Electricity Packet 4
Series and Parallel Circuits
Honors Physics
Reiter 2021
There are 2 basic ways to connect resistors in a circuit: series and parallel. First, we’ll learn pure series and
pure parallel concepts. Then, we’ll combine series and parallel connections, creating more complex circuits.
What does it mean to be connected in series?
Resistors are connected in series when they look like this:
etc for more than 3
You can think of it spatially in terms of “end – to – end” connection. A better way to think of it is to see that
there is only one path for current to flow through these resistors. All the charges must flow through all three
resistors.
What does a series circuit look like?
+
R2
R1
R3
VT
Like this
-
What are the current (meaning i) relationships in a series circuit?
When thinking of current in series and parallel circuits, it is helpful to consider the flow of charge as
analogous to the flow of water in pipes. Since there is only one path for the current to flow through in a
series circuit, this current must be the same in all three resistors. Just like if this were a bunch of pipes, the
amount of flowing water would be the same everywhere.
i1
i2
iT
+
iT = i1 = i2 = i3
i3
R2
R1
where: iT is the total current coming out of the battery
i1 is the current through R1
etc
R3
VT
-
What are the voltage relationships in a series circuit?
As usual, voltage is a bit trickier than current. Here’s how to think of it. The battery picks up charges from a
voltage level of 0 volts at its - side and pumps them up to a higher total voltage level, VT. Let’s say VT
is 12 volts in this case (like in your car). So the battery raises the energy level of the charges to 12 volts at
its + side. Then, the battery pushes the charges through all three resistors, one after the other. When the
charges get through the third resistor, they must be back to a voltage level of 0 volts, since that’s a wire along
the bottom of the circuit and voltage level doesn’t change in a wire. So if the charges start at a voltage level
of 12 volts at the beginning of the first resistor, and end at a voltage level of 0 volts after the third resistor,
they must lose all of that 12 volts along the way. They lose some of it in the first resistor, some more in the
second resistor, and the rest in the third resistor. So the voltage change across all three resistors must add up
to the battery voltage, 12 volts in this case.
+
+ ∆VR1 -
+ ∆VR2 -
R1
R2
R3
VT
-
VT = ∆VR1 + ∆VR2 + ∆VR3
+
∆VR3
-
where: VT is the voltage level at the + side of the battery
∆VR1 is the voltage change across R1
etc
1
What does it mean to be connected in parallel?
Resistors are connected in parallel when they look like either of these two configurations:
etc for more than 3
You can think of it spatially in terms of a “ladder” connection. A better way to think of it is to see that there
are multiple paths for current to flow through these resistors. This time, the moving charges (ie the current)
must split into different paths, some flowing though one resistor, some through the next, and the remaining
charges through the last resistor.
What does a parallel circuit look like?
+
Like this:
R1
VT
R2
R3
-
What are the current (meaning i) relationships in a parallel circuit?
Thinking of current as the flow of water in pipes, the total current coming out of the battery must split into
three paths (or pipes), one for each resistor. Since pipes don’t leak, if you add up the current though each
resistor, you must get the total current coming out of the battery.
i1
iT
+
R2
R1
VT
iT = i1 + i2 + i3
i3
i2
where:
iT is the total current coming out of the battery
i1 is the current through R1
etc
R3
-
What are the voltage relationships in a parallel circuit?
Parallel resistors all must have the same voltage change across them. The key to understanding voltage
relationships in parallel circuits is to remember that voltage level doesn’t change in a given section of wire.
So if our battery is again 12 volts, all wire which has an uninterrupted path to the + side of the battery is also
at a voltage level of 12 volts. Similarly, all wire which has an uninterrupted path to the – side of the battery
thus has a voltage level of 0 volts. Meaning, the voltage levels of the parallel circuit look like this:
= a voltage level of 12 volts
+
VT = 12 volts
-
+
∆VR1
-
+
R1
∆VR2
-
+
R2
∆VR3
-
R3
= a voltage level of 0 volts
Thus, all three resistors have a high side voltage level of 12 volts and a low side voltage level of 0 volts,
making:
∆VR1 = ∆VR2 =∆VR3 In this particular circuit, that ∆V is the same as the battery voltage, 12 volts
2
How do I combine resistors in series?
If you have a group of resistors in series, you can create a single equivalent resistor which has the same
effect on the rest of the circuit as the original configuration of multiple series resistors. You do so merely by
adding up the series resistors.
