Electronics
Parallel
Resistive
Circuits
Part 2
1
Copyright © Texas Education Agency, 2014. All rights reserved.
More Complex Parallel Circuits


Let’s do a parallel circuit analysis with three
branches
Follow the same analysis process as before
VS
R1
R2
R3
2
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More Complex Parallel Circuits



Path 1 has current I1
Apply Kirchhoff’s Law to this current loop
+ VS – VR1 = 0 or
I1
VS

R1
R2
VS = VR1 , so from Ohm’s Law I1 =
R3
VS
R1
3
Copyright © Texas Education Agency, 2014. All rights reserved.
More Complex Parallel Circuits



Path 2 has current I2
Apply Kirchhoff’s Law to this current loop
+ VS – VR2 = 0 or
I2
VS

R1
R2
VS = VR2 , so from Ohm’s Law I2 =
R3
VS
R2
4
Copyright © Texas Education Agency, 2014. All rights reserved.
More Complex Parallel Circuits



Path 3 has current I3
Apply Kirchhoff’s Law to this current loop
+ VS – VR3 = 0 or
I3
VS

R1
R2
VS = VR3 , so from Ohm’s Law I3 =
R3
VS
R3
5
Copyright © Texas Education Agency, 2014. All rights reserved.
Parallel Circuit Equations
I T = I1 + I2 + I 3
VS = VR1 = VR2 = VR3
1
RT
=
1
R1
+
1
R2
1
+
R3
6
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Parallel Circuit Equations
I T = I1 + I2 + I3
VS = VR1 = VR2 = VR3
1
RT
=
1
R1
+
1
R2
1
+
R3
(current adds)
(voltage is the same)
(resistance is more
complex, but it basically
divides)
These three formulas (plus Ohm’s Law)
form a “tool kit” to analyze parallel circuits
7
Copyright © Texas Education Agency, 2014. All rights reserved.
Understanding Resistance in a
Parallel Circuit


Resistance looks a little more complicated, so
let’s examine it more closely
Consider the following circuit
S1
S2
S3
VS
L1

L2
L3
Each switch is open, each light is off
8
Copyright © Texas Education Agency, 2014. All rights reserved.
Understanding Resistance in a
Parallel Circuit



Close S1 and L1 comes on
We get current I1 from the battery
Each light is identical
S1
S2
S3
VS
L1

L2
L3
Total current = I1 , total resistance = R1
9
Copyright © Texas Education Agency, 2014. All rights reserved.
Understanding Resistance in a
Parallel Circuit



Next close S2 and S3, L2 and L3 come on
We get additional current I2 ,I3 from the battery
Total current = I1 + I2 + I3, triple the current
S1
S2
S3
VS
L1

L2
L3
This means total resistance must be cut to one
third
10
Copyright © Texas Education Agency, 2014. All rights reserved.
Do the Math

Use the following formula:
1
RT

=
1
R1
+
1
R2
1
+
R3
Assume R1 = R2 = R3 = 30 Ω
1
RT
=
1
RT
= .0333 + .0333 + .0333 = .1
1
RT
1
R1
+
1
R2
+
1
R3
=
1
1
1
+ +
30
30
30
1
or
= .1
RT = = 10 Ω
.1
11
Copyright © Texas Education Agency, 2014. All rights reserved.
An Easier Way





Now lets look at an easier way to calculate total
resistance
Use the
button on your calculator
On some calculators it will be a
button
This button does the hard part of the math for
you by calculating the inverse of a value
Let’s try using this button on the previous
example
12
Copyright © Texas Education Agency, 2014. All rights reserved.
Calculating Total Resistance

Let’s try this first with two resistors
1
RT

=
1
R1
+
1
R2
1
1
= +
30
30
Using the TI-83 buttons, perform the following
3
0
+
3
0
ENTER
ENTER
13
Copyright © Texas Education Agency, 2014. All rights reserved.
Calculating Total Resistance

Let’s try this first with two resistors
1
RT


+
1
R2
1
1
= +
30
30
Using the TI-83 buttons, perform the following
3

=
1
R1
0
+
3
0
ENTER
ENTER
If you did it right, 15 will be displayed
You only enter resistance values and calculator
functions
14
Copyright © Texas Education Agency, 2014. All rights reserved.
Calculating Total Resistance

