Calculate the equivalent resistance between $A$ and $B$.

In the circuit shown in the figure,if no current flows through the galvanometer when the key $K$ is closed,the bridge is balanced. The balancing condition for the bridge is

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

The Wheatstone bridge shown in the diagram is balanced. If $P_3$ is the power dissipated by $R_3$ and $P_1$ is the power dissipated by $R_1$,then the ratio $\frac{P_3}{P_1}$ is

Kirchhoff's first law $(\sum i = 0)$ and second law $(\sum iR = \sum E)$,where the symbols have their usual meanings,are respectively based on:

Find the current flowing through the resistance $R_1$ of the circuit shown in the figure,if the resistances are equal to $R_1=10 \Omega, R_2=20 \Omega$,and $R_3=30 \Omega$,and the potentials of points $1, 2$,and $3$ are equal to $\phi_1=10 \text{ V}, \phi_2=6 \text{ V}$,and $\phi_3=5 \text{ V}$. (in $\text{ A}$)