Download Parallel Circuits

Parallel Circuits
The rules of OHM’s law still hold true.
The formulas you need to know are:
1/RTOTAL = 1/R1 + 1/R2 + 1/R3 …
ITOTAL = I1+ I2+ I3 …
VTOTAL= V1= V2= V3 …
Voltage: VTOTAL= V1= V2= V3 …
Remember Voltage is the Potential Voltage
Difference between 2 points
A
D
E
R3 1k
R1 1k
V1 5
C
R2 1k
B
G
F
H
Measuring the above circuit with a Multimeter
will give the following readings:
First set your meter to DC Volts
If you place the positive lead (red) at point A, B,
C or D and measure to points E, F, G, or H with
the common lead (black) you will get the same
reading of 15V
V1 5
H
R1 1k
B
E
C
D
R3 1k
A
R2 1k
To simplify the above circuit it could also be
drawn as follows
F
G
You can see that each branch comes from the
Power Supply and returns to the Power Supply
Essentially you are measuring across the Power
Supply in the above Diagram
Current: ITOTAL = I1+ I2+ I3 …
In a Parallel Circuit the Amperage coming into
the parallel branches is divided amongst the
branches depending on the resistance of the
branch.
Using OHM’s Law we will be able to see how
different resistances will react together
What we know:
The Voltage over each branch will be the same
We know V=IR (OHM’s Law)
So we can see that if there are 3 branches with
the same resistance in each branch then the
Current will be the same in each branch
A
+
+
A
R3 1k
V1 15
R2 1k
R1 1k
AM1
+
AM2
A
+
AM3
A
AM4
AM5
A
+
Voltage over all branches = 15V
Resistance of all branches = 1kΩ = 1000Ω
Current = ?
V=IR
15V = I x 1000Ω
15/1000 = I
I = .015A = 15mA = AM2 = AM3 = AM4
Therefore there would be 15mA of Current
traveling through each branch
Using the formula ITOTAL = I1+ I2+ I3 … We would be
able to see that before the branches the Current
would be
ITOTAL = .015 + .015 + .015
ITOTAL = .045A = 45mA = AM1 = AM5
However if the resistance is different then:
The branch with the highest resistance will
have the lowest current
The branch with the lowest resistance will have
the highest current
The Current in all branches will add up to the
total current before the branches and after the
branches.
Always think of Current as the thing that moves.
Current reacts very much the same as water in a
hose.
Now lets look at this
A
+
+
A
R3 5k
V1 10
R2 2k
R1 1k
AM1
+
AM2
A
+
AM3
A
AM4
AM5
A
+
Again we can see with OHM’s Law
V = IR
For R1
10V = I x 1000Ω
10/1000 = I
I = .01A = 10mA = A2
A
+
+
A
R3 5k
V1 10
R2 2k
R1 1k
AM1
+
AM2
A
+
AM3
A
AM4
AM5
A
+
Again we can see with OHM’s Law
V = IR
For R2
10V = I x 2000Ω
10/2000 = I
I = .005A = 5mA = A3
A
+
+
A
R3 5k
V1 10
R2 2k
R1 1k
AM1
+
AM2
A
+
AM3
A
AM4
AM5
A
+
We can see with OHM’s Law
V = IR
For R3
10V = I x 5000Ω
10/5000 = I
I = .002A = 2mA = A4
Therefore we would know that AM1 and AM5
would be equal to the sum of the 3 branches
ITOTAL = I1+ I2+ I3
ITOTAL = 10mA + 5mA + 2mA
ITOTAL = 17mA = AM1 = AM5
There are more curve balls but first lets look at
Resistance and its Formulas
Resistance: 1/RTOTAL = 1/R1 + 1/R2 + 1/R3 …
R2 1k
V1 10
R1 1k
Before using the above circuits lets see how this
works with just 2 branches using the same value
resistors
1/RTOTAL = 1/R1 + 1/R2
1/RTOTAL = 1/1000Ω + 1/1000Ω
1/RTOTAL = 2/1000Ω
1/RTOTAL = 1/500Ω
1/RTOTAL = .002
RTOTAL = 1/.002
RTOTAL = 500Ω
It is true to say that if you have multiple
branches with equal resistance then RTOTAL is
equal to the value of resistance divided by the
number of branches
R2 5k
V1 10
R1 1k
Now let’s look at Resistors of different values
1/RTOTAL = 1/R1 + 1/R2
1/RTOTAL = 1/1000Ω + 1/5000Ω
1/RTOTAL = 6/5000Ω
1/RTOTAL = .0012
RTOTAL = 1/.0012
RTOTAL = 833Ω
RTOTAL will always be less than the resistance of
the lowest branch.
How you know this will be true is to go back to
the Current formula for a parallel circuit
Current: ITOTAL = I1+ I2+ I3 …
R1 1kΩ or 1000Ω
R2 5kΩ or 5000Ω
RTOTAL or 833Ω
V=IR
10V = I x 1000Ω
10V/1000Ω = I
I = .01A or 10mA
V=IR
10V = I x 5000Ω
10V/5000Ω = I
I = .002A = 2mA
V=IR
10V = I x 833Ω
10V/833Ω = I
I = .012A =12mA
The above chart shows how the formulas
confirm the results
Practice Parallel Circuits #1
1. Find the overall Voltage, Current and Resistance. Also what would the Current be
through the 2 branches (resistors).
V1 15
R1 1k
V1 = ____________
V2 = ____________
VTOTAL =____________
R2 1.5k
I1___________
I2___________
ITOTAL =_____________
RTOTAL =_____________
2. Find the values for the following circuit
V1 12
R1 1k
V1 = ____________
V2 = ____________
V3 = ____________
VTOTAL =____________
R2 2M
R3 500
I1___________
I2___________
I3___________
ITOTAL =_____________
RTOTAL =_____________
R6 600
R4 600
V1 12
R5 600
3. Find the values for the following circuit
R1 1.5k
V1 = ____________ I1___________
V2 = ____________ I2___________
V3 = ____________ I3___________
R2 2M
R3 3k
V4 = ____________ I4___________
V5 = ____________ I5___________
V6 = ____________ I6___________
Before you can figure out the Totals you must simplify the circuit
1/RTOTAL(1st branch) = 1/R1 + 1/R2 + 1/R3 RTOTAL(1st branch) =___________________
1/RTOTAL(2nd
branch)
= 1/R4 + 1/R5 + 1/R6
RTOTAL(2nd
branch)
1/RTOTAL(whole circuit) = 1/RTOTAL(1st branch) + 1/RTOTAL(2nd
VTOTAL =____________
=___________________
branch)
ITOTAL =_____________
RTOTAL =_____________
4. Find the values for the following circuit
A
B
V1 12
R4 2k
R5 5k
R1 1.5k
R6 4M
R2 2M
R3 3k
V1 = ____________ I1___________
V2 = ____________ I2___________
V3 = ____________ I3___________
V4 = ____________ I4___________
V5 = ____________ I5___________
V6 = ____________ I6___________
1/RTOTAL(1st branch) = 1/R1 + 1/R2 + 1/R3
RTOTAL(1st branch) =___________________
1/RTOTAL(2nd
RTOTAL(2nd
branch)
= 1/R4 + 1/R5 + 1/R6
branch)
1/RTOTAL(whole circuit) = 1/RTOTAL(1st branch) + 1/RTOTAL(2nd
VTOTAL =____________
ITOTAL =_____________
=___________________
branch)
RTOTAL =_____________
If Amp meters were placed in the above circuit at points A and B what would the
readings be A=______________ B=__________________