A voltmeter reads the potential difference across the terminals of an old battery as $1.40 \mathrm{V}$ while a potentiometer reads its voltage as $1.55 \mathrm{V}$. The voltmeter resistance is $280 \mathrm{\Omega }$.

The figure shows a balanced Wheatstone bridge.

Two voltmeters and two resistances are connected as shown in the figure. On closing the switch $S$, what will be the effect on the readings of the voltmeters?

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.