Transcript More basic electricity

Kirchhoff’s Rules
When series and parallel
combinations aren’t enough
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Some circuits have resistors which are
neither in series nor parallel
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They can still be analyzed, but one
uses Kirchhoff’s rules.
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Not in series
The 1-k resistor is not in series with the 2.2-k since the
some of the current that went through the 1-k might go
through the 3-k instead of the 2.2-k resistor.
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Not in parallel
The 1-k resistor is not in parallel with the 1.5-k since their
bottoms are not connected simply by wire, instead that 3-k
lies in between.
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Kirchhoff’s Node Rule
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A node is a point at which wires meet.
“What goes in, must come out.”
Recall currents have directions, some currents will point
into the node, some away from it.
The sum of the current(s) coming into a node must equal
the sum of the current(s) leaving that node.
I1 + I2 = I3
The node rule is about currents!
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I1 
 I2
 I3
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Kirchhoff’s Loop Rule 1
“If you go around in a circle, you get back
to where you started.”
 If you trace through a circuit keeping track
of the voltage level, it must return to its
original value when you complete the
circuit
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Sum of voltage gains = Sum of voltage losses
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Batteries (Gain or Loss)
Loop direction
Whether a battery is a gain or a loss
depends on the direction in which you are
tracing through the circuit
Loop direction

Gain
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Loss
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Resistors (Gain or Loss)
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Gain
I
Current direction
Current direction
Loss
I
Loop direction
Whether a resistor is a gain or a loss
depends on whether the trace direction
and the current direction coincide or not.
Loop direction
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Branch version
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Neither Series Nor Parallel
I1.5 
I1 
I3 
I2.2 
I1.7 
Assign current variables to each branch. Draw
loops such that each current element is included
in at least one loop.
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Apply Current (Node) Rule
I1.5 
I1 
*
I3 
*
I1-I3 
I1.5+I3 
*Node rule applied.
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Three Loops
Voltage Gains = Voltage Losses
 5 = 1 • I1 + 2.2 • (I1 – I3)
 1 • I1 + 3 • I3 = 1.5 • I1.5
 2.2 • (I1 – I3) = 3 • I3 + 1.7 • (I1.5 + I3)

 Units:
Voltages are in V, currents in mA,
resistances in k
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5 = 1 • I1 + 2.2 • (I1 – I3)
I1.5 
I1 
I3 
I1-I3 
I1.5+I3 
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1 • I1 + 3 • I3 = 1.5 • I1.5
I1.5 
I1 
I3 
I1-I3 
I1.5+I3 
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2.2 • (I1 – I3) = 3 • I3 + 1.7 • (I1.5 + I3)
I1.5 
I1 
I3 
I1-I3 
I1.5+I3 
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Simplified Equations
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5 = 3.2 • I1 - 2.2 • I3
I1 = 1.5 • I1.5 - 3 • I3
0 = -2.2 • I1 + 1.7 • I1.5 + 6.9 • I3
 Substitute
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middle equation into others
5 = 3.2 • (1.5 • I1.5 - 3 • I3) - 2.2 • I3
0 = -2.2 • (1.5 • I1.5 - 3 • I3) + 1.7 • I1.5 + 6.9 • I3
 Multiply
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out parentheses and combine like terms.
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Solving for I3
5 = 4.8 • I1.5 - 11.8 • I3
 0 = - 1.6 I1.5 + 13.5 • I3
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 Solve
the second equation for I1.5 and
substitute that result into the first
5 = 4.8 • (8.4375 I3 ) - 11.8 • I3
 5 = 28.7 • I3
 I3  0.174 mA

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Comparison with Simulation
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Other currents
Return to substitution results to find other
currents.
 I1.5 = 8.4375 I3 = 1.468 mA
 I1 = 1.5 • I1.5 - 3 • I3
 I1 = 1.5 • (1.468) - 3 • (0.174)
 I1 = 1.68 mA

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Loop version
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Neither Series Nor Parallel
JB
JA
JC
Draw loops such that each current element is included
in at least one loop. Assign current variables to each
loop. Current direction and lop direction are the same.
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Loop equations
5 = 1  (JA - JB) + 2.2  (JA - JC)
 0 = 1 (JB - JA) + 1.5  JB + 3 (JB - JC)
 0 = 2.2 (JC - JA) + 3 (JC - JB) + 1.7 JC

 “Distribute” the parentheses
5 = 3.2JA – 1 JB - 2.2 JC
 0 = -1 JA + 5.5 JB – 3
JC
 0 = -2.2JA – 3
JB + 6.9 JC

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Algebra
JC = (2.2/6.9)JA + (3/6.9)JB
 JC = 0.3188 JA + 0.4348 JB
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5 = 3.2 JA – 1 JB - 2.2 (0.3188 JA + 0.4348 JB)
0 = -1 JA + 5.5 JB – 3 (0.3188 JA + 0.4348 JB)
5 = 2.4986 JA – 1.9566 JB
0 = -1.9564 JA + 4.1956 JB
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More algebra
JB = (1.9564/4.1956) JA
 JB = 0.4663 JA

5 = 2.4986 JA – 1.9566 (0.4663 JA)
 5 = 1.5862 JA
 JA = 3.1522 mA

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Other loop currents
JB = 0.4663 JA = 0.4663 (3.1522 mA)
 JB = 1.4699 mA
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JC = 0.3188 JA + 0.4348 JB
 JC = 0.3188 (3.1522) + 0.4348 (1.4699)
 JC = 1.644 mA

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Branch Variables
I1.5 
I1 
I3 
I2.2 
I1.7 
Assign current variables to each branch. Draw
loops such that each current element is included
in at least one loop.
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Loop Variables
JB
JA
JC
Draw loops such that each current element is included
in at least one loop. Assign current variables to each
loop. Current direction and lop direction are the same.
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Branch Currents from Loop currents
= JA – JB = 3.1522 – 1.4699 = 1.6823 mA
I1.5 = JB = 1.4699 mA
 I1

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Matrix equation
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Loop equations as matrix equation
5 = 3.2JA – 1 JB - 2.2 JC
 0 = -1 JA + 5.5 JB – 3
JC
 0 = -2.2JA – 3
JB + 6.9 JC

 3.2  1  2.2  J A  5
  1 5.5  3   J   0

 B   
 2.2  3 6.9   J C  0
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Enter matrix in Excel, highlight a
region the same size as the matrix.
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In the formula bar, enter
=MINVERSE(range) where range is the set
of cells corresponding to the matrix (e.g.
B1:D3). Then hit Crtl+Shift+Enter
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Result of matrix inversion
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Prepare the “voltage vector”, then highlight a range the
same size as the vector and enter =MMULT(range1,range2)
where range1 is the inverse matrix and range2 is the voltage
vector. Then Ctrl-Shift-Enter.
Voltage vector
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Results of Matrix Multiplication
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