High School

An inductor with an inductance of [tex]4.00 \, \text{H}[/tex] and a resistance of [tex]8.00 \, \Omega[/tex] is connected to the terminals of a battery with an emf of [tex]6.00 \, \text{V}[/tex] and negligible internal resistance. When the current has reached its final steady-state value, how much energy is stored in the inductor?

Answer :

The energy stored in the inductor is 1.125 Joules at a steady state.

When an inductor reaches its steady-state current, the energy stored in the inductor can be calculated using the formula:

  • E = (1/2) × L × I², where L is the inductance and I is the current.

The steady-state current (I) can be calculated as the emf (V) divided by the resistance (R):

  • I = V / R = 6.00 V / 8.00 Ω = 0.75 A

Now, substituting the values into the energy formula:

  • E = (1/2) × 4.00 H × (0.75 A)² = (1/2) × 4.00 H × 0.5625 A² = 1.125 J

Therefore, the energy stored in the inductor is 1.125 Joules.

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