Two resistors of $200\Omega$ and $400\Omega$ are connected in series with a battery of $100\text{ V}$. $A$ bulb rated at $200\text{ V}, 100\text{ W}$ is connected across the $400\Omega$ resistance. The potential drop across the bulb is . . . . . . $\text{V}$.

$R_1$ and $R_2$ are two resistors made of different materials. The temperature coefficient of resistance for $R_1$ is $\alpha$ and for $R_2$ is $-\beta$. If the resistance of the series combination of $R_1$ and $R_2$ does not change with temperature,then the ratio of the resistances of the two wires is:

In the circuit shown in the figure,the currents in different branches and the value of one resistor are given. Then,the potential at point $B$ with respect to point $A$ is $.......V$.

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

For the circuit shown in the figure,calculate the resistance between the points $A$ and $B$.

In the circuits shown below, the readings of the voltmeters and the ammeters will be: