page 19
•These methods are quite well suited to matrix solutions
•Lets consider a simple problem,
Find Vo |
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10Ω |
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20Ω |
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I1 |
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30V |
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40Ω |
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40Ω |
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After adding the mesh currents (and directions) we may write equations,
∑
∑
VI1 |
= 0 = – 30 + 10I1 + 40( I1 – I2) + 5I1 |
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30 = 55I1 – 40I2 |
(1) |
VI2 |
= 0 = 20I2 + 40I2 + 30I2 + 40( I2 – I1) |
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0 = 40I1 – 130I2 |
(2) |
Solve the equation matrix for I2, (I will use Cramer’s Rule)
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55 |
30 |
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55 |
–40 |
30 |
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40 |
0 |
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55( 0) |
– 30( 40) |
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40 |
–130 |
0 |
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I2 = --------------------------------- |
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------------------------------------------------------ |
= 0.216A |
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55 |
( –130) |
– ( –40) ( 40) |
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–40 |
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40 |
–130 |
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Finally calculate the voltage across the 40 ohm resistor,
VO = 40I2 = 8.6V
3.1.4 More Advanced Applications
page 20
3.1.4.1 - Voltage Dividers
• The voltage divider is a very common and useful circuit configuration. Consider the circuit below, we add a current loop, and assume there is no current out at Vo,
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R1 |
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Vs |
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First, sum the voltages about the loop, |
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V = – Vs + IR1 + IR2 |
= 0 |
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I = |
Vs |
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----------------- |
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R1 + R2 |
Next, find the output voltage, based on the current, |
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page 21
ASIDE: variable resistors are often used as voltage dividers. As the wiper travels along the resistor the output voltage changes.
Vs +
Vo
3.1.4.2 - The Wheatstone Bridge
• The wheatstone bridge is a very common engineering tool for magnifying and measuring signals. In this circuit a supply voltage Vs is used to power the circuit. Resistors R1 and R2 are generally equal, Rx is a resistance to be measured, and R3 is a tuning resistor. An ammeter is shown in the center, and resistor R3 is varied until the current in the center Ig is zero.
page 22
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Rx
R3
For practice try to show the relationship below holds for a balanced bridge (ie, Ig=0),
Rx = |
R2R3 |
----------- |
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R1 |
3.1.4.3 - Tee-To-Pi (Y to Delta) Conversion
•It is fairly common to use a model of a circuit. This model can then be transformed or modified as required.
•A very common model and conversion is the Tee to Pi conversion in electronics. A similar conversion is done for power circuits called delta to y.
page 23
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• We can find equivalent resistors considering that, |
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page 24
Rac |
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1 |
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R------------------------------a + Rb + Rc |
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R------------------------------a + Rb + Rc |
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Rb |
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To find R1, (1)-(3)+(2) |
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Rb( Rc + Ra) – Ra( Rb + Rc) + Rc( Ra + Rb) |
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Ra + Rb + Rc |
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RbR |
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– RaR |
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– RaR |
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+ Rc R |
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+ RcR |
b = 2R1 |
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Ra + Rb + Rc |
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R------------------------------a + Rb + Rc |
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Likewise, |
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R------------------------------a |
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R3 |
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R------------------------------a |
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• To find the equivalents the other way,