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```Single Node-Pair Circuits
and
Current Division
Lecture 9
1
Example: 2 Light Bulbs in
Parallel
I
I1
R1
I2
R2
+
V
-
How do we find I1 and I2?
Lecture 9
2
Apply KCL at the Top Node
I1 + I 2 = I
I1 
V
I2 
V
R1
R2
Lecture 9
3
Solve for V
 1
1 

I 

 V 


R1 R 2
R
R
2 
 1
V
V
V  I
1
1
R1

1
 I
R1 R 2
R1  R 2
R2
Lecture 9
4
Equivalent Resistance
If we wish to replace the two parallel resistors
with a single resistor whose voltage-current
relationship is the same, the equivalent
resistor has a value of:
R eq 
R1 R 2
R1  R 2
Lecture 9
5
Now to find I1
I1 
V
 I
R1
R2
R1  R 2
• This is the current divider formula.
• It tells us how to divide the current through
parallel resistors.
Lecture 9
6
What is the formula for I2?
Lecture 9
7
Example: 2 Light Bulbs in
Parallel
1.17A
I1
144W
I2
360W
+
V
-
Find I1 and I2
Lecture 9
8
I1 and I2
I 1  1 . 17 A
I 2  1 . 17 A
360 W
144 W  360 W
144 W
144 W  360 W
Lecture 9
 0 . 836 A
 0 . 334 A
9
Example: 3 Light Bulbs in
Parallel
I
I1
R1
I2
R2
I3
R3
+
V
-
How do we find I1, I2, and I3?
Lecture 9
10
Apply KCL at the Top Node
I1 + I2 + I3 = I
I1 
V
I2 
R1
I3 
V
R2
V
R3
Lecture 9
11
Solve for V
 1
1
1 

I 


 V 



R1 R 2
R3
R
R
R
2
3 
 1
V
V
V  I
V
1
1
R1
1

R2
Lecture 9

1
R3
12
Req
R eq 
1
1
R1
1

R2
Lecture 9

1
R3
13
No Current Divider!
• We cannot make a simple current divider
equation for three or more parallel resistors.
• We have to solve for V, then solve for the
current(s) of interest.
Lecture 9
14
Example: 3 Light Bulbs in
Parallel
1.67A
I1
144W
I2
360W
I3
240W
+
V
-
Find I1
Lecture 9
15
Compute V, then I1
R eq 
1
1
144 W

1
360 W

1
 72 W
240 W
V  IR eq  1 . 67 A  72 W  120 V
I1 
V
R1

120 V
144 W
Lecture 9
 0 . 833 A
16
More Than One Source
Is1
I1
R1
Is2
I2
R2
+
V
-
How do we find I1 or I2?
Lecture 9
17
Apply KCL at the Top Node
I1 + I2 = Is1 - Is2
I s1  I s 2
 1
1 



 V 


R1 R 2
R
R
2 
 1
V
V
V   I s1  I s 2 
Lecture 9
R1 R 2
R1  R 2
18
Multiple Current Sources
• We find an equivalent current source by
algebraically summing current sources.
• We find an equivalent resistance.
• We find V as equivalent I times equivalent
R.
• We then find any necessary currents using
Ohm’s law.
Lecture 9
19
```
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