JEE MAIN - Physics (2022 - 27th July Morning Shift - No. 12)
A direct current of $$4 \mathrm{~A}$$ and an alternating current of peak value $$4 \mathrm{~A}$$ flow through resistance of $$3\, \Omega$$ and $$2\,\Omega$$ respectively. The ratio of heat produced in the two resistances in same interval of time will be :
3 : 2
3 : 1
3 : 4
4 : 3
Explanation
Ratio = $${{i_1^2{R_1}} \over {{{\left( {{{{i_2}} \over {\sqrt 2 }}} \right)}^2}{R_2}}} = {{{4^2} \times 3} \over {{{\left( {{4 \over {\sqrt 2 }}} \right)}^2} \times 2}}$$
$$\Rightarrow$$ Ratio = 3 : 1
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