Combined Series and Parallel Circuits
Objectives:
1. Calculate the equivalent resistance, current, and
voltage of series and parallel circuits.
2. Calculate the equivalent resistance of circuits
combining series and parallel connections.
3. To understand the origins of both of Kirchhoff's
rules and how to use them to solve a circuit
problem.
4. Solve circuit problems.
Resistance and Current
Series Circuit
• Equivalent resistance is equal to the sum of all the
resistance in the circuit.
• Circuit current is equal to the voltage source divided
by the equivalent resistance.
Req = R1 + R2 + R3 + Rn...
I = Vsource / Req
Resistance and Current
Parallel Circuit
• The reciprocal of the equivalent resistance is equal to the
sum of the reciprocals of the individual resistances.
• The total current is the sum of all the currents.
• The potential difference across each resistor is the same
1/ Req = 1/ R1 + 1/ R2 + 1/ Rn...
I = I1 + I 2 + I n.....
I1 = Vsource / R1
Household Circuits
Why do the lights dim when the hair
dryer goes on?
Small resistance
from wiring
This is called a combination series and parallel circuit
Series and Parallel Circuits
1. Draw a diagram of the circuit
2. Find any resistors in parallel. They must have the same
potential difference across them. Calculate the single
equivalent resistance of a resistor that can replace them.
3. Are any resistors (including the parallel equivalent
resistor) in series? Resistors in series have one and only
one current path through them. Calculate the new single
equivalent resistance that can replace them. Draw a new
schematic diagram using that resistor.
4. Repeat steps 2 and 3 until you can reduce the current to a
single resistor. Find the total circuit current. Then go
backwards to find the currents through and the voltages
across individual resistors.
Kirchhoff’s Rules
Gustav Kirchhoff - 1845
1. The sum of the currents entering any junction
must equal the sum of the currents leaving
that junction. (junction rule)
2. The sum of the potential differences across all
the elements around any closed circuit loop
must equal zero. (loop rule)
In this example you will notice that 8 Amps
of current enter the junction and 3 and 5
Amps leave the junction. This makes a total
of 8 Amps entering and 8 Amps leaving.
In this example you will notice 8 Amps and
1 Amp entering the junction and 9 Amps
leaving. This makes a total of 9 Amps
entering and 9 Amps leaving.
In this example you will notice 8 Amps and
1 Amp entering the junction while 7 Amps
and 2 Amps leave. This makes a total of 9
Amps entering and 9 Amps leaving.
This is a simple circuit showing the potential differences
across the source and the resistor. According to
Kirchhoff's 2nd law the sum of the potential differences
will be zero.
This diagram shows the potentials in the little circles
and then shows the potential differences off to the
side. Notice that the potential difference is actually the
difference between one potential and another. Moving
from a low potential to a high potential is considered a
potential rise or positive potential difference. Moving
from a high potential to a lower potential is considered a
potential drop or negative potential difference.
This animation shows the same circuit as above but only
looks at the potential differences as you go around the
loop. Again, Kirchhoff's 2nd law says the sum of the
potential differences has to be zero.
Now lets try some problems
Don’t wait to get totally lost. Ask
your questions as they come to you.
#1
Series Circuit
Rt = R1 + R2 + R3 + …
Rt = 4 + 6 + 3 + 1 =
14 Ω
I=V÷R
I = 40 ÷ 14 =
2.86 amps
#2
Series Circuit
Rt = R1 + R2 + R3 + …
Rt = 5 + 4 + 12 =
21 Ω
I=V÷R
I = 10 ÷ 21 =
0.476 amps
#3
Series Circuit
Rt = R1 + R2 + R3 + …
Rt = 3 + 1 + 7 =
11 Ω
I=V÷R
I = 120 ÷ 11 =
10.9 amps
#4
Series Circuit
Rt = R1 + R2 + R3 + …
Rt = 5 + 1 + 6 + 3 + 4 + 1 =
20 Ω
I=V÷R
I = 9 ÷ 20 =
0.45 amps
#5
Series Circuit
Rt = R1 + R2 + R3 + …
Rt = 12 + 20 + 5 =
37 Ω
I=V÷R
I = 60 ÷ 37 =
1.62 amps
#6
Parallel Circuit
1/Rt = 1/R1 + 1/R2 + 1/R3 + …
1/Rt = 1/2 + 1/2 + 1/2 = 1.5
Rt = 0.667 Ω
I=V÷R
I = 6 ÷ 0.667 =
9.00 amps
#7
Parallel Circuit
1/Rt = 1/R1 + 1/R2 + 1/R3 + …
1/Rt = 1/6 + 1/8 + 1/4 = 0.542
Rt = 1.85 Ω
I=V÷R
I = 120 ÷ 1.85 =
64.9 amps
#8
Parallel Circuit
1/Rt = 1/R1 + 1/R2 + 1/R3 + …
1/Rt =1/2.5 + 1/6 + 1/1 = 1.57
Rt = 0.638 Ω
I=V÷R
I = 14 ÷ 0.638 =
21.9 amps
Now let’s put ‘em together
Simplify diagram in steps to a single resistor
Calculate total resistance and current for the whole
circuit
Then work backwards to find voltages and currents at
individual resistors
It takes time and care to do this right
DON’T TRY TO RUSH THROUGH IT!
