In the circuit shown in figure, the cell is ideal with emf $9 \mathrm{V}$, If the resistance of the coil of galvanometer is $1 \mathrm{\Omega },$ then

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.

In the circuit shown, battery, ammeter and voltmeter are ideal and the switch $S$ is initially closed as shown. When switch $S$ is opened,

All the resistance in the given Wheatstone bridge have different values and the current through the galvanometer is zero. The current through the galvanometer will still be zero if,