For the circuit shown in the figure,the current $I$ will be .......$A$.

Two batteries with e.m.f $12\ V$ and $13\ V$ are connected in parallel across a load resistor of $10\,\Omega$. The internal resistances of the two batteries are $1\,\Omega$ and $2\,\Omega$ respectively. The voltage across the load lies between

In the following network, $I_1 = -0.4 \text{ A}, I_4 = 1 \text{ A}, I_5 = 0.4 \text{ A}$. The values of $I_2, I_3$ and $I_6$ are respectively

$A$ $16\, \Omega$ wire is bent to form a square loop. $A$ $9\, V$ supply having an internal resistance of $1\, \Omega$ is connected across one of its sides. The potential drop across the diagonals of the square loop is $.......\, \times 10^{-1}\, V$.

In the circuit shown in the given figure,the resistances $R_1$ and $R_2$ are respectively:

For the circuit given below,the Kirchhoff's loop rule for the loop $BCDEB$ is given by the equation: