What work does the current in the electric motor do in 30 s, if the voltage of 220V and the current in the motor is 0.1A?

The work that the current must perform in the taken electric motor can be calculated using the Joule-Lenz law: Q = U * I * t, where U is the operating voltage (U = 220 V); I is the current in the taken motor (I = 0.1 A); t is the duration of the engine operation (t = 30 s).

Let’s perform the calculation: Q = U * I * t = 220 * 0.1 * 30 = 660 J.

Answer: For 30 seconds of operation of the taken electric motor, the current will perform work equal to 660 J.