Now try it for a circuit with three resistors
1
RT

=
1
R1
+
1
R2
+
1
R3
=
1
1
1
+ +
30
30
30
Using the TI-83 buttons, perform the following
3
+
0
3
0
3
ENTER
+
0
ENTER
15
Copyright © Texas Education Agency, 2014. All rights reserved.
Calculating Total Resistance

Now try it for a circuit with three resistors
1
RT

=
+
1
R2
+
1
R3
=
1
1
1
+ +
30
30
30
Using the TI-83 buttons, perform the following
3
+
0
3

1
R1
0
3
ENTER
+
0
ENTER
If you did it right, 10 will be displayed
16
Copyright © Texas Education Agency, 2014. All rights reserved.
Example Problem 1

For the following circuit, calculate RT and IT
R1 =
300 Ω
VS =
15 V

R2 =
200 Ω
Begin by writing down the equations we need
1
RT
=
1
R1
+
1
R2
and
IT =
VT
RT
17
Copyright © Texas Education Agency, 2014. All rights reserved.
Example Problem 1
1
RT
3
0
=
1
R1
1
+
R2
+
0
ENTER
1
1
=
+
300
200
2
0
0
ENTER
18
Copyright © Texas Education Agency, 2014. All rights reserved.
Example Problem 1
1
RT
3
0
=
1
R1
1
+
R2
+
0
ENTER
1
1
=
+
300
200
2
0
0
ENTER
RT = 120 Ω
19
Copyright © Texas Education Agency, 2014. All rights reserved.
Example Problem 1
1
RT
3
0
=
1
R1
1
+
R2
+
0
ENTER
1
1
=
+
300
200
2
0
0
ENTER
RT = 120 Ω
IT =
VT
RT
=
15 V
= 0.125
120 Ω
A
20
Copyright © Texas Education Agency, 2014. All rights reserved.
Example Problem 2

For the following circuit, calculate RT and IT
VS =
14.5 V

R2 =
2.2 kΩ
R3 =
1.8 kΩ
Begin by writing down the equations we need
1
RT

R1 =
1.4 kΩ
=
1
R1
1
+
R2
1
+
R3
and
IT =
VT
RT
We have enough information to solve these
21
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Example Problem 2
1
RT
=
1
R1
+
1
R2
+
1
R3
=
1
1
1
+
+
1.4 kΩ
2.2 kΩ
1.8 kΩ
22
Copyright © Texas Education Agency, 2014. All rights reserved.
Example Problem 2
1
RT
=
1
R1
+
1
R2
+
1
R3
=
1
1
1
+
+
1.4 kΩ
2.2 kΩ
1.8 kΩ
RT = 580 Ω
23
Copyright © Texas Education Agency, 2014. All rights reserved.
Example Problem 2
1
RT
=
1
R1
+
1
R2
+
1
R3
=
1
1
1
+
+
1.4 kΩ
2.2 𝑘Ω
1.8 𝑘Ω
RT = 580 Ω
IT =
VT
RT
=
14.5 V
=
580 Ω
24
Copyright © Texas Education Agency, 2014. All rights reserved.
Example Problem 2
1
RT
=
1
R1
+
1
R2
+
1
R3
=
1
1
1
+
+
1.4 kΩ
2.2 kΩ
1.8 kΩ
RT = 580 Ω
IT =
VT
RT
=
14.5 V
= 0.025
580 Ω
A
IT = 25 mA
25
Copyright © Texas Education Agency, 2014. All rights reserved.
Example Problem 3

For the following circuit, calculate VS
IT = 106 mA
VS =
?
R1 =
330 Ω
R2 =
560 Ω
26
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Example Problem 3

For the following circuit, calculate VS
IT = 106 mA
VS =
?

R1 =
330 Ω
R2 =
560 Ω
Write the equation that solves the problem
VS = VT = IT • RT
27
Copyright © Texas Education Agency, 2014. All rights reserved.
Example Problem 3

For the following circuit, calculate VS
IT = 106 mA
VS =
?

