1. Voltage and Current Division
Topic 8
Voltage and Current Division
(3.3 & 3.4)
 
Symbol & MT Extra included

2. The Voltage Divider v1 + R1 -vs v1 v2 + -
Here we apply a source to a couple of resistors in series i
By KVL
That is, some of the source voltage appears across R1 and some of it across R2.
Thus the voltages are given by
But how much does each resistor get?
The same current i flows through all the resistors
& So
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Division

3. The Voltage Divider R1 R2 vs v1 v2 + - i
The voltage is divided proportionally between the two resistors according to their proportion of the total resistance If
R1 is
v1 is
R2 is
v2 is
of total resistance
of vs
R1=R2
R2=2R1 ⅓ ⅓ ⅔ ⅔
R1=1KΩ R2=9KΩ
10%
10%
90%
90%
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Division

4. Loading a Divider
A first year Engineering student is asked to get 3 volts out of a 9 volt battery in order to drive a 500Ω load
What reading does the student get when the load is connected?
Reasoning that a voltage divider should do the trick, the student gets a 1 KΩ resistor and a 2 KΩ resistor then hooks them up as follows: 9v
1KΩ
2KΩ
500Ω 9v
1KΩ
2KΩ v +
Hooking a voltmeter across the 1K resistor, the student confirms that there are 3 volts.
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Division

5. Current Divider Resistors in series give you voltage division +vr + -
Resistors in parallel give you current division i1 i2
Now it is the source current which divides to flow through the two resistors So
As the resistors are in parallel, they have the same voltage across them, vr
Therefore
By Ohm’s Law
By KCL
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6. Discussion is R1 R2 i1 i2 vr + -
Notice that once again we have a proportionality relationship
The more resistance (to flow) the less current that flows through it If
R1 is
i1 is
R2 is
i2 is
of total resistance
of is
R1=R2
R2=2R1 ⅓ ⅔ ⅔ ⅓
R1=1KΩ R2=9KΩ
10%
90%
90%
10%
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Division

7. Water Flow Analogy Big resistance (to flow) = narrower pipe
= less flow
small resistance (to flow) = wider pipe
= more flow
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8. Current Divider Using Conductanceis R1 R2 i1 i2 vr + -
Recall that
“is defined as”
G is conductance which measures the ease of flow
Multiply top & bottom by G1G2
Which gives us the same kind of proportionality as exists for voltage dividers and resistance
Similarly
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9. Assesment 3.2 25KΩ (a) Find the no-load value of vo.
(no-load means RL is out of there)
75KΩ vo + - RL
200v
Using the voltage divider relationship
We can just divide out the KΩ So
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Division

10. Mentally, we have done this
Assesment 3.2
25KΩ
(b) Find vo when RL is 150KΩ
RL is in parallel with the 75KΩ resistor
75KΩ vo + - RL
200v
So we can use the voltage divider relationship to write
25KΩ
75KΩ|| RL
200v vo + -
Mentally, we have done this So
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Division

11. General Case of Voltage Division
Circuit R1 R2 Rj
Rn-1 Rn v + -
We have a circuit of some kind that produces a voltage v across its terminals vj + - i
We then apply a string of resistors across it in series
How much voltage appears across Rj?
A single current i flows through the n resistors
Thus
By KVL So
where
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Division

12. General Case of Voltage Division
Circuit R1 R2 Rj
Rn-1 Rn v + - vj + - + i vj -
j’th resistance
voltage across the j’th resistor
the original voltage
equals
over
times
the equivalent series resistance
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Division

13. General Case of Current Division
Circuit v + - R1 R2 Rj
Rn-1 Rn
The resistors all experience the same voltage v
Some circuit that produces a current i
By KCL
Has n parallel resistors hooked across it Or
where
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Division

14. General Case of Current Division
Circuit v + - ij R1 R2 Rj
Rn-1 Rn
Then the current through the j’th resistor is
Given that
Clearly, the voltage across all the resistors is
Notice the symmetry with the voltage division case
current division
voltage division
resistors in parallel
resistors in series
inversely proportional
directly proportional
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15. Which comes first? R1 Both series and parallel
Parallel operator is multiplicative, series is additive, so parallel has algebraic precedence
Is it R1 in series with R2 first and then R3 in parallel? R2 vo + - R3 vs
No, because R1 and R2 do not share the same current.
For R1 to be in series with R2 there can be no current split here
Does this mean
Another way to look at it or
It means this
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Division

16. Assesment 3.4 40Ω 50Ω 70Ω 20Ω 30Ω 10Ω 60V vo + -
Use voltage division to determine the voltage vo across the 40Ω resistor a)
Now it’s easy
40Ω
70Ω
20Ω
30Ω
60Ω
60V vo + -
40Ω
70Ω
10Ω
60V vo + -
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Division

17. Assesment 3.4 40Ω 50Ω 70Ω 20Ω 30Ω 10Ω 60V vo + -
Use vo to find the current in the 40Ω resistor then use current division to determine the current in the 30Ω resistor b)
By Ohm’s Law
40Ω
70Ω
20Ω
30Ω
60Ω
60V
500mA
Using our first mental reduction model
we can see the current splits three ways.
The current through the 30Ω resistor is given by
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18. The current through the 50Ω resistor
Assesment 3.4
40Ω
50Ω
70Ω
20Ω
30Ω
10Ω
60V vo + -
i50
How much power is absorbed by the 50Ω resistor c)
The current through the 50Ω resistor
40Ω
70Ω
20Ω
30Ω
60Ω
60V
500mA
i60
Is the same as the current through the 60Ω resistor in our first reduction
Proceeding as in part b) So
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19. A Perspective on the Parallel Relationship
Serial relationship is dominated by the largest resistor
For series resistors
It should be pretty obvious that
Similarly for parallel resistors
You could hold R2 steady and vary R1 with the same result
Let’s just look at a pair
Notice, the combination is always less than either resistor
Similarly for R1
Parallel relationship is dominated by the smallest resistor
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Division