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ELECTRICITY 5: More on series circuits and parallel circuits, circuit diagrams,
measurements & calculations
Doc Brown's Physics Revision
Notes
Suitable for GCSE/IGCSE Physics/Science courses or
their equivalent
This page will answer help you answer questions e.g. What is the formula for Ohm's Law relating
potential difference, current flow and resistance? What is the difference between series and
parallel in electrical circuits? How do you calculate potential differences,
resistances, current flows using Ohm's Law?
Sub-index for this page
1.
Introduction
and a simple series circuit
2.
Calculations involving a
series circuit
3.
The total resistance of
a circuit by adding identical resistors together in series
4.
Investigating
simple parallel circuit
5.
Looking at the total resistance
in a parallel circuit
6.
The total
resistance of a circuit on adding identical resistors together in parallel
7.
Extra QUESTIONS on series and parallel circuits
See also
APPENDIX 1 for a summary of all
the electricity
equations you may need.
1. Introduction - what you
need to learn about a SERIES CIRCUITS
AND
Watch out for different styles of circuit diagrams -
follow the wires and follow the logic!
e.g. look at the two styles on the circuit 34 and 35
diagrams, the first two circuits I'm analysing on this page!
Make sure you study circuit 34 before
circuit 35 - important rules and reminders are first introduced for this
page.
NOTE: Everything you need to know
about series and parallel is included in the first two examples.
There are some questions
on series and parallel circuits at the end.
A
simple series circuit (compare with the
parallel circuit 35
further down)
In a series circuit all the different
components are connected 'end to end' between the terminals of the power
supply e.g. the +ve and -ve terminals of a dc supply (as in the circuit
diagram 34 below).
In a series circuit, if you break the circuit at any
point i.e. disconnect a component, everything stops working - no
complete circuit, therefore no current flow.
You can include a variable resistor
in the circuit to vary the p.d. and current to make a wide range of
readings to verify the pattern of resistance - which should stay
constant if they don't heat up according to
Ohm's Law.
BUT, you can investigate the relative
ratio of the potential differences across the two resistors.
 SERIES circuit
R1 and R2 are resistors
(e.g. any device/resistance) wired consecutively end to end, this is
what wiring in series means. I'm assuming the circuit is
powered by a 12.0 V d.c. supply from batteries or power pack.
Reminder: If any part of a
series circuit is disconnected, then the whole circuit fails - no
current can flow
Three ammeters, A1, A2
and A3 are wired in series with everything else -
giving 3 readings I1, I2 and I3
A (amps).
Ammeters are always wired in
series with any component whose current is being measured.
Voltmeters V1 and V2
are wired in
parallel to measure the potential difference (p.d.) across each resistor
- component.
Reminder: Wiring in parallel means each component is
independently connected to the positive and negative terminals of the power
supply - on separate loops ...
... AND this is how you must
always wire a voltmeter in a circuit.
... noting (i) that a
voltmeter has a VERY high resistance, so virtually no current
flows through it, and so the voltmeter will not affect the ammeter
readings,
and (ii) an ammeter is always
wired in series in any part of a circuit (series or parallel
circuits), and has a very LOW resistance, so there virtually
no potential difference across it and so the ammeter will not affect
the voltmeter readings.
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2.
Calculations involving a
series circuit
Using some basic rules and Ohm's Law (V = IR)
I'll
show you how to work out all sorts of things from just only the three pieces
of data give.
(a) The total resistance of the two resistors
in series (2.0 Ω and 6.0 Ω)
When resistance are wired in series, you add them up to
get the total resistance.
Rtotal = R1 + R2 =
6.0 + 2.0 =
8.0 Ω
(ohms)
 (b) The
current flowing in the circuit
We know the total p.d. (Vtotal) across the
resistors part of the circuit and we know the total resistance.
Vtotal = 12.0 V
(marked on power supply) and Rtotal = 8.0 Ω
(calculated in (a)).
Also, in a series circuit, the current is the same at
any point in the circuit.
