School Physics notes: Series & parallel circuits - design explained & calculations

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?

1. Introduction and a simple series circuit

2.

4. Investigating simple parallel circuit

5. Looking at the total resistance in a parallel circuit

7. Extra QUESTIONS on series and parallel circuits

See also APPENDIX 1 for a summary of all the electricity equations you may need.

- what you need to learn about a SERIES CIRCUITS

• Know and understand that for components connected in series:

• the total resistance is the sum of the resistances of each component,

• the same value of current passes through each component in a series circuit - identical current at any point in a series circuit,

• the total potential difference of the supply is shared between the components - so each component has a different p.d. unless two resistors are identical, so they would have the same p.d. across them.

• Reminder: Two lamps wired in series, the two cells and ammeter are also wired in series.

• Know and understand that the potential difference provided by cells connected in series is the sum of the potential differences of each cell (cells should be connected + - + - etc., so non cancel each other out).

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.

TOP OF PAGE and sub-index 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.

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3. How to investigate the effect on the total resistance of a circuit by adding identical resistors together 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)

• Know and understand that for components connected in parallel:

• the potential difference across each component is the same,

• the current flow is split, but the total current through the whole circuit is the sum of the currents through the separate components,

• and the current flow through each resistor is different, unless any of the resistors have an identical resistance - in which case they will both experience the same current flow.

• each component wired in parallel can act independently, i.e. each one can be switched on and off irrespective of the others AND if a component fails, the others still work!

• Reminder: 1 ammeter, 1 switch, 2 cells in series with pairs of ammeters and bulbs wired in parallel.

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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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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6. How to investigate the effect on the total resistance of a circuit by adding identical resistors together 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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