REq series = R1 + R2 + R3 etc
Example 1
What is the equivalent single resistor circuit of the following series circuit?
iT
+
R2 = 3 Ω
R1 = 2 Ω
R3 = 4 Ω
VT
-
You must pattern recognize the resistors as being
connected in series. Then you calculate the equivalent
single resistance as follows:
Original Circuit
REq series = R1 + R2 + R3
REq = 2 + 3 + 4
REq = 9 Ω
iT
+
REq
9Ω
VT
-
Equivalent Circuit
Meaning the battery cannot tell the difference between the original circuit and
the equivalent one, which means: The total current iT coming out of battery is
the same in the equivalent circuit as in the original circuit.
This is important. We make use of this fact all the time in problem solving
How do I combine resistors in parallel?
If you have a group of resistors in parallel, you can again create a single equivalent resistor which has the
same effect on the rest of the circuit as the original configuration of multiple parallel resistors. But it’s a bit
more complicated this time. This time, you must add the reciprocals of the individual resistors to get the
reciprocal of the equivalent resistor.
1
1
1
1
=
+
+
𝑒𝑡𝑐
𝑅𝐸𝑞 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙
𝑅1 𝑅2 𝑅3
Example 2
What is the equivalent single resistor circuit of the following parallel circuit?
Original Circuit
iT
+
VT
-
R1
2Ω
R2
3Ω
R3
4Ω
You must pattern recognize the resistors as being
connected in parallel. Then you calculate the equivalent
single resistance as follows:
1
𝑅𝐸𝑞 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙
1
iT
+
𝑅𝐸𝑞 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙
REq = 0.92 Ω
VT
-
1
𝑅𝐸𝑞 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙
=
13
12
=
=
1
1
𝑅1
1
2
1
+𝑅 +𝑅
2
1
3
1
+3+4
but that’s 𝑅
1
𝐸𝑞 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙
Common
Mistake!!!
, not REq
Equivalent Circuit
𝑅𝐸𝑞 =
12
13
= 0.92 Ω
Meaning the total current iT coming out of battery is the same in the equivalent circuit as in the original.
3
Current into a node must equal current out of a node
One more concept is helpful when doing circuit problems. We’ve already kind of covered it, but now we
will explicitly state it. At any place in a circuit where current can split, the total current coming into that
place, called a node, must equal the total current going out. In other words, current doesn’t leak, but it can
split. Let’s look at our parallel circuit again.
iT
iT
+
VT
A iA
i1 A
R1
R2
R3
-
Look at node A. As usual, the current coming out of the battery is iT meaning the total current being
supplied to the circuit. Since there is no branch path for the current to take between the battery and Node A,
all of iT must show up at Node A since current doesn’t leak.
But at Node A, there is a choice of two different paths for the current to take. The total current splits and
some of it (i1) goes through R1 while the rest of it continues along toward R2 and R3. I called the current that
continues along toward R2 and R3, iA. (There’s nothing special about that name, you could call it anything
you want.) Here’s the concept: iT must equal i1 + iA. Current still doesn’t leak, but it can split into different
paths (or pipes).
Circuits Problem Solving Procedure
A typical problem is: Given a circuit, find the current through, or voltage change across a given resistor.
The basic problem solving procedure goes like this.
Build an Ohm’s Law table. (I’ll show how in the next example)
Combine resistors using series and parallel techniques to get to a single equivalent resistor REq
Use REq to get iT (or occasionally VT, but iT is much more likely)
Use iT together with concepts of current and voltage in parallel and series along with current at
nodes to fill in Ohm’s Law table (this is the hard part)
5. Your answer is in your Ohm’s Law table.
1.
2.
3.
4.
See next page for an example.
4
Example 3 In the following circuit, what is the current through R2 and what is the voltage change across R1?
Step 1: Build an Ohm’s Law table as follows:
+
R1 = 5 Ω
R2 = 3 Ω
VT = 16 v
-
Create 3 columns: ∆V, i, and R . You will build a line for
each individual resistor in the circuit as well as for the single
equivalent resistor REq. Fill in the individual resistors
∆V (volts)
R (Ω)
R1 = 5
R2= 3
REq
i (A)
Next, the ∆V for the REq line is the total voltage,
which is the battery voltage. So fill it in next.
VT = 16
You would
only make
one table
and fill it in
as you go.
Step 2: Calculate the single equivalent resistor REq.