So, using Ohm's Law: V = IR, so I = Vtotal/Rtotal
= 12.0/8.0 =
1.5
A (amps)
Therefore readings: I1 = I2 = I3
= 1.5 A
(c) We can now calculate the p.d
across each resistor using Ohm's Law V = IR
V1 = 1.5 x 6.0 =
9.0 V
and V2 = 1.5 x 2.0 =
3.0 V
Noting the total p.d. of the circuit
is shared between the various components of the system,
AND the p.d. across a resistor
is proportional to the resistance value,
(check out ... 6 Ω and 9 V
for resistor 1, similarly 2 Ω and 3 V for resistor 2)
so, bigger the
resistance, the greater the potential difference across it,
AND the p.d. values for components
wired in series add up to the maximum source p.d.
(which is 12.0 V here)
So, Vtotal = V1
+ V2 = 12.0 = 9.0 + 3.0 (which gives you a check on
the previous calculation!)
When two or more resistances are
wired in series they all have to share the total p.d. of the circuit.
The higher the value of the
resistance, the bigger its share of the total source p.d. in other
words the p.d. is always shared between resistances/components wired in
series.
The p.d. across each resistor must be
lower than Vtotal, BUT, the current is still the same around
all points of the circuit.
However, as you increase the number
of resistances in the series, the current is reduced (I = V/R).
If you carry out the experiment, the
readings should match in principle the theoretical values above.
If you don't know the value of a
resistance, using the above type of circuit, you can measure the current
lowing through it and the p.d. across it and then use Ohm's Law R = V/I to
calculate it.
See also 3.
Ohm's Law, experimental investigations of
resistance, I-V graphs and calculations using I = V/R
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and sub-index
3. How to investigate the effect on the total resistance of
a circuit by adding identical resistors together in series
(a) Adding resistors in series
You set up a relatively simple circuit of
a power supply, ammeter and one or more identical resistors, all wired in
series.
You start by putting one resistor in
the circuit and measuring the current flowing through it (I) and the
p.d. across it (V).
Then, add a second resistor in series
and remeasure the p.d. and current again. Just repeat by adding more
resistors and for each pair of readings calculate the total resistance
(R = V/I).
From your recorded results plot a
graph of total resistance versus number of resistors in series.
You find the graph is linear
proving the rule that: Rtotal = R1 + R2
+ R3 etc. for resistors wired in series.
In other words the total
resistance steadily increases the more resistors you connect in series,
and you should notice that the
current flowing steadily decreases as the total resistance increases.
Water flow analogy: The longer the
thin pipe the greater the friction is to oppose the flow.
(b)
comparison with parallel
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4. Investigating
simple parallel circuit (compare with the
series circuit 34
above)
In an exam you might not be told whether the
components are wired in series or parallel - you are expected to know!
In a parallel circuit, each component is
connected separately to the terminals of the power supply e.g. the +ve and
-ve terminals of a dc supply (as in the circuit diagram 35 below).
If you disconnect one component, e.g. one
of the resistors, the other components will keep working.
That is why most electrical appliances
include parallel circuit systems and enables you to switch individual things
'on and off' without affecting other devices.
Practically everything in a car is
wired in parallel and is independently controllable e.g. heater,
windscreen wiper, headlamps etc. It would be bad news if one thing
failed and the rest did the same!
In reality, most circuits are a mixture of series and
parallel wiring.
 PARALLEL circuit
You can include a variable resistor in
the circuit to vary the p.d. and current and make a varied series of
readings to verify the pattern of resistance - which should stay constant if
they don't heat up ...
... and check that the potential
difference across each resistor is the same.
R1 and R2 are resistors
(e.g. any device/resistance) wired in parallel and again I'm assuming the circuit is
powered by a 12.0 V d.c. supply from batteries or power pack.
Reminder: Apart from the
resistors wired in parallel, if any other part of the circuit is
disconnected, then the whole circuit fails - no current can flow
Reminder: Wiring in parallel means each component is
independently connected to the positive and negative terminals of the power
supply - in this case two resistors on separate loops ...
... therefore if either
resistor 1 or 2 fails or removed-disconnected braking that part of the
circuit, the other resistor will continue to function. This is
one reason why parallel circuits tend to be more useful than series
circuits.
 It
also means you can control each component independently of the other.
For example, look at circuit 14 on
the right (ignoring the ammeters!).
This is a simple way to control two
lamps wired in parallel.
