12. ENG 282 ch6 X

# 12. ENG 282 ch6 X - GeorgeHaines,Instructor Chapter6.16.5...

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BASIC ELECTRIC CIRCUITS BASIC ELECTRIC CIRCUITS George Haines, Instructor George Haines, Instructor Pima Community College Chapter 6.1 - 6.5

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Today Today 6.1 The Inductor 6.2 The Capacitor 6.3 Series-Parallel Combinations of Inductance  and Capacitance 6.4 Mutual Inductance 6.5 A Closer Look at Mutual Inductance
Chapter 6 Chapter 6 Inductance, Capacitance, and Mutual Inductance Inductance, Capacitance, and Mutual Inductance 6.1 The Inductor v = L di/dt i.e. the voltage is proportional to the rate of change in current. L is measured in henrys (H), and is represented as a coiled wire, a conductor linking a magnetic field. v(t) t t o t o i(t) t t o arcing Electric power linemen can be killed. Opening a current carrying line can be dangerous.

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Example 6.1 a) Sketch the current waveform for Figure 6.2 b) At what instant of time is the current maximum? di/dt = 10(-5e -5t + e -5t ) = 0 when t = 1/5 sec. c) Express v(t) across the terminals of the 100 mH inductor. V(t) = L di/dt = (0.1)e -5t (1-5t) = e -5t (1 – 5t) V d) Sketch the voltage waveform. Figure 6.4 e) Are the voltage and current at maximum at the same time? No. v is proportional to di(t)/dt not i(t). f) At what instant of time does the voltage change polarity? At 0.2 sec. when di/dt passes through 0. g) Is there ever an instantaneous change in voltage across the inductor? Yes, at t = 0. The voltage can change instantaneously across an inductor
Current in an Inductor in Terms of the Voltage Across the Inductor v(t) = L di/dt therefore v(t) dt = L(di/dt) dt v dt = L di L ∫ i(t0) t dx = ∫ t0 t v dτ => i(t) = (1/L)∫ t0 t v d τ + i(t 0 ) or simply: v = L di/dt = the inductor v-I equation, and i(t) = (1/L) ∫ 0 t vd τ + i(t 0 ) = the inductor i-v equation

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Example 6.2 v(t) = 20t e -10t V for t > 0 Assume v = 0 for t < 0 a) Sketch the voltage as a function of time. b) Find the inductor current as a function of time. i = (1/0.1) ∫ 0 t 20e -10τ dτ + 0 = 200{[(-e -10τ )/100][10τ + 1]}| 0 t = 2(1 – 10ie -10t – 10t) A for t > 0 c) Sketch the current as a function of time.
Power and Energy in the Inductor v = L di/dt therefore: p = vi = Li di/dt = the power stored in an inductor p = v[(1/L)∫ to t vdτ + i(t o )] p = dw/dt = Li di/dt Therefore: dw = Li di and ∫ 0 w dx = L∫ 0 i ydy => w = ½ Li 2 = the energy in an inductor

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a) Plot i, v, p, w versus time. b) In what time interval is energy being stored in the inductor? Increasing energy curve occurs when energy is being stored: 0 to 0.2 s (note also time when p > 0). c) In what time is energy being extracted from the inductor? Decreasing energy curve (p < 0): 0.2 s to ∞. d) What is the maximum energy stored in the inductor? Note w is a maximum when current is a
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## This note was uploaded on 10/27/2010 for the course ECE 220 taught by Professor Strickland during the Spring '08 term at Arizona.

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12. ENG 282 ch6 X - GeorgeHaines,Instructor Chapter6.16.5...

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