When the resistance $R$ (indicated in the figure below) is changed from $1 \, k\Omega$ to $10 \, k\Omega$,the current flowing through the resistance $R'$ does not change. What is the value of the resistor $R'$?

Consider the following circuit shown below. All the resistors are identical. The ratio of $I / I^{\prime}$ is

The Wheatstone bridge principle is used to measure the specific resistance $(S_1)$ of a given wire,having length $L$ and radius $r$. If $X$ is the resistance of the wire,then the specific resistance is: $S_1 = X \left( \frac{\pi r^2}{L} \right)$. If the length of the wire is doubled,then the value of the specific resistance will be:

The value of current $i_{1}$ flowing from $A$ to $C$ in the circuit diagram is $.......A$

$A$ Wheatstone bridge is initially at room temperature and all arms of the bridge have the same value of resistance $(R_1=R_2=R_3=R_4=R)$. When $R_3$ is heated to some temperature, its resistance value increases by $10 \%$. The potential difference $(V_a - V_b)$ (after $R_3$ is heated) is . . . . . . $V$. Given the total voltage is $40 \text{ V}$.

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