∆V (volts)
I (A)
You must be able to pattern recognize this as a series circuit
and recall that resistors add in a series circuit.
VT= 16
So REq = R1 + R2
REq = 5 Ω + 3 Ω = 8 Ω Enter 8 Ω for REq in your table
Step 3: Calculate iT.
R (Ω)
R1 = 5
R2 = 3
REq = 8
IREq
+
iT
REq = 8 Ω
VT = 16 volts
-
You have simplified your circuit to this:
Equivalent Circuit
The whole reason you do this is so you can calculate iT. Here’s the logic of how to do this. (See Pkt 3, Ex 2)
iT is the total current leaving the battery. Since there are no different paths for those charges to take between
battery and REq, all of iT goes through REq, meaning iT = iReq.
Also, VT = ∆VREq. (see Electricity Packet 3, Example 2 (pg 6) for the logic if this statement confuses you)
Therefore you can use Ohm’s Law as follows:
∆VREq = (iREq)x(REq)
16 volts = (iT)x(8 Ω)
iT = 2 A
Put it in the table
Step 4:
∆V (volts)
VT = 16
I (A)
iT = 2
R (Ω)
R1 = 5
R2 = 3
REq = 8
Use iT together with concepts of current and voltage in parallel and series along with current
at nodes to fill in Ohm’s Law table
Up to this point, the first 3 steps are pretty much the same each time. The only difference is how to calculate
REq. But now, you have to think. Go back and look to the original circuit. It’s a series circuit. Recall from
page 1 of this packet that the current (i) is the same everywhere in a series circuit. (go back and re-read that
part if this confuses you). Therefore, iT = iR1 = iR2
∆V (volts)
I (A)
R (Ω)
iR1 = 2
R1 = 5
Since you just found iT to be 2 A,
iR2 = 2
R2 = 3
you can now fill in iR1 and iR2 as 2 A also
VT = 16
iT = 2
REq = 8
Since you now have the current through each of R1 and R2, use Ohm’s Law on each line of the table to
calculate the change of voltage across R1 and R2.
∆V (volts)
I (A)
R (Ω)
∆VR1 = (iR1)x(R1) and ∆VR2 = (iR2)x(R2)
∆V
=
10
i
=
2
R
R1
R1
1=5
∆VR1 = (2)x(5)
∆VR2 = (2)x(3)
∆VR2 = 6
iR2 = 2
R2 = 3
∆VR1 = 10 volts
∆VR2 = 6 volts
Finish the table!
VT = 16
iT = 2
REq = 8
Now, read your answer from the table: the current through R1
is 2 A and the voltage change across R2 is 6 volts.
5
Example 4 In the following circuit, what is the current through R2 and what is the voltage change across R1?
Step 1: Build an Ohm’s Law table
+
VT
16 volts -
R1
5Ω
∆V (volts)
R2
3Ω
i (A)
VT = 16
Step 2: Calculate the single equivalent resistor REq.
∆V (volts)
You must be able to pattern recognize this as a parallel circuit
and recall that resistors add in reciprocal in a parallel circuit.
VT = 16
1
𝑅𝐸𝑞 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙
=
1
𝑅1
1
2
Step 3: Calculate iT.
Step 4:
1
1
+ 𝑅 = 5 + 3 ≅ 0.53
R (Ω)
R1 = 5
R2 = 3
REq
R (Ω)
R1 = 5
R2 = 3
REq = 1.88
I (A)
1
1
Be careful here. 0.53 is not REq. It is 𝑅 . So 𝑅𝐸𝑞 = 0.53 ≅ 1.88 Ω
∆VREq = (iREq)x(REq)
16 volts = (iT)x(1.88 Ω)
iT = 8.51 A
Put it in the table
𝐸𝑞
∆V (volts)
VT= 16
I (A)
iT = 8.51
R (Ω)
R1 = 5
R2 = 3
REq = 1.88
Use iT together with concepts of current and voltage in parallel and series along with current
at nodes to fill in Ohm’s Law table
So far, this has been much the same as the previous example, except that you found REq slightly differently.
But this step is where the unique thinking occurs. Again, go back and look to the original circuit. This time,
it’s a parallel circuit. Recall from page 2 of this packet that the voltage change is the same for resistors in
parallel (again, go back and re-read that part if this confuses you). Therefore, ∆VR1 = ∆VR2 =VT
Since you VT is 16 V, so is ∆VR1 and ∆VR2
Put them in the table.
