A current of 2.5 A flows along a wire spiral, the resistance of which
| September 16, 2021 | education
A current of 2.5 A flows along a wire spiral, the resistance of which in a heated state is 48 Ohms. How much heat will this spiral release in 2.5 minutes?
The given tasks: R (resistance of the taken wire spiral in a heated state) = 48 Ohm; I (coil current) = 2.5 A; t (operating time) = 2.5 minutes (in SI t = 150 s).
The heat that the taken wire spiral will release in 2.5 minutes of operation is determined by the formula: Q = I ^ 2 * R * t.
Calculation: Q = 2.52 * 48 * 150 = 45 * 103 J (45 kJ).
Answer: B) the wire spiral taken will release 45 kJ of heat.
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