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QJEEM22Ph4848: JEE Main 2022 Physics - Class 12 JEEM22Ph4848 Solution: A resistor develops 300 J of thermal energy in 15 s, when a…

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.

Step-by-Step Verified Solution

300=I2R×15

⇒R=5Ω

Now I22Rt2

=9×5×10

=450J

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