∆V (volts)
∆VR1 = 16
∆VR2 = 16
VT = 16
I (A)
iT = 8.51
R (Ω)
R1 = 5
R2 = 3
REq = 1.88
Since you now have the voltage change across each of R1 and R2, use Ohm’s Law on each line of the table to
calculate the current through R1 and R2.
∆VR1 = (iR1)x(R1)
and ∆VR2 = (iR2)x(R2)
∆V (volts)
I (A)
R (Ω)
16 = (iR1)x(5)
16 = (iR2)x(3)
∆VR1 = 16
iR1 = 3.2
R1 = 5
iR1= 3.2 As
iR2 = 5.33 A
Finish the table!
∆VR2 = 16
iR2 = 5.33
R2 = 3
Now, read your answer from the table: The current through
VT = 16
iT = 8.51
REq = 1.88
R2 is 5.33 A, and the voltage change across R1 is 16 volts.
Note: You could have done this problem without ever finding REq. In fact, you don’t even need the table.
You could look at this circuit and apply your knowledge of parallel circuit theory and realize that ∆VR1 =
∆VR2 = 16 volts just by looking at the original circuit. That’s the answer to the 2nd part of the question. You
could then realize that since you have the voltage change across R1 and the value of R1 itself, you can apply
Ohm’s Law to R1 and calculate the current through R1 directly. That is a perfectly valid approach.
Most people prefer to work with the table. It provides you with a structured approach, which is especially
helpful as the problems get more complex, as in the next example…
6
Example 5 In the following circuit, what is the current through R2 and what is the voltage change across R1?
Step 1: Build an Ohm’s Law table
R1 = 2.6 Ω
+
R2
4Ω
VT
18 volts -
∆V (volts)
R3
6Ω
R (Ω)
R1 = 2.6
R2 = 4
R3 = 6
REq
i (A)
VT = 18
Step 2: Calculate the single equivalent resistor REq.
This circuit is neither purely series, nor purely parallel. You must take 2 steps to find REq. First, recognize
R2 and R3 as a parallel combination and combine them into a REq parallel. Then, realize this REq parallel is now in
series with R1, and so you combine those two by adding them together.
1
𝑅𝐸𝑞 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙
=
1
𝑅2
1
1
1
1
+ 𝑅 = 4 + 6 ≅ 0.417 So 𝑅𝐸𝑞 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 = 0.417 ≅ 2.4 Ω
3
so now the circuit looks like this:
VT
18 volts
+
R1 = 2.6 Ω
REq parallel
2.4 Ω
-
Step 3: Calculate iT.
Step 4:
∆V (volts)
then, R1 + REq parallel
are in series, so
REq = R1 + REq parallel
REq = 2.6 Ω + 2.4 Ω = 5 Ω
VT = 18
i (A)
iT = 3.6
R (Ω)
R1 = 2.6
R2 = 4
R3 = 6
REq = 5
∆VREq = (iREq)x(REq)
18 volts = (iT)x(5 Ω)
iT = 3.6 A
Use iT together with concepts of current and voltage in parallel and series along with current
at nodes to fill in Ohm’s Law table
Again, this is the unique thinking step. Looking at the original circuit, we just determined that the total
current leaving the battery is 3.6 A (that’s what iT is: the total current leaving the battery). Since there is no
other path for the charges to flow between the battery and R1, all of that iT must also flow through R1,
meaning iR1 = iT = 3.6 A. So you can put that in the table, but you can’t say the same thing about R2 and R3.
When the current leaves R1, some of it splits off and goes through R2, and the rest of it goes through R3, so
you can’t call iR2 or iR3 3.6 A. But once you have 2 numbers on a line, you can get the 3rd with Ohm’s Law:
so
∆VR1=(iR1)(R1)
∆V (volts)
i (A)
R (Ω)
∆VR1=(3.6)(2.6) = 9.36 volts. So line 1 is done
∆V = 9.36 i = 3.6
R = 2.6
R1
R1
1
R2 = 4
R3 = 6
REq = 5
Now, here’s how you need to think next. This part is kind of new.
Since the left or “high” side of R1 is at a voltage level of 18 volts,
VT = 18
iT = 3.6
and there is a voltage change of 9.36 volts across R1, then the
voltage level on the right or “low” side of R1 is found as follows: (See Packet 3, Example 1, pg 5, if needed)
∆VR1= VR1 high - VR1 low
so: 9.36 = 18 - VR1 low
making: VR1 low = 8.64 volts.