After closing switch s3 (power supply
on) you can close s1 or s2 to select the lamp to be lit b1 or b2.
Actually, the vast majority of
circuits of practical use include a mixture of series and parallel
wiring.
Ammeters: Each resistor R1
and R2 is wired in series with its own ammeter A1
and A2.
Two ammeters, A3,
and A4 are wired in series with everything
else.
Therefore the possible ammeter
readings are: I1, I2 and
I3
A (amps).
Ammeters are always wired in
series with any component whose current is being measured.
Voltmeters V1 and V2
are wired in parallel with each resistor to measure the potential difference (p.d.) across
it.
Reminder: Wiring in parallel means each component is
independently connected to the positive and negative terminals of the power
supply - i.e. on separate loops ...
AND this is how you must always
wire a voltmeter in a circuit,
and ammeters are always wired
in series, at any point in any circuit.
Using some basic rules and Ohm's Law (V = IR) I'll
show you how to work out all sorts of things from just only the three pieces
of data give.
(a)
The current flowing through each resistor and the rest of the circuit
Since the resistors are wired in
parallel, they have the same maximum source p.d. across them (V1
= V2 = 12.0 V
in this case).
(You couldn't say this for the series
circuit where the total p.d. is divided up between the resistors.)
Another consequence, of wiring
components in parallel, is that the current is split between the
two or more component sections ('loops') of that part of the
circuit.
Calculations using Ohm's Law equation:
I
= V/R
I1 = Vtotal/R1
= 12.0/6.0 =
2.0
A and I2
= Vtotal/R2 = 12.0/2.0 =
6.0
A
You can add together the separate currents in
the parallel part of the circuit to obtain the total current, which must
be the same in the rest of the circuit.
I3 = I4 = I1
+ I2 = 2.0 + 6.0 =
8.0
A
Note:
(i) The current is shared
between the two branches in the circuit.
(ii) The larger the resistance,
the smaller the current:
The 2 ohm allows a flow of 6.0 A
and the 6 ohm allows 2.0 A.
(iii) Comparison with the series
circuit 34 shows that the two same resistances wired in parallel,
offer a much smaller resistance than when they are wired in series
and the current is greater.
(iv) If the two resistors R1
and R2 had
the same resistance, ammeter readings I1 and I2 would be
equal, for example if they
were two identical bulbs, they would glow with the same brightness.
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5.
Looking at the total resistance
in a parallel circuit
We know the p.d. is 12.0 V and
we now know from (a) the total current flowing is 8.0 A.
So, again using Ohm's Law: V = IR, R
= V/I = 12.0/8.0 =
1.5 Ω
Note:
(i) When two resistors
are wired in parallel, their combined resistance is much less than
when wired in series - in fact it is lower than the lowest
individual resistance.
In the series circuit 34, the
total resistance was 8 ohms, here in circuit 35 when the
resistors are wired in
parallel, it is only 1.5 ohms and less than any of the
individual resistances (2 ohms and 6 ohms).
The water pipe analogy helps
here - think of the water being able to pass through two pipes of
similar diameter.
(ii)
NOT ON THE GCSE
specification, (as far as I know), is the formula to calculate the
total resistance of resistors in parallel.
The reciprocal of the total
resistance = the total of all the reciprocals of the individual
resistances added together.
1/Rtotal = 1/R1
+ 1/R2 ... etc.
e.g. for the example above a 2
ohm resistor was wired in parallel with a 6 ohm resistor:
1/Rtotal = 1/R1 + 1/R2
1/Rtotal = 1/2 + 1/6 = 4/6 =
2/3 = 1/1.5, so Rtotal = 1.5/1 =
1.5 Ω
and note that the total
resistance is less than the smallest resistance in the circuit.
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and sub-index
6. How to investigate the effect on the total resistance of
a circuit by adding identical resistors together in parallel
(b) Adding resistors in parallel
You set up a relatively simple
circuit of a power supply, ammeter and one or more identical resistors,
all wired in parallel. You can use filament bulbs as the resistors.
You start by putting one resistor in
the circuit and measuring the current flowing through it (I) and the
p.d. across it (V).
Then, add a second resistor in
parallel and remeasure the p.d. and current again. Just repeat by adding
more resistors and for each pair of readings calculate the total
resistance (R = V/I).
