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ECE201 Lect-51 Single-Node-Pair Circuits (2.4); Sinusoids (7.1); Dr. S. M. Goodnick September 5, 2002.

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Presentation on theme: "ECE201 Lect-51 Single-Node-Pair Circuits (2.4); Sinusoids (7.1); Dr. S. M. Goodnick September 5, 2002."— Presentation transcript:

1 ECE201 Lect-51 Single-Node-Pair Circuits (2.4); Sinusoids (7.1); Dr. S. M. Goodnick September 5, 2002

2 ECE201 Lect-52 Example: 3 Light Bulbs in Parallel How do we find I 1, I 2, and I 3 ? I R2R2 V + – R1R1 I1I1 I2I2 R3R3 I3I3

3 ECE201 Lect-53 Apply KCL at the Top Node I= I 1 + I 2 + I 3

4 ECE201 Lect-54 Solve for V

5 ECE201 Lect-55 R eq Which is the familiar equation for parallel resistors:

6 ECE201 Lect-56 Current Divider This leads to a current divider equation for three or more parallel resistors. For 2 parallel resistors, it reduces to a simple form. Note this equation’s similarity to the voltage divider equation.

7 ECE201 Lect-57 Is2Is2 V R1R1 R2R2 + – I1I1 I2I2 Example: More Than One Source How do we find I 1 or I 2 ? Is1Is1

8 ECE201 Lect-58 Apply KCL at the Top Node I 1 + I 2 = I s1 - I s2

9 ECE201 Lect-59 Multiple Current Sources We find an equivalent current source by algebraically summing current sources. As before, we find an equivalent resistance. We find V as equivalent I times equivalent R. We then find any necessary currents using Ohm’s law.

10 ECE201 Lect-510 In General: Current Division Consider N resistors in parallel: Special Case (2 resistors in parallel)

11 ECE201 Lect-511 Class Examples Learning Extension E2.11

12 ECE201 Lect-512 Sinusoids: Introduction Any steady-state voltage or current in a linear circuit with a sinusoidal source is a sinusoid. –This is a consequence of the nature of particular solutions for sinusoidal forcing functions. –All steady-state voltages and currents have the same frequency as the source.

13 ECE201 Lect-513 Introduction (cont.) In order to find a steady-state voltage or current, all we need to know is its magnitude and its phase relative to the source (we already know its frequency). Usually, an AC steady-state voltage or current is given by the particular solution to a differential equation.

14 ECE201 Lect-514 The Good News! We do not have to find this differential equation from the circuit, nor do we have to solve it. Instead, we use the concepts of phasors and complex impedances. Phasors and complex impedances convert problems involving differential equations into simple circuit analysis problems.

15 ECE201 Lect-515 Phasors A phasor is a complex number that represents the magnitude and phase of a sinusoidal voltage or current. Remember, for AC steady-state analysis, this is all we need---we already know the frequency of any voltage or current.

16 ECE201 Lect-516 Complex Impedance Complex impedance describes the relationship between the voltage across an element (expressed as a phasor) and the current through the element (expressed as a phasor). Impedance is a complex number. Impedance depends on frequency.

17 ECE201 Lect-517 Complex Impedance (cont.) Phasors and complex impedance allow us to use Ohm’s law with complex numbers to compute current from voltage, and voltage from current.

18 ECE201 Lect-518 Sinusoids Period: T –Time necessary to go through one cycle Frequency: f = 1/T –Cycles per second (Hz) Angular frequency (rads/sec):  = 2  f Amplitude: V M

19 ECE201 Lect-519 Example What is the amplitude, period, frequency, and angular (radian) frequency of this sinusoid?

20 ECE201 Lect-520 Phase

21 ECE201 Lect-521 Leading and Lagging Phase x 1 (t) leads x 2 (t) by  -  x 2 (t) lags x 1 (t) by  -  On the preceding plot, which signals lead and which signals lag?

22 ECE201 Lect-522 Class Examples Learning Extension E7.1 Learning Extension E7.2


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