R1 =
330 Ω
R2 =
560 Ω
Write the equation that solves the problem
VS = VT = IT • RT

Look for what is needed to solve this equation

We have IT, we need RT
Copyright © Texas Education Agency, 2014. All rights reserved.
28
Example 3 Solution

Write the equation for RT
1
RT
=
1
R1
+
1
R2
1
1
=
+
330 Ω
560 Ω
29
Copyright © Texas Education Agency, 2014. All rights reserved.
Example 3 Solution

Write the equation for RT
1
RT
=
1
R1
+
1
R2
1
1
=
+
330 Ω
560 Ω
RT = 207.6 Ω
30
Copyright © Texas Education Agency, 2014. All rights reserved.
Example 3 Solution

Write the equation for RT
1
RT
=
1
R1
+
1
R2
1
1
=
+
330 Ω
560 Ω
RT = 207.6 Ω

Plug this value into the first equation
VS = VT = IT • RT = .106 A • 207.6 Ω
31
Copyright © Texas Education Agency, 2014. All rights reserved.
Example 3 Solution

Write the equation for RT
1
RT
=
1
R1
+
1
R2
1
1
=
+
330 Ω
560 Ω
RT = 207.6 Ω

Plug this value into the first equation
VS = VT = IT • RT = .106 A • 207.6 Ω
VS = 22 V
32
Copyright © Texas Education Agency, 2014. All rights reserved.
Example Problem 4

Calculate R1
IT = 18 mA
VS =
20 V
R1 =
?Ω
I1 = ?
R2 =
?Ω
I2 = 10 mA
33
Copyright © Texas Education Agency, 2014. All rights reserved.
Example Problem 4

Calculate R1
IT = 18 mA
VS =
20 V

R1 =
?Ω
I1 = ?
R2 =
?Ω
I2 = 10 mA
Write the formula for R1
34
Copyright © Texas Education Agency, 2014. All rights reserved.
Example Problem 4

Calculate R1
IT = 18 mA
VS =
20 V


R1 =
?Ω
I1 = ?
R2 =
?Ω
I2 = 10 mA
Write the formula for R1
R1 = V1 I1 = VS I1
Now we need a formula that solves for I1
35
Copyright © Texas Education Agency, 2014. All rights reserved.
Example Problem 4

Calculate R1
IT = 18 mA
VS =
20 V


R1 =
?Ω
I1 = ?
R2 =
?Ω
I2 = 10 mA
Write the formula for R1
R1 = V1 I1 = VS I1
Now we need a formula that solves for I1
IT = I1 + I2
36
Copyright © Texas Education Agency, 2014. All rights reserved.
Problem 4 Solution

To calculate I1,
IT = I1 + I2 or I1 = IT – I2
IT = 18 mA
VS =
20 V
R1 =
?Ω
I1 = ?
R2 =
?Ω
I2 = 10 mA
I1 = 18 mA – 10 mA = 8 mA
37
Copyright © Texas Education Agency, 2014. All rights reserved.
Problem 4 Solution

To calculate I1,
IT = I1 + I2 or I1 = IT – I2
IT = 18 mA
VS =
20 V

R1 =
?Ω
I1 = ?
R2 =
?Ω
I2 = 10 mA
I1 = 18 mA – 10 mA = 8 mA
Plug into first formula
R1 = V1
I1 =
20 V
8 mA =
38
Copyright © Texas Education Agency, 2014. All rights reserved.
Problem 4 Solution

To calculate I1,
IT = I1 + I2 or I1 = IT – I2
IT = 18 mA
VS =
20 V

R1 =
?Ω
I1 = ?
R2 =
?Ω
I2 = 10 mA
I1 = 18 mA – 10 mA = 8 mA
Plug into first formula
R1 = V1
I1 =
20 V
8 mA = 2.5
kΩ
39
Copyright © Texas Education Agency, 2014. All rights reserved.
Example Problem 5

For the following circuit calculate R1
IT = 3.36 mA
VS =
38 V
R1 =
?Ω
R2 =
26 kΩ
40
Copyright © Texas Education Agency, 2014. All rights reserved.
Example Problem 5

For the following circuit calculate R1
IT = 3.36 mA
R1 =
?Ω
VS =
38 V

Write the equation that solves the problem

Note: there is more than one equation for R1
R1 = V1

R2 =
26 kΩ
VS
=
I1
I1
Look for what is needed to solve this equation
41
Copyright © Texas Education Agency, 2014. All rights reserved.
Example Problem 5

