JEE MAIN - Physics (2021 - 16th March Evening Shift - No. 12)
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.
1000 J
2000 J
1500 J
500 J
Explanation
$${H_1} = i_1^2R\Delta t$$
$${H_2} = i_2^2R\Delta t$$
$$ \Rightarrow {{{H_1}} \over {{H_2}}} = {{i_1^2} \over {i_2^2}}$$
$$ \Rightarrow {{500} \over {{H_2}}} = {\left( {{1 \over 2}} \right)^2}$$
$$ \Rightarrow {H_2} = 2000J$$
$${H_2} = i_2^2R\Delta t$$
$$ \Rightarrow {{{H_1}} \over {{H_2}}} = {{i_1^2} \over {i_2^2}}$$
$$ \Rightarrow {{500} \over {{H_2}}} = {\left( {{1 \over 2}} \right)^2}$$
$$ \Rightarrow {H_2} = 2000J$$
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