phy 202 (blum)1 combinations of resistors series, parallel and kirchhoff
TRANSCRIPT
PHY 202 (Blum) 1
Combinations of Resistors
Series, Parallel and Kirchhoff
PHY 202 (Blum) 2
Simplifying circuits using series and parallel equivalent
resistances
PHY 202 (Blum) 3
Analyzing a combination of resistors circuit
Look for resistors which are in series (the current passing through one must pass through the other) and replace them with the equivalent resistance (Req = R1 + R2).
Look for resistors which are in parallel (both the tops and bottoms are connected by wire and only wire) and replace them with the equivalent resistance (1/Req = 1/R1 + 1/R2).
Repeat as much as possible.
PHY 202 (Blum) 4
Look for series combinations
Req=3k
Req=3.6 k
PHY 202 (Blum) 5
Look for parallel combinations
Req = 1.8947 k
Req = 1.1244 k
PHY 202 (Blum) 6
Look for series combinations
Req = 6.0191 k
PHY 202 (Blum) 7
Look for parallel combinations
Req = 2.1314 k
PHY 202 (Blum) 8
Look for series combinations
Req = 5.1314 k
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Equivalent Resistance
I = V/R = (5 V)/(5.1314 k) = 0.9744 mA
PHY 202 (Blum) 10
Backwards 1
V= (3)(.9744) = 2.9232
V= (2.1314)(.9744) = 2.0768
PHY 202 (Blum) 11
Backwards 2
V = 2.0768=I (3.3)I=0.629mA
V = 2.0768=I (6.0191)I=0.345mA
PHY 202 (Blum) 12
Backwards 3
V=(.345)(1.1244)=0.388
V=(.345)(1.8947)=0.654
V=(.345)(3)=1.035
PHY 202 (Blum) 13
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
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
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 I1 I2
I3The node rule is about currents!
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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
Sum of voltage gains = Sum of voltage losses
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Batteries (Gain or Loss)
Whether a battery is a gain or a loss depends on the direction in which you are tracing through the circuit
Gain Loss
Loo
p di
rect
ion
Loo
p di
rect
ion
20 PHY 202 (Blum)
Resistors (Gain or Loss)
Whether a resistor is a gain or a loss depends on whether the trace direction and the current direction coincide or not. I I
Loss Gain
Loo
p di
rect
ion
Loo
p di
rect
ion
Cur
rent
dir
ectio
n
Cur
rent
dir
ectio
n
PHY 202 (Blum) 21
Branch version
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Neither Series Nor Parallel
I1
I2.2
I1.5
I1.7
I3
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
I1-I3
I1.5
I1.5+I3
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
I1-I3
I1.5
I1.5+I3
I3
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1 • I1 + 3 • I3 = 1.5 • I1.5
I1
I1-I3
I1.5
I1.5+I3
I3
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2.2 • (I1 – I3) = 3 • I3 + 1.7 • (I1.5 + I3)
I1
I1-I3
I1.5
I1.5+I3
I3
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Simplified Equations
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 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 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 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
PHY 202 (Blum) 32
Loop version
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Neither Series Nor Parallel
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.
JA
JB
JC
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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 = 3.2 JA – 1 JB - 2.2 JC 0 = -1 JA + 5.5 JB – 3 JC 0 = -2.2JA – 3 JB + 6.9JC
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Algebra
JC = (2.2/6.9)JA + (3/6.9)JB
JC = 0.3188 JA + 0.4348 JB
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
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
I2.2
I1.5
I1.7
I3
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
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.
JA
JB
JC
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Branch Currents from Loop currents
I1 = JA – JB = 3.1522 – 1.4699 = 1.6823 mA
I1.5 = JB = 1.4699 mA
PHY 202 (Blum) 41
Matrix equation
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Loop equations as matrix equation
5 = 3.2 JA – 1 JB - 2.2 JC 0 = -1 JA + 5.5 JB – 3 JC 0 = -2.2JA – 3 JB + 6.9JC
0
0
5
9.632.2
35.51
2.212.3
C
B
A
J
J
J
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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