The current in the coil was 3A. When the coil was short-circuited, 0.9J of heat was released in it.

The current in the coil was 3A. When the coil was short-circuited, 0.9J of heat was released in it. What is the inductance of the coil.

When the coil was short-circuited, then the amount of heat Q = 0.9 J. was released in it. Since the energy of the magnetic field of the coil is calculated by the formula Wк = (L • i ^ 2): 2, and it is released in the form of a known amount of heat. then from here we can find L = 2Wk: (i ^ 2). Since the current in the coil was 3 A, then in order to determine the inductance of the coil, we substitute the values of physical quantities from the condition of the problem and make calculations: L = 2 • 0.9 J: ((3A) ^ 2); L = 0.2 H.
Answer: the inductance of this coil is L = 0.2 H.