Two cities are $$150\,km$$ apart. Electric power is sent from one city to another city through copper wires. The fall of potential per $$km$$ is $$8\,V$$ and the average resistance per $$km$$ is $$0.5\,\Omega .$$ The power loss in the wire is
A.
$$19.2\,W$$
B.
$$19.2\,kW$$
C.
$$19.2\,J$$
D.
$$12.2\,kW$$
Answer :
$$19.2\,kW$$
Solution :
Potential drop between two cities is $$ = 150 \times 8 = 1200\,V$$
Average resistance of total wire $$ = 0.5 \times 150 = 75\,\Omega $$
So, power loss $$P = \frac{{{V^2}}}{R} = \frac{{1200 \times 1200}}{{75}} = 19200\,W$$
$$ = 19.2\,kW$$
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.