From your recorded results plot a
graph of total resistance versus number of resistors in
parallel.
You should find the graph differs
from the series graph in two ways.
(i) Its a curve, not linear,
(ii) and most significantly, the
total resistance dramatically decreases the more resistors you have
wired in parallel.
Therefore the more resistors
wired in parallel, the lower the total resistance.
This is the complete opposite
effect, compared to resistances wired in series.
Water flow analogy: There are more
'shorter' pipes for the water to flow through offers less resistance so
the total water flow can increase.
compared to series
7. QUESTIONS on series and parallel circuits
Q1
From circuit diagram 38 deduce ...
(a) What is the total resistance of
the circuit?
The resistors are in series so Rtotal
= 5.0 + 8.5 = 13.5 Ω =
14 Ω (2 sf)
(b) Predict what current the ammeter
should show.
From Ohm's Law equation: V = IR,
so I = V/R = 12.0/13.5 =
0.89 A
(2 sf)
(c) If a third resistance of 3.5 Ω
is connected between the 5 and 8.5 ohm resistors, what will be the total
resistance and current flow?
The resistors are in series so Rtotal
= 5.0 + 8.5 + 3.5 = 13.5 Ω =
17 Ω (2 sf)
From Ohm's Law equation: V =
IR, so I = V/R = 12/17 =
0.71 A
(2 sf)
(d) What third resistance would have
to added to the circuit to reduce the current flow to around 0.2 A?
Rtotal = V/I =
12/0.2 = 60 Ω
R3 = Rtotal
- R1 - R2 = 60 -5.0 - 8.5 = 46.5 Ω =
47 Ω
(2 sf)
Q2
Study the diagram of circuit 36 carefully
and for each question part explain or justify your answer.
(a) Predict the reading on voltmeter
V2?
When components are wired in
series you can add up the individual p.d. values to give the total.
Therefore: Vtotal = V1
+ V2 = 20.0 = 12.0 + V2, so V2 =
20.0 - 12.0 =
8.0 V
(b) What current will flow through
each resistor?
The ammeter in the main circuit
reads 5.0 A.
Since everything is wired in
series, the current at any point in the circuit is the same.
Therefore
5.0 A flows
through each resistor.
(c)
Calculate the resistance of each resistor.
From Ohm's Law equation: R = V/I
R1 = V1/5.0
= 12.0/5.0 = 2.4 Ω
R2 = V2/5.0
= 8.0.0/5.0 =
1.6 Ω
(d) What is the total resistance?
Since they are wired in series,
you can add up the consecutive resistances.
Rtotal = R1
+ R2 = 2.4 + 1.6 =
4.0 Ω
Note:
(i) The larger the resistance
the greater the p.d. across it.
(ii) A numerical check on the
R values. Rtotal = Vtotal/Itotal
= 20.0/5.0 =
4.0 Ω
Q3
Study the diagram of circuit 37 carefully
and for each question part, explain or justify your answer.
(a) Predict the ammeter reading for A2.
The total current running through the parallel
section of the circuit must equal that running through the rest of
the circuit.
So, A3 = A1 + A2,
so A2 = A3 - A1
= 5.0 - 2.0 = 3.0 A
(b)
Calculate the value of the parallel resistances R1 and
R2.
Resistors wired in parallel have the same p.d.
across them, so using Ohm's Law equation ...
R1 = V1/I1 =
20.0/2.0 = 10.0 Ω
R2 = V2/I2 =
20.0/3.0 =
6.7 Ω
(c) Calculate the total resistance of the resistor
section of the circuit.
BUT not by adding them up!
You have to use the total current
flowing and the potential difference of the battery.
Rtotal = Vtotal/Itotal
= 20.0/5.0 =
4.0 Ω
Note:
(i) The larger the resistance the
lower the current running through it (c).
(ii) The total resistance is
actually much less than any of the individual resistances in
parallel.
If the resistors were wired
in series, the total resistance would 16.7 Ω.
Again, the water pipe analogy
helps here - think of the water being able to through two pipes
of similar diameter.
Q4 If a circuit contains
two resistors of 5.0 Ω and two of 3.5 Ω, what is the
total resistance in the circuit?
