents resistance (in ohm) between $A$ and $D$.