Two ideal batteries and two ammeters are arranged as shown in the figure.

In the above figure, if the polarity of $E_2$ is reversed, then

In the figure, the voltmeter is not ideal. If the voltmeter is removed from $R_1$, and then put across $R_2$, what will be the effect on current $I$? Given $R_1>R_2$

Three ammeters $A, B$ and $C$ of resistances $R_A, R_B$ and $R_C,$ respectively, are joined as shown in the figure.

When some potential difference is applied across the terminals $T_1$ and $T_2,$ their readings are $I_A, I_B$ and $I_C,$ respectively, then

A battery of emf ${\mathrm{\epsilon }}_0=5 \mathrm{V}$ and internal resistance $5 \mathrm{\Omega }$ is connected across a long uniform wire $AB$ of length $1 \mathrm{m}$ and resistance per unit length $5 \mathrm{\Omega } {\mathrm{m}}^{-1}$. Two cells of ${\mathrm{\epsilon }}_1=1 \mathrm{V}$ and ${\mathrm{\epsilon }}_2=2 \mathrm{V}$are connected as shown in the figure.

A circuit is shown below:

A circuit contains an ideal battery, three resistors and two ideal ammeters. The ammeters read $0.2 \mathrm{A}$ and $0.3 \mathrm{A}$. After two of the resistors are switched, the readings of the ammeters did not change. Find the battery current.