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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 I V IR L t L must have units of Ohmsseconds From the Construction Inductance N A L 2 N = number of turns on the coil = permeability of the core (henrys/m) A = cross sectional area (m2) l = length of core (m) L = inductance in henrys Relative Permeability Many texts and handbooks publish Km, where = Km o o = permeability of free space = 4p X 10-7 Wb/A Ex: Compute L for the following coil: N = 100 turns l = 25 X 10-3 m A = 1.3 X 10-4 m2 Km = 400 (steel) N 2 K m o A L (100)2 ( 400 )( 4pX 10 7 )( 1.3 X 10 4 ) 25 X 10 3 0.0261H 26.1mH Time Dependence E t / I (1 e ) R , = 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=IR) At first I = 0, so V = IR = 0 As current rises the voltage drop across the resistor (IR) gets greater, leaving less voltage to be dropped through the coil. Voltage V Ee t / Inductors in Series Kirchhoff’s Voltage Law E VL1 VL2 VL3 L1 LT = L1 + L2 + L3 I I I L2 L3 t t t I ( L1 L2 L3 ) t I LT t Inductors in Parallel E VL1 VL2 VL3 The analysis is difficult in a dc circuit since the voltage drains to zero, but the result is… 1 1 1 1 LT L1 L2 L3 Real Inductors Inductors have… Internal Resistance Internal Capacitance between windings So a real inductor in a circuit looks like… Example The equivalent circuit is 1 1 1 RT 4 k 8 k RT 8 k 3 Continued LT 20 X 10 3 H 7.5 s 3 RT 2.66 X 10 E I ( 1 e t / ) 3.76 mA( 1 e t / 7.5 s ) RT 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 C1 and L1 acts as if it’s not there Solve Circuit R1 and R2 are in series, so… E 10 V I 2A R1 R2 5 For the path ABCD IR1 + IR2 = E VR1 VR2 E Notice that R2 and C1 are in parallel, so VR2 is the voltage drop across the capacitor also. VR2 E VR1 10 V IR1 10 V - (2A)(2 ) 6V Stored Energy Capacitor WC = ½CV2 Inductor WL = ½LI2