A coil of wire with a resistance of 0.45 $\varOmega$ has a self-inductance of 0.083 H. If a 6.0-V battery is connected across the ends of the coil and the current in the circuit reaches an equilibrium value, what is the stored energy in the inductor?
A)7.4 J
B)4.6 J
C)1.6 J
D)5.1 J
E)3.4 J

Question

Quizplus · Accepted Answer

The stored energy in an inductor is given by the equation U = 1/2 * L * I^2, where U is the stored energy, L is the inductance, and I is the current. In this problem, we are given L = 0.083 H and I = 6.0 V / 0.45 $\varOmega$ = 13.3 A. Plugging these values into the equation, we get U = 1/2 * 0.083 H * (13.3 A)^2 = 7.4 J.

A Coil of Wire with a Resistance of 0 Ω\varOmegaΩ Has a Self-Inductance of 0

A Coil of Wire with a Resistance of 0 $\varOmega$ Has a Self-Inductance of 0