Two wires of the same metal have same length, but their cross-sections are in the ratio $$3:1.$$ They are joined in series. The resistance of thicker wire is $$10\,\Omega .$$ The total resistance of the combination will be
A.
$$10\,\Omega $$
B.
$$20\,\Omega $$
C.
$$40\,\Omega $$
D.
$$100\,\Omega $$
Answer :
$$40\,\Omega $$
Solution :
For the same length and same material,
$$\eqalign{
& \frac{{{R_2}}}{{{R_1}}} = \frac{{{A_1}}}{{{A_2}}} = \frac{3}{1} \cr
& {\text{or}}\,\,{R_2} = 3{R_1} \cr} $$
The resistance of thick wire,
$${R_1} = 10\,\Omega $$
The resistance of thin wire $$ = 3{R_1} = 3 \times 10 = 30\,\Omega $$
Total resistance $$ = 10 + 30 = 40\,\Omega $$
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.