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Combinations of Resistors
Series, Parallel and Kirchhoff
PHY 202 (Blum)
1
Simplifying circuits using series and
parallel equivalent resistances
PHY 202 (Blum)
2
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)
3
Look for series combinations
Req=3k
Req=3.6 k
PHY 202 (Blum)
4
Look for parallel combinations
Req = 1.8947 k
Req = 1.1244 k
PHY 202 (Blum)
5
Look for series combinations
Req = 6.0191 k
PHY 202 (Blum)
6
Look for parallel combinations
Req = 2.1314 k
PHY 202 (Blum)
7
Look for series combinations
Req = 5.1314 k
PHY 202 (Blum)
8
Equivalent Resistance
I = V/R = (5 V)/(5.1314 k) = 0.9744 mA
PHY 202 (Blum)
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Backwards 1
V= (3)(.9744) = 2.9232
V= (2.1314)(.9744) = 2.0768
PHY 202 (Blum)
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Backwards 2
V = 2.0768=I (3.3)
I=0.629mA
PHY 202 (Blum)
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
PHY 202 (Blum)
12
Kirchhoff’s Rules
When series and parallel
combinations aren’t enough
PHY 202 (Blum)
13
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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PHY 202 (Blum)
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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PHY 202 (Blum)
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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PHY 202 (Blum)
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.
 I2
I1 
I 1 + I 2 = I3
The node rule is about currents!
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 I3
PHY 202 (Blum)
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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PHY 202 (Blum)
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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PHY 202 (Blum)
Resistors (Gain or Loss)
Loop direction
Current direction
Loss
I
Gain
20
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

PHY 202 (Blum)
Branch version
PHY 202 (Blum)
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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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PHY 202 (Blum)
Apply Current (Node) Rule
I1.5 
I1 
*
I3 
*
I1-I3 
I1.5+I3 
*Node rule applied.
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PHY 202 (Blum)
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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PHY 202 (Blum)
5 = 1 • I1 + 2.2 • (I1 – I3)
I1.5 
I1 
I3 
I1-I3 
I1.5+I3 
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PHY 202 (Blum)
1 • I1 + 3 • I3 = 1.5 • I1.5
I1.5 
I1 
I3 
I1-I3 
I1.5+I3 
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PHY 202 (Blum)
2.2 • (I1 – I3) = 3 • I3 + 1.7 •
(I1.5 + I3)
I1.5 
I1 
I3 
I1-I3 
I1.5+I3 
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PHY 202 (Blum)
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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PHY 202 (Blum)
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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PHY 202 (Blum)
Comparison with Simulation
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PHY 202 (Blum)
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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PHY 202 (Blum)
Loop version
PHY 202 (Blum)
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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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PHY 202 (Blum)
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 + 6.9 JC

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PHY 202 (Blum)
Algebra
 JC
 JC




= (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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PHY 202 (Blum)
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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PHY 202 (Blum)
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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PHY 202 (Blum)
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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PHY 202 (Blum)
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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PHY 202 (Blum)
Branch Currents from Loop currents


I1 = JA – JB = 3.1522 – 1.4699 = 1.6823 mA
I1.5 = JB = 1.4699 mA
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PHY 202 (Blum)
Matrix equation
PHY 202 (Blum)
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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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PHY 202 (Blum)
Enter matrix in Excel, highlight a region the
same size as the matrix.
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PHY 202 (Blum)
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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PHY 202 (Blum)
Result of matrix inversion
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PHY 202 (Blum)
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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PHY 202 (Blum)
Results of Matrix Multiplication
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PHY 202 (Blum)