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Published byStephen Mills Modified over 9 years ago
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Inductors
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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
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Simple RL Circuit L must have units of Ohms seconds
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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
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Relative Permeability Many texts and handbooks publish K m, where = K m o o = permeability of free space = 4 X 10 -7 Wb/A Ex: Compute L for the following coil: N = 100 turns A = 1.3 X 10 -4 m 2 l = 25 X 10 -3 m K m = 400 (steel)
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Time Dependence = L/R This is the same way that voltage varied in the capacitor Try it!
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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.
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Voltage
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Inductors in Series Kirchhoff’s Voltage Law L T = L 1 + L 2 + L 3
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Inductors in Parallel The analysis is difficult in a dc circuit since the voltage drains to zero, but the result is…
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Real Inductors Inductors have… Internal Resistance Internal Capacitance between windings So a real inductor in a circuit looks like…
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Example The equivalent circuit is
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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.
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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
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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.
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Stored Energy Capacitor W C = ½CV 2 Inductor W L = ½LI 2
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