Transcript More basic electricity

Combinations of Resistors
Series, Parallel and Kirchhoff
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Simplifying circuits using series and
parallel equivalent resistances
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Analyzing a combination of
resistors circuit
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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.
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Look for series combinations
Req=3k
Req=3.6 k
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Look for parallel combinations
Req = 1.8947 k
Req = 1.1244 k
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Look for series combinations
Req = 6.0191 k
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Look for parallel combinations
Req = 2.1314 k
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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
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Backwards 1
V= (3)(.9744) = 2.9232
V= (2.1314)(.9744) = 2.0768
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Backwards 2
V = 2.0768=I (3.3)
I=0.629mA
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V = 2.0768=I (6.0191)
I=0.345mA
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Backwards 3
V=(.345)(3)=1.035
V=(.345)(1.8947)=0.654
V=(.345)(1.1244)=0.388
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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
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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.
 I2
I1 
I 1 + I 2 = I3
The node rule is about currents!
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 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

Loss
Gain
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Resistors (Gain or Loss)
Loop direction
Current direction
Loss
I
Gain
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I
Current 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 
I1.7 
I2.2 
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)
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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
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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
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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
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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
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 “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 + 6.9 JC
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Algebra
 JC
 JC
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= (2.2/6.9)JA + (3/6.9)JB
= 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
I1.5 
I1 
I3 
I1.7 
I2.2 
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
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I1 = JA – JB = 3.1522 – 1.4699 = 1.6823 mA
I1.5 = JB = 1.4699 mA
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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 + 6.9 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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