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ECE201 Lect-51 Phasor Relationships for Circuit Elements (8.4); Impedance and Admittance (8.5) Dr. Holbert February 1, 2006
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ECE201 Lect-52 Phasor Relationships for Circuit Elements Phasors allow us to express current-voltage relationships for inductors and capacitors much like we express the current-voltage relationship for a resistor. A complex exponential is the mathematical tool needed to obtain this relationship.
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ECE201 Lect-53 I-V Relationship for a Resistor Suppose that i(t) is a sinusoid: i(t) = I M e j( t+ Find v(t) R v(t) + – i(t)
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ECE201 Lect-54 Computing the Voltage
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ECE201 Lect-55 Class Example Learning Extension E8.5
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ECE201 Lect-56 I-V Relationship for a Capacitor Suppose that v(t) is a sinusoid: v(t) = V M e j( t+ Find i(t) C v(t) + – i(t)
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ECE201 Lect-57 Computing the Current
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ECE201 Lect-58 Phasor Relationship Represent v(t) and i(t) as phasors: V = V M I = j C V The derivative in the relationship between v(t) and i(t) becomes a multiplication by j in the relationship between V and I.
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ECE201 Lect-59 Example v(t) = 120V cos(377t + 30 ) C = 2 F What is V? What is I? What is i(t)?
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ECE201 Lect-510 Class Example Learning Extension E8.7
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ECE201 Lect-511 I-V Relationship for an Inductor V = j L I L v(t) + – i(t)
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ECE201 Lect-512 Example i(t) = 1 A cos(2 9.1510 7 t + 30 ) L = 1 H What is I? What is V? What is v(t)?
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ECE201 Lect-513 Class Example Learning Extension E8.6
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ECE201 Lect-514 Circuit Element Phasor Relations (ELI and ICE man)
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ECE201 Lect-515 Impedance AC steady-state analysis using phasors allows us to express the relationship between current and voltage using a formula that looks likes Ohm’s law: V = I Z Z is called impedance (units of ohms, )
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ECE201 Lect-516 Impedance Resistor:V = I R –The impedance is Z R = R Inductor:V = I j L –The impedance is Z L = j L
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ECE201 Lect-517 Impedance Capacitor: –The impedance is Z C = 1/j C
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ECE201 Lect-518 Some Thoughts on Impedance Impedance depends on the frequency, f Impedance is (often) a complex number. Impedance is not a phasor (why?). Impedance allows us to use the same solution techniques for AC steady state as we use for DC steady state.
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ECE201 Lect-519 Impedance Example: Single Loop Circuit 20k +–+– 1F1F10V 0 VCVC + – = 377 Find V C
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ECE201 Lect-520 Impedance Example How do we find V C ? First compute impedances for resistor and capacitor: Z R = 20k = 20k 0 Z C = 1/j (377·1 F) = 2.65k -90
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ECE201 Lect-521 Impedance Example 20k 0 +–+– 2.65k -90 10V 0 VCVC + –
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ECE201 Lect-522 Impedance Example Now use the voltage divider to find V C :
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ECE201 Lect-523 Low Pass Filter: A Single Node-pair Circuit Find v(t) for =2 3000 1k 0.1 F 5mA 0 + – V
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ECE201 Lect-524 Find Impedances 1k -j530 5mA 0 + – V
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ECE201 Lect-525 Find the Equivalent Impedance 5mA 0 + – VZ eq
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ECE201 Lect-526 Parallel Impedances
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ECE201 Lect-527 Computing V
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ECE201 Lect-528 Impedance Summary
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ECE201 Lect-529 Class Examples Learning Extension E8.8 Learning Extension E8.9
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