#9
The 2 Ω and 3 Ω resistors are in series with one
another
They combine to form a 5 Ω resistor
The three resistors are hooked up in parallel with
each other.
1/Rt = 1/R1 + 1/R2 + 1/R3 + …
1/Rt =1/5 + 1/1 + 1/6 = 1.37
Rt = 0.730 Ω
The circuit now looks like this
I = V/R = 120/.730 =
I = 164 amps
#10
Combine the 3 Ω and the 7 Ω resistors that are in
series with one another to make a 10 Ω resistor
Combine the 1 Ω and the 2 Ω resistor that are in
series with one another to make a 3 Ω resistor
Then re-draw the circuit
It should look like this
The three resistors are hooked up in parallel with each
other.
1/Rt = 1/R1 + 1/R2 + 1/R3 + …
1/Rt =1/10 + 1/4 + 1/3 = 0.683
Rt = 1.46 Ω
The circuit now looks like this
I = V/R = 40/1.46 =
I = 27.4 amps
#11
Combine the resistors hooked up in series with
one another and re-draw the circuit
It should look like this
The three resistors are hooked up in parallel with each
other.
1/Rt = 1/R1 + 1/R2 + 1/R3 + …
1/Rt =1/4 + 1/4 + 1/10 = 0.600
Rt = 1.67 Ω
The circuit now looks like this
I = V/R = 220/1.67 =
I = 132 amps
#12
What type of circuit is this?
Find the total resistance
Find the total current
The ___________ is the same for all
devices in a series circuit
Therefore I3 =
To find V2 we need to know…
V = IR so V2 = 1.17(4)
Series
12 Ω
1.17 amps
Current
1.17 amps
R2 and I2
4.68 V
#13
What type of circuit is this?
Find the total resistance
Find the total current
The ___________ is the same for all
lines in a parallel circuit
Therefore V3 =
To find I3 we need to know…
I = V/R so I3 = 120(2.5)
Parallel
1.11 Ω
108 amps
Voltage
120 V
V3 & R3
48 amps
#14
R1 Ω
R3 Ω
R5 Ω
R2 Ω
R4 Ω
R6 Ω
110 V
Is this a series or a parallel circuit?
Combine the resistors in series first.
Then re-draw the circuit.
It should look something like this
Both
Now find the total resistance
Do we need the total current? If
so, what is it?
How is resistor #5 hooked up?
Those two resistors combine to
form a resistor that is _____ Ω
How is that 5 Ω resistor hooked up
to the circuit?
2.4 Ω
No
In series with #6
5Ω
In parallel with the
other combined
resistors
More about Resistor #5
What is the same for all the lines in a
parallel circuit?
What is the voltage across the 5 Ω
resistor?
What is the current through that 5 Ω
resistor?
What is the current through resistor #5?
Why?
V = IR V = 22(3)
Voltage
110 V
22 amps
22 amps
Series w/ #6
66 V
Let’s look at Resistor #4
How is resistor #4 hooked up?
Those two resistors combine to
form a resistor that is __ Ω
How is that 11 Ω resistor
hooked up to the circuit?
What is the same for all the
lines in a parallel circuit?
What is the voltage across the
11 Ω resistor?
In series with #3
11 Ω
Parallel with the other
combined resistors
Voltage
110 V
More about Resistor #4
What is the current through
that 11 Ω resistor?
What is the current through
resistor #4?
Why?
10 amps
10 amps
Series w/ #3
Now move on to the rest of the packet.
We have the answers at all the problems.
We have the solutions to #17 - #22.
#23 is extra credit (it is really hard)