Rtotal = sum of all the
resistances in series.
Rtotal = 5.0 + 5.0
+ 3.5 + 3.5 =
17.0 Ω
Q5 Three 1.5 V
batteries are wired in series with two identical bulbs wired in series.
(a) What is the total p.d. across the
bulbs?
1.5 x 3 =
4.5 V (added up in series)
(b) What is the p.d. across each bulb?
When wired in series, the total p.d.
is divided up between the resistances.
Since the bulb resistances are the
same, each will have the same p.d.
The p.d. across each bulb will be
4.5/2 = 2.3 V
(2sf)
Q6 Suppose a lamp, a buzzer and
three 1.5 volt batteries were wired in series with an ammeter and switch
closed (circuit 46 below).
The resistance of the lamp is 10.0 Ω
and the resistance of the buzzer is 5.0 Ω.
(a) What is the total p.d. across the
lamp and buzzer?
1.5 x 3 =
4.5 V
(b) What is the total resistance of
the lamp and buzzer?
They are wired in series, so you
can just add them up.
Rtotal = Rlamp
+ Rbuzzer = 10 + 5 =
15 Ω
(c) What reading would you expect on
the ammeter? and what current flows through each component?
They are both wired in series, so
same current passes through everything.
From Ohm's Law: I = V / R
= 4.5 / 15 =
0.30 A
(d) Calculate the p.d. across (i) the
lamp, and (ii) the buzzer.
From Ohm's Law: V = I x R
(i) lamp p.d.: V = 0.3 x 10 =
3.0 V
(ii) buzzer p.d.: V = 0.3 x
5.0 =
1.5 V
(e) How could you quickly check if
you had made an error in (d)?
Its a series circuit, so the
total p.d. should be 4.5 = 3.0 + 1.5
(f) If the 10 Ω lamp was replaced by
a 20 Ω lamp, what might you notice in the performance of he circuit?
(i) the lamp would glow dimmer
because of the greater resistance reducing the current flow.
(ii) although its resistance is
unchanged, the buzzer wouldn't sound as loud because of the overall
greater resistance in the circuit reducing the current flow.
(g) Suppose we now rewire the10.0 Ω
lamp and the 5.0 Ω buzzer in parallel (circuit 47 below).
The ammeter gave a reading of 1.35 A.
(i) Why is the current flow
greater than when the lamp and buzzer were wired in series?
Resistors wired in parallel
offer a smaller resistance to current flow.
Two parallel wires ('pipes')
are available
for the current to flow through.
(ii) What is the total resistance
of the lamp and buzzer? Compare with the series circuit 46 above.
R = V / I = 4.5 / 1.35 =
3.33 Ω
(3 sf)
Confirms your deduction in
(i), the resistance is much less than 15 Ω, in fact it is less
than any of the individual resistances, characteristic of a
parallel circuit compared to the series circuits involving the
same components.
(iii) If the current through the
lamp is 0.45 A, what current flows through the buzzer?
For a parallel circuit: Itotal
= Ilamp + Ibuzzer = 1.35 A
Therefore Ibuzzer
= 1.35 - 0.45 =
0.90 A
From the information you
can also calculate this from Ohm's Law
I = V / R = 4.5 / 5.0 =
0.90 A
(iv) If a second lamp, wired in
parallel with the first lamp, was added to the circuit, what changes
might, or might not happen in the p.d. across the resistors and
maximum current flow?
The p.d. across the resistors
remains the same.
However, the current flow
will increase because you now have a 3rd pathway for the current
to flow through, and the total resistance is reduced.
(h) Explain the use of a component
you add could to the circuit to vary simultaneously the brightness of
the lamp and the volume of sound from the buzzer, and what is its
symbol?
You can add a variable resistor
 to
the circuit, wired in series with the parallel sections of the lamp
and buzzer.
By increasing/decreasing the
resistance you can decrease/increase the brightness of the bulb and
loudness of the buzzer.
Q7 A 12 Ω resistor is
wired in series with a 36 Ω and connected to a 24 V power supply.
(a) What is the potential difference
across each resistor?
The p.d. across a resistor in
a series circuit is proportional to its resistance.
