The current $I$ in the circuit shown is:

Current measured by the ammeter $(A)$ in the given circuit when no current flows through the $10\,\Omega$ resistance is . . . . . . $A$.

Two batteries,one with an $emf$ of $3\,V$ and internal resistance $1\,\Omega$,and the other with an $emf$ of $15\,V$ and internal resistance $2\,\Omega$,are connected in series with a resistance $R$ as shown in the figure. If the potential difference between points $a$ and $b$ is zero,find the resistance $R$ in $\Omega$.

Find the potential difference $(V_A - V_B)$ for the given circuit.

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

In the following electrical network,the value of $I$ is (in $A$)