Resistances are joined as shown in the figure. In the balanced condition, the current $I$ drawn from the battery is (in $\text{ A}$)

As the switch $S$ is closed in the circuit shown in the figure,the current passed through it is .................. $A$.

If the Wheatstone bridge with four resistors $R_1, R_2, R_3$ and $R_4$ is balanced,then the correct expression is

The current through the $1 \Omega$ resistance in the following circuit is (in $A$)

In the given part of a circuit,the potential at point $B$ is zero. Then the potentials at $A$ and $C$ respectively are

The four arms of a Wheatstone bridge (Figure) have the following resistances: $AB = 100 \; \Omega$,$BC = 10 \; \Omega$,$CD = 5 \; \Omega$,and $DA = 60 \; \Omega$. $A$ galvanometer of $15 \; \Omega$ resistance is connected across $BD$. Calculate the current through the galvanometer when a potential difference of $10 \; V$ is maintained across $AC$.