This is the voltage level on the right or “low” side of R1 It is also the voltage level on the top or “high” side
of R2. The bottom or “low” side of R2 is 0 volts (see pg 2, this packet). Thus the voltage change across R2 is
∆VR2= VR2 high - VR2 low = 8.64 – 0 = 8.64 volts. And since R3 is in parallel with R2, ∆VR3 is also 8.64 volts.
That give you two numbers on the remaining lines, meaning you can use Ohm’s Law to finish the table
7
`
∆VR2=(iR2)(R2)
8.64=(iR2)(4)
iR2= 2.16 A
∆VR3=(iR3)(R3)
8.64=(iR3)(6)
iR2= 1.44 A
∆V (volts)
∆VR1= 9.36
∆VR2= 8.64
∆VR3= 8.64
VT = 18
i (A)
iR1 = 3.6
iR2= 2.16
iR2= 1.44
iT = 3.6
R (Ω)
R1 = 2.6
R2 = 4
R3 = 6
REq = 5
The question asked for voltage change across R1 and current through R2. You can read those off from the
table. (You could have answered the first part a few steps ago)
This is a hard problem. It pulls together all of the circuits concepts we’ve covered. If you understand all of
it the first time you look at it, you’re doing well. Most people need a few laps around the track on this one.
But it, or a similar version, is a classic quiz / test / final exam problem.
Again, you don’t continually redraw the Ohm’s Law table as I have been doing in these examples. You draw
it once, and fill it in as you go, similar to 10 item data tables in projectile motion days.
Most people need to read this packet more than once.
8
Practice Questions
R1
1. What is the equivalent single resistor?
200 Ω
+
R2
400 Ω
VT
-
R1
2. What is the equivalent single resistor?
R2
3Ω
17 Ω
+
R3
6Ω
VT
-
3. What is the equivalent single resistor?
+
R1
12 Ω
VT
R2
18 Ω
-
4. What is the equivalent single resistor?
R1
1Ω
+
VT
R3
3Ω
R2
2Ω
-
R1
5. What is the equivalent single resistor?
8Ω
+
18 volts
R2
5Ω
R3
9Ω
-
R1 7 Ω
R3
6. What is the equivalent single resistor?
VT
+
8Ω
R2 5 Ω
R4
2Ω
-
9
7. In a pure series combination of resistors, is REq >, <, =, or unable to determine compared to any
individual resistor in the combination?
8. In a pure parallel combination of resistors, is REq >, <, =, or unable to determine compared to any
individual resistor in the combination?
9. What is the current through R1
R1 = 2 Ω
+
R2 = 4 Ω
VT = 5 v
-
10. Consider the following circuit
a. What is the voltage change across R2
b. What is the voltage level between R1 and R2
11. What is the voltage change across R1
1.5
volts
R1 = 3 Ω
30 +
volts
-
R3
7Ω
R1
12 Ω
+
R2 = 2 Ω
R2
18 Ω
-
12. What is the current through R3
7
volts
+
-
R1
2Ω
R2
3Ω
R3
4Ω
10
13. What is the voltage change across R2
R1 = 6 Ω
+
R2
3Ω
12 volts
-
R3
6Ω
R1 = 8 Ω
14. What is the current through R1?
+
12 volts
R3
4Ω
R2 = 6 Ω
-
15. What is the voltage level at Node A?
R1 4 Ω
A
R3
18 volts
+
8Ω
R2 4 Ω
R4
2Ω
-
11
R2 = 4 Ω
16. What is the voltage change across R2?
R1
6Ω
R3
2Ω
R4
3Ω
R5
6Ω
R6
3Ω
+
30 volts
-
R7
5Ω
17. What is the current through R3?
3Ω
10 Ω
4Ω
R3 = 6 Ω
6Ω
+
48 volts
-
5Ω
Answers:
1. 600 Ω
2. 26 Ω
3. 7.2 Ω
4. 0.55 Ω
5. 11.21 Ω
6. 12.92 Ω
9. 0.83 A
10. a. 5 volts
11. 1.5 volts
12. 1.75 A
13. 3 volts
14. 0.69 A
15. 15 volts
16. 10 volts
17. 0.33 A
3Ω
b. 22.5 volts
12