The total resistance is 48 Ω
For the 12 Ω resistor: p.d. = 24
x 12 / 48 = 6 V
For the 36 Ω resistor: p.d. = 24
x 36 / 48 = 18 V
Maths check: Vtot = V12Ω
+ V36Ω =
24 V
(b) What single resistor could you
use to replace the original pair?
For a series circuit: Rtot
= 12 + 36 = 48 Ω, therefore
a single 48 Ω resistor will do.
(c) What current flows through each
resistor?
Since they are wired in series,
they both experience the same current flow.
I = V / R = 24 / 48 =
0.50 A
Q8
For future questions -
lots of
examples of basic circuits
TOP OF PAGE
and sub-index
Practical work
Wiring two resistors in series or
parallel using the same power source p.d. and measuring the current and p.d.
at various points in the circuit. Compare the results with lots of
calculations of and/or current and resistance.
APPENDIX 1: Important definitions, descriptions,
formulae and
units
Note: You may/may
not (but don't worry!), have come across all of these terms, it depends
on how far your studies have got. In your course, you might not need
every formula - that's up to you to find out.
V
the potential difference (p.d., commonly called
'voltage') is the driving potential that moves the electrical charge around
a circuit - usually electrons.
Potential difference is the work done in
moving a unit of charge.
It indicates how much energy is transferred
per unit charge when a charge moves between two points in a circuit
e.g. between the terminals of a battery.
The p.d. across any part of a circuit is measured in volts,
V.
I
the current is rate of flow of electrical charge in
coulombs/second (C/s), measured in amperes (amps, A).
The quantity of electric charge transferred in
a give time = current flow in amps x time elapsed in seconds
Formula connection:
Q = It,
I = Q/t, t = Q/I, Q = electrical charge moved in
coulombs (C), time t (s)
R
the resistance in a circuit, measured in ohms (Ω).
A resistance slows down the flow of electrical charge
- it opposes the flow of electrical charge.
Formula connection:
V = IR,
I = V/R, R = V/I (This is the formula for
Ohm's Law)
P
is
the power delivered by a circuit = the
rate of energy
transfer (J/s) and is measured in watts (W).
Formula connection:
P = IV,
I = P/V, V = P/I also
P = I2R
(see also P = E/t below)
E = QV,
the energy transferred by the quantity of electric charge by a potential
difference of V volts.
energy transferred (joules) =
quantity of electric charge (coulombs) x potential difference
(volts)
Q =
E/V, V = E/Q, E = energy transfer in joules (J),
Q = electrical charge moved (C), V = p.d. (V)
E = Pt,
P = E/t, t = E/P, where P = power (W), E
= energy transferred (J), t = time taken (s)
Energy transferred in joules = power in watts
x time in seconds
Formula connection: Since E = Pt and P = IV,
energy transferred E =
IVt
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What next?
Electricity and
magnetism revision
notes index
1.
Usefulness of electricity, safety, energy transfer, cost & power calculations, P = IV = I2R,
E = Pt, E=IVt
2.
Electrical circuits and how to draw them, circuit symbols, parallel
circuits, series circuits explained
3. Ohm's Law, experimental investigations of
resistance, I-V graphs, calculations V = IR, Q = It, E = QV
4. Circuit devices and how are they used? (e.g.
thermistor and LDR), relevant graphs gcse physics revision
5. More on series and parallel circuits,
circuit diagrams, measurements and calculations
gcse physics
6. The 'National Grid' power supply, environmental
issues, use of transformers
gcse
physics revision notes
7.
Comparison of methods of generating electricity
gcse
physics revision notes (energy 6)
8. Static electricity and electric fields, uses
and dangers of static electricity gcse
physics revision notes
9.
Magnetism
- magnetic materials - temporary (induced) and permanent magnets - uses gcse
physics
10.
Electromagnetism, solenoid coils, uses of electromagnets gcse
physics revision notes
11. Motor effect of an electric current,
electric motor, loudspeaker, Fleming's left-hand rule, F = BIL
12.
Generator effect, applications e.g. generators
generating electricity and microphone
gcse
physics
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Two lamps wired in series, the two cells and ammeter are also wired
in 
1 ammeter, 1 switch, 2 cells in series with
pairs of ammeters and bulbs wired in 
