Verified Solution • 2025–26 Session moderate

QJEEM21Pha9ab: JEE Main 2021 Physics - Class 12 JEEM21Pha9ab Solution: A resistor develops 500 J of thermal energy in 20 s when a…

A resistor develops 500 J of thermal energy in 20 s when a current of 1.5A is passed through it. If the current is increased from 1.5A to 3A, what will be the energy developed in 20 s.

Multiple Choice Options

  • A 1000 J
  • B 2000 J
  • C 1500 J
  • D 500 J

Step-by-Step Verified Solution

H1=i12RΔt{H_1} = i_1^2R\Delta t

H2=i22RΔt{H_2} = i_2^2R\Delta t

⇒H1H2=i12i22\Rightarrow {{{H_1}} \over {{H_2}}} = {{i_1^2} \over {i_2^2}}

⇒500H2=(12)2\Rightarrow {{500} \over {{H_2}}} = {\left( {{1 \over 2}} \right)^2}

⇒H2=2000J\Rightarrow {H_2} = 2000J

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