The current inside a copper voltameter:

As shown in the figure,a network of resistors is connected to a battery of $24\,V$ with an internal resistance of $3\,\Omega$. The currents through the resistors $R_4$ and $R_5$ are $I_4$ and $I_5$ respectively. The values of $I_4$ and $I_5$ are:

What current will flow through the $2 k\Omega$ resistor in the circuit shown in the figure (in $mA$)?

In the ladder network shown, the current through the resistor $3\, \Omega$ is $0.25\, A$. The input voltage $V$ is equal to: (in $V$)

The galvanometer deflection,when key $K_1$ is closed but $K_2$ is open,equals $\theta_0$ (see figure). On closing $K_2$ also and adjusting $R_2$ to $5\,\Omega$,the deflection in the galvanometer becomes $\frac{\theta_0}{5}$. The resistance of the galvanometer is,then,given by [Neglect the internal resistance of battery]: .................. $\Omega$

In the following circuit,if $V_A - V_B = 6\,V$,then the value of resistance $R$ (in $\Omega$) is ............... $\Omega$.