An ideal inductor of inductance 40 mH is connected in series with a resistor of 5 across a 10 V DC battery. What is the magnetic energy stored in the inductor when the circuit reaches steady state?

Step-by-step solution

In a DC circuit at steady state, an ideal inductor acts as a short circuit (zero resistance). The steady current i in the circuit is determined entirely by the battery voltage and the resistance. Using Ohm's law: i = V R = 10 5 = 2 A The magnetic energy U stored in the inductor is: U = 1 2 L i^2 Substitute L = 40 10⁻³ H and i = 2 A : U = 1 2 (40 10⁻³) (2)^2 U = 20 10⁻³ 4 U = 80 10⁻³ J = 80 mJ Assuming no energy is stored because the current is constant (and thus induced emf is zero) incorrectly yields 0 mJ . Forget