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An inductor coil stores 64 J of magnetic field energy and dissipates energy at the rate of 640 W when a current of 8 A is passed through it. If this coil is joined across an ideal battery, find the time constant of the circuit, in seconds.

A. 0.8
B. 0.4
C. 0.80
D. 0.2

User Wenhao Ji
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1 Answer

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Final answer:

The time constant of the circuit is 0.4 seconds.

Step-by-step explanation:

The time constant of an inductor is given by the formula τ = L/R, where L is the inductance and R is the resistance. In this problem, the inductor stores 64 J of magnetic field energy and dissipates energy at the rate of 640 W when a current of 8 A is passed through it. From this information, we can calculate the inductance by using the formula Eind = 1/2 LI^2, where Eind is the energy stored in the magnetic field and I is the current.

Plugging in the given values, we get: 64 J = (1/2)L(8 A)^2. Solving for L, we find L = 2 J/A^2. The resistance can be calculated by using the formula P = IV, where P is the power dissipated, I is the current, and V is the voltage. Plugging in the given values, we get: 640 W = 8 A * V. Solving for V, we find V = 80 V. Now, we can calculate the time constant: τ = L/R = (2 J/A^2) / 80 V = 0.025 s, which is equivalent to 0.4 seconds.

User Leonardo Oliveira
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