single-node-pair circuits (2.4); sinusoids (7.1);
DESCRIPTION
Single-Node-Pair Circuits (2.4); Sinusoids (7.1);. Dr. S. M. Goodnick September 5, 2002. Example: 3 Light Bulbs in Parallel. How do we find I 1 , I 2 , and I 3 ?. +. I 1. I 2. I 3. I. R 1. R 2. R 3. V. –. Apply KCL at the Top Node. I= I 1 + I 2 + I 3. Solve for V. R eq. - PowerPoint PPT PresentationTRANSCRIPT
ECE201 Lect-5 1
Single-Node-Pair Circuits (2.4); Sinusoids (7.1);
Dr. S. M. Goodnick
September 5, 2002
ECE201 Lect-5 2
Example: 3 Light Bulbs in Parallel
How do we find I1, I2, and I3?
I R2 V
+
–
R1
I1 I2
R3
I3
ECE201 Lect-5 3
Apply KCL at the Top Node
I= I1 + I2 + I3
11 R
VI
22 R
VI
33 R
VI
ECE201 Lect-5 4
Solve for V
321321
111
RRRV
R
V
R
V
R
VI
321
1111
RRR
IV
ECE201 Lect-5 5
Req
321
1111
RRR
Req
iMpar RRRRR
11111
21
Which is the familiar equation for parallel resistors:
ECE201 Lect-5 6
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.
j
parSR R
RII
j
ECE201 Lect-5 7
Is2 VR1 R2
+
–
I1 I2
Example: More Than One Source
How do we find I1 or I2?
Is1
ECE201 Lect-5 8
Apply KCL at the Top Node
I1 + I2 = Is1 - Is2
212121
11
RRV
R
V
R
VII ss
21
2121 RR
RRIIV ss
ECE201 Lect-5 9
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.
ECE201 Lect-5 10
In General: Current Division
Consider N resistors in parallel:
Special Case (2 resistors in parallel)
iNpar
j
parSR
RRRRR
R
Rtiti
kj
11111
)()(
21
21
2)()(1 RR
Rtiti SR
ECE201 Lect-5 11
Class Examples
• Learning Extension E2.11
ECE201 Lect-5 12
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.
ECE201 Lect-5 13
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.
ECE201 Lect-5 14
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.
ECE201 Lect-5 15
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.
ECE201 Lect-5 16
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.
ECE201 Lect-5 17
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.
ECE201 Lect-5 18
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: VM
tVtv M cos)(
ECE201 Lect-5 19
Example
What is the amplitude, period, frequency, and angular (radian) frequency of this sinusoid?
-8
-6
-4
-2
0
2
4
6
8
0 0.01 0.02 0.03 0.04 0.05
ECE201 Lect-5 20
Phase
-8
-6
-4
-2
0
2
4
6
8
0 0.01 0.02 0.03 0.04 0.05
ECE201 Lect-5 21
Leading and Lagging Phase
x1(t) leads x2(t) by -x2(t) lags x1(t) by -
On the preceding plot, which signals lead and which signals lag?
tXtx M cos)(11
tXtx M cos)(22
ECE201 Lect-5 22
Class Examples
• Learning Extension E7.1
• Learning Extension E7.2