A wire spiral, the resistance of which in a heated state is 55 Ohm, is connected to a network with a voltage of 127 V
| September 3, 2021 | education
A wire spiral, the resistance of which in a heated state is 55 Ohm, is connected to a network with a voltage of 127 V. How much heat does this spiral generate in 1 minute?
To determine the heat released by a given wire spiral, we apply the formula: Q = U ^ 2 * t / R.
Values of variables: U – mains voltage (U = 127 V); t is the time of connecting the wire spiral to the network (t = 1 min; in SI t = 60 s); R is the resistance of the coil (R = 55 Ohm).
Let’s make a calculation: Q = U ^ 2 * t / R = 1272 * 60/55 ≈ 17.6 * 10 ^ 3 J (17.6 kJ).
Answer: A given spiral wire will release 17.6 kJ of heat in 1 minute.
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