JEE MAIN - Physics (2021 - 31st August Evening Shift - No. 29)
A resistor dissipates 192 J of energy in 1s when a current of 4A is passed through it. Now, when the current is doubled, the amount of thermal energy dissipated in 5s in _________ J.
Answer
3840
Explanation
E = i2Rt
192 = 16 (R) (1)
R = 12$$\Omega$$
E1 = (8)2 (12) (5)
= 3840 J
192 = 16 (R) (1)
R = 12$$\Omega$$
E1 = (8)2 (12) (5)
= 3840 J
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