What amount of heat will be released in 30 minutes by a 20 Ohm wire spiral with a current of 5 A?
| June 16, 2021 | education
t = 30 min = 1800 s.
R = 20 ohms.
I = 5 A.
Q -?
The amount of heat Q that is released in a conductor with a current is determined by the Joule-Lenz law: Q = I * U * t, where I is the current in the conductor, U is the voltage at the ends of the conductor, t is the time the current passes through the conductor.
We express the voltage U from Ohm’s law for a section of the circuit: U = I * R, where R is the resistance of the conductor.
Q = I ^ 2 * R * t.
Q = (5 A) ^ 2 * 20 Ohm * 1800 s = 900000 J.
Answer: in a conductor with a current, heat energy Q = 900000 J.
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