Inductors

Energy Storage Current passing through a coil causes a magnetic field  Energy is stored in the field  Similar to the energy stored by capacitors We saw a charging time for a capacitor An inductor takes time to store energy also

Simple RL Circuit L must have units of Ohms  seconds

From the Construction Inductance N = number of turns on the coil  = permeability of the core (henrys/m) A = cross sectional area (m 2 ) l = length of core (m) L = inductance in henrys

Relative Permeability Many texts and handbooks publish K m, where  = K m  o  o = permeability of free space = 4  X Wb/A Ex: Compute L for the following coil:  N = 100 turns A = 1.3 X m 2  l = 25 X m K m = 400 (steel)

Time Dependence  = L/R This is the same way that voltage varied in the capacitor Try it!

Notes The final current (E/R) doesn’t depend on L  There is no voltage drop across the inductor after the full current has been established  The coil then acts as a short circuit (as if it weren’t there) The inductance depends on the change of current (once I is established,  I /  t → 0 and V= I R) At first I = 0, so V = I R = 0  As current rises the voltage drop across the resistor ( I R) gets greater, leaving less voltage to be dropped through the coil.

Voltage

Inductors in Series Kirchhoff’s Voltage Law L T = L 1 + L 2 + L 3

Inductors in Parallel The analysis is difficult in a dc circuit since the voltage drains to zero, but the result is…

Real Inductors Inductors have…  Internal Resistance  Internal Capacitance between windings So a real inductor in a circuit looks like…

Example The equivalent circuit is

Continued Comparing inductors to capacitors After about 5 , the current has reached a maximum for the coil and zero for a capacitor. The coil acts as a short, while the capacitor acts like an open circuit.

Sample RLC Circuit After about 5 , the equivalent circuit is No current flows through C 1 and L 1 acts as if it’s not there

Solve Circuit R 1 and R 2 are in series, so… For the path ABCD IR 1 + IR 2 = E Notice that R 2 and C 1 are in parallel, so V R2 is the voltage drop across the capacitor also.

Stored Energy Capacitor  W C = ½CV 2 Inductor  W L = ½LI 2