A current of $2\mathrm{mA}$ was passed through an unknown resistor which dissipated a power of $4.4\mathrm{ }\mathrm{W}.$ Dissipated power when an ideal power supply of $11\mathrm{V}$ is connected across it is:

The Wheatstone bridge shown in fig. here gets balanced when the carbon resistor used as ${\mathrm{R}}_1$ has the colour code (orange, Red, Brown). The resistors ${\mathrm{R}}_2$ and ${\mathrm{R}}_4$ are $80 \mathrm{\Omega }$ and $40 \mathrm{\Omega },$ respectively. Assuming that the colour code for the carbon resistors given their accurate values, the colour code for the carbon resistor, used as ${\mathrm{R}}_3,$ would be :

The actual value of resistance $R,$ shown in the figure is $30 \mathrm{\Omega }.$ This is measured in an experiment as shown using the standard formula $R=\frac{V}{I},$ where $V$ and $I$ are the readings of the voltmeter and ammeter, respectively. If the measured value of $R$ is $5\%$ less, then the internal resistance of the voltmeter is

In a Wheatstone bridge (see figure), resistances $P$ and $Q$ are approximately equal. When $R=400 \mathrm{\Omega }$, the bridge is equal. When $R=400 \mathrm{\Omega }$, the bridge is balanced. On interchanging $P$ and $Q$, the value of $R$ for balance is $405 \mathrm{\Omega }$. The value of $X$ is close to,

The resistance of the meter bridge $AB$ in given figure is $4 \mathrm{\Omega }.$ With a cell of emf $\epsilon =0.5 \mathrm{V}$ and rheostat resistance $R_h=2 \mathrm{\Omega }$ the null point is obtained at some point $J.$ When the cell is replaced by another one of emf $\epsilon ={\epsilon }_2$ the same null point $J$ is found for $R_h=6 \mathrm{\Omega }.$ The emf ${\epsilon }_2$ is

In the experimental set--up of metre bridge shown in the figure, the null point is obtained at a distance of $40 \mathrm{cm}$ from $A$. If a $10 \mathrm{\Omega }$ resistor is connected in series with $R_1,$ the null point shifts by $10 \mathrm{cm}.$ The resistance that should be connected in parallel with $\left(R_1+10\right) \mathrm{\Omega }$ such that the null point shifts back to its initial position is

In the circuit shown, the potential difference between $\mathrm{A}$ and $\mathrm{B}$ is :