The current flowing in the given circuit is $0.1\,A$. The potential difference between the points $X$ and $Y$ is ................ $V$.

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 circuit given, $E = 6.0 \, V$, $R_1 = 100 \, \Omega$, $R_2 = R_3 = 50 \, \Omega$, and $R_4 = 75 \, \Omega$. The equivalent resistance of the circuit, in $\Omega$, is:

In the given circuit,the internal resistance of the $18\,V$ cell is negligible. If $R_1 = 400\,\Omega$,$R_3 = 100\,\Omega$,and $R_4 = 500\,\Omega$,and the reading of an ideal voltmeter across $R_4$ is $5\,V$,then the value of $R_2$ will be ........... $\Omega$.

$A$ current of $2 \, A$ flows in an electric circuit as shown in the figure. The potential difference $(V_R - V_S)$, in volts ($V_R$ and $V_S$ are potentials at $R$ and $S$ respectively) is:

An electric bulb of $60 \, W, 120 \, V$ is to be connected to a $220 \, V$ source. What resistance should be connected in series with the bulb so that the bulb glows properly?