A potentiometer arrangement is shown in fig. The driver cell has emf $$e$$ and internal resistance $$r.$$ The resistance of potentiometer wire $$AB$$ is $$R.$$ $$F$$ is the cell of $$em\,\frac{e}{3}$$ and internal resistance $$\frac{r}{3}.$$ Balance point $$\left( J \right)$$ can be obtained for all finite values of
A.
$$R > \frac{r}{2}$$
B.
$$R < \frac{r}{2}$$
C.
$$R > \frac{r}{3}$$
D.
$$R < \frac{r}{3}$$
Answer :
$$R > \frac{r}{2}$$
Solution :
Current in $$AB = I = \frac{e}{{R + r}}$$
potential difference across $$AB = IR = \frac{{eR}}{{R + r}}$$
To obtain balanced point,
$$\frac{{eR}}{{R + r}} > \frac{e}{3}\,\,{\text{or}}\,\,R > \frac{r}{2}$$
Releted MCQ Question on Electrostatics and Magnetism >> Electric Current
Releted Question 1
The temperature coefficient of resistance of a wire is 0.00125 per $$^ \circ C$$ At $$300\,K,$$ its resistance is $$1\,ohm.$$ This resistance of the wire will be $$2\,ohm$$ at.
The electrostatic field due to a point charge depends on the distance $$r$$ as $$\frac{1}{{{r^2}}}.$$ Indicate which of the following quantities shows same dependence on $$r.$$
A.
Intensity of light from a point source.
B.
Electrostatic potential due to a point charge.
C.
Electrostatic potential at a distance r from the centre of a charged metallic sphere. Given $$r$$ < radius of the sphere.