JEE MAIN - Physics (2022 - 25th June Morning Shift - No. 24)
A resistor develops 300 J of thermal energy in 15 s, when a current of 2 A is passed through it. If the current increases to 3 A, the energy developed in 10 s is ____________ J.
Answer
450
Explanation
$$300 = {I^2}R \times 15$$
$$ \Rightarrow R = 5\,\Omega $$
Now $$I_2^2R{t_2}$$
$$ = 9 \times 5 \times 10$$
$$ = 450\,J$$
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