A cell has an emf $$1.5\,V.$$ When connected across an external resistance of $$2\,\Omega ,$$ the terminal potential difference falls to $$1.0\,V.$$ The internal resistance of the cell is
A.
$$2\,\Omega $$
B.
$$1.5\,\Omega $$
C.
$$1.0\,\Omega $$
D.
$$0.5\,\Omega $$
Answer :
$$1.0\,\Omega $$
Solution :
Internal resistance of the cell is given by
$$r = \left( {\frac{{E - V}}{V}} \right)R$$
Given, $$E = 1.5\,V,V = 1.0\,V,R = 2\,\Omega $$
$$\therefore r = \left( {\frac{{1.5 - 1.0}}{{1.0}}} \right) \times 2 = \frac{{0.5}}{{1.0}} \times 2 = 1\,\Omega $$
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.