A $$220\,volt,1000\,watt$$ bulb is connected across a $$110\,volt$$ mains supply . The power consumed will be
A.
$$750\,watt$$
B.
$$500\,watt$$
C.
$$250\,watt$$
D.
$$1000\,watt$$
Answer :
$$250\,watt$$
Solution :
We know that $$R = \frac{{V_{rated}^2}}{{{P_{rated}}}} = \frac{{{{\left( {220} \right)}^2}}}{{1000}}$$
When this bulb is connected to $$110\,volt$$ mains supply we get
$$P = \frac{{{V^2}}}{R} = \frac{{{{\left( {110} \right)}^2} \times 1000}}{{{{\left( {220} \right)}^2}}} = \frac{{1000}}{4} = 250W$$
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.