The electric motor of the tram for 30 minutes at a voltage of 500V consumes 118,800,000 joules of energy
| September 8, 2021 | education
The electric motor of the tram for 30 minutes at a voltage of 500V consumes 118,800,000 joules of energy to determine the current flowing through the winding of the electric motor.
Initial data: t (operating time of the tram electric motor) = 30 min (1800 s); U (motor voltage) = 500 V; Q (energy consumed by the tram electric motor during operation) = 118,800,000 J.
We express the strength of the current flowing through the winding of the electric motor of the tram from the formula (Joule-Lenz law): Q = U * I * t, whence I = Q / (U * t).
Calculation: I = 118800000 / (500 * 1800) = 132 A.
Answer: A current of 132 A flows through the tram motor winding.
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