For an L-R circuit, the time constant equals:

twice the ratio of the energy stored in the magnetic field to the rate of energy dissipation in the resistance
the ratio of the energy stored in the magnetic field to the rate of energy dissipation in the resistance
half the ratio of the energy stored in the magnetic field to the rate of energy dissipation in the resistance
the square of the ratio of the energy stored in the magnetic field to the rate of energy dissipation in the resistance

Correct answer: twice the ratio of the energy stored in the magnetic field to the rate of energy dissipation in the resistance

Solution

tau = L/R. Energy stored in inductor U = (1/2)L*I²; power dissipated P = I²*R. So U/P = (1/2 L I²)/(I² R) = L/(2R) = tau/2. Therefore tau = 2*(U/P), i.e. twice the ratio of stored magnetic energy to the rate of dissipation.

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