JEE MAIN - Physics (2003 - No. 13)
A $$220$$ volt, $$1000$$ watt bulb is connected across a $$110$$ $$volt$$ mains supply. The power consumed will be
$$750$$ watt
$$500$$ watt
$$250$$ watt
$$1000$$ watt
Explanation
We know that $$R = {{V_{rated}^2} \over {{P_{rated}}}} = {{{{\left( {220} \right)}^2}} \over {1000}}$$
When this bulb is connected to $$110$$ volt mains supply we get
$$P = {{{V^2}} \over R} = {{{{\left( {110} \right)}^2} \times 1000} \over {{{\left( {220} \right)}^2}}} = {{1000} \over 4} = 250W$$
When this bulb is connected to $$110$$ volt mains supply we get
$$P = {{{V^2}} \over R} = {{{{\left( {110} \right)}^2} \times 1000} \over {{{\left( {220} \right)}^2}}} = {{1000} \over 4} = 250W$$
Comments (0)
