A potentiometer wire $\mathrm{AB}$ having length $L$ and resistance $12 r$ is joined to a cell $D$ of emf $\epsilon$ and internal resistance $r.$ A cell $C$ having emf $\frac{\epsilon }{2}$ and internal resistance $3 r$ is connected. The length $\mathrm{AJ}$ at which the galvanometer as shown in figure shows no deflection is

In the given circuit the cells have zero internal resistance. The currents (in Amperes) passing through resistance $R_1$ and $R_2$ respectively, are

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