For the following circuit calculate R1
IT = 3.36 mA
VS =
38 V

R1 =
?Ω
R2 =
26 kΩ
Now we need to solve for I1 R1 = V1
VS
=
I1
I1
42
Copyright © Texas Education Agency, 2014. All rights reserved.
Example Problem 5

For the following circuit calculate R1
IT = 3.36 mA
VS =
38 V


R1 =
?Ω
R2 =
26 kΩ
Now we need to solve for I1 R1 = V1
What equation has I1 in it?
VS
=
I1
I1
43
Copyright © Texas Education Agency, 2014. All rights reserved.
Example Problem 5

For the following circuit calculate R1
IT = 3.36 mA
R1 =
?Ω
VS =
38 V


Now we need to solve for I1 R1 = V1
What equation has I1 in it?
IT = I1 + I2

R2 =
26 kΩ
or
VS
=
I1
I1
I 1 = IT - I 2
We have IT, can we solve for I2?
44
Copyright © Texas Education Agency, 2014. All rights reserved.
Problem 5 Solution
V2
I2 =
R2
VS
=
R2
38 V
=
26 kΩ
45
Copyright © Texas Education Agency, 2014. All rights reserved.
Problem 5 Solution
V2
I2 =
R2
=
VS
R2
38 V
=
26 kΩ
I2 = 0.00146 A = 1.46 mA
46
Copyright © Texas Education Agency, 2014. All rights reserved.
Problem 5 Solution
V2
I2 =
R2
=
VS
R2
38 V
=
26 kΩ
I2 = 0.00146 A = 1.46 mA

Now that we have calculated the first value we
need, work back through the steps one by one
I1 = IT – I2 = 3.36 mA – 1.46 mA
47
Copyright © Texas Education Agency, 2014. All rights reserved.
Problem 5 Solution
V2
I2 =
R2
=
VS
R2
38 V
=
26 kΩ
I2 = 0.00146 A = 1.46 mA

Now that we have calculated the first value we
need, work back through the steps one by one
I1 = IT – I2 = 3.36 mA – 1.46 mA
I1 = 1.9 mA
48
Copyright © Texas Education Agency, 2014. All rights reserved.
Problem 5 Solution
V2
I2 =
R2
=
VS
R2
38 V
=
26 kΩ
I2 = 0.00146 A = 1.46 mA

Now that we have calculated the first value we
need, work back through the steps one by one
I1 = IT – I2 = 3.36 mA – 1.46 mA
I1 = 1.9 mA
R1 = VS
38 V
=
I1
1.9 mA
49
Copyright © Texas Education Agency, 2014. All rights reserved.
Problem 5 Solution
V2
I2 =
R2
=
VS
R2
38 V
=
26 kΩ
I2 = 0.00146 A = 1.46 mA

Now that we have calculated the first value we
need, work back through the steps one by one
I1 = IT – I2 = 3.36 mA – 1.46 mA
I1 = 1.9 mA
R1 = VS
38 V
=
I1
1.9 mA
50
Copyright © Texas Education Agency, 2014. All rights reserved.
Alternate Problem 5 Solution

Here is another formula to solve for R1
1
RT

=
+
1
R2
or
1
R1
=
1
RT
1
R2
−
We can solve for RT with Ohm’s Law
RT =

1
R1
VT
IT
=
38 V
= 11309.5
3.36 mA
Ω
Plug into above equation to solve for R1
11309.5
--
26000
ENTER
ENTER
51
Copyright © Texas Education Agency, 2014. All rights reserved.
Parallel Circuit Equations
For more than three resistors
I T = I 1 + I2 + I3 + I 4 + …
VS = VR1 = VR2 = VR3 = VR4 = …
1
RT
=
1
R1
+
1
R2
1
+
R3
1
+
R3
+…
52
Copyright © Texas Education Agency, 2014. All rights reserved.
Parallel Circuit Equations
For more than three resistors
I T = I 1 + I2 + I3 + I 4 + …
VS = VR1 = VR2 = VR3 = VR4 = …
1
RT
=
1
𝑅1
+
1
𝑅2
1
+
𝑅3
1
+
𝑅3
+…
Just keep adding terms for each new parallel path
53
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What’s Next?
Practice
 Practice
 Practice

54
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