Main aims

Adding resistors in series increases the effective resistance

Adding resistors in parallel decreases the effective resistance

Lesson Summary

• Demonstration: Combining lamps in series and parallel (10 minutes)
• Discussion: Deriving formulae (20 minutes)
• Worked example: Adding resistors (15 minutes)
• Student experiment: Predicting and measuring resistance (20 minutes)
• Student questions: Practice with the formulae (30 minutes)

Demonstration: Combining lamps in series and parallel

Ask students to recall the equation which defines resistance. ( R = VI where V is the pd across a component and I is the current through it.) Connect 1, 2 and then 3 lamps in series across a supply (constant pd) showing them the reduction in current at the same pd The ratio VI has increased: adding resistors in series increases overall resistance. You may wish to calculate the resistance each time but beware that, since the temperature of the lamps will be different at different currents you are unlikely to get a simple ratio of resistances.

Repeat the experiment but this time add lamps in parallel. The current increases and the effective (load) resistance decreases. If you calculate the resistance values you should get 1:1/2:1/3 as you add the lamps.

Episode 114-1: Connecting lamps in parallel and in series (Word, 34 KB)

Discussion: Deriving formulae

Now tackle the theory. You are trying to find the single resistor R_total which will have the same resistance as two or more resistors R₁ , R₂ etc in series. Derive the equations for series and parallel resistance combinations. The starting points are:

Series resistors have the same current but the pds add, so:

V_total = I × R_total

IR_total = IR₁ + IR₁

divide by I to get:

R_total = R₁ + R₁

Parallel resistors have all have the same pd across them but the currents add, so:

I_total = VR_total

I_total = I₁ + I₂

VR_total = VR₁ + VR₂

then divide by V to get the familiar formula.

1R_total = 1R₁ + 1R₂

Beware that the parallel formula is usually quoted as 1R_total so they still have to take the reciprocal to get the value of R_total. It is unusual (at A-level) for questions to involve more than two resistors in parallel so it is worth pointing out that the parallel formula for two resistors can be rearranged to give:

R_total = R₁R₂R₁  +  R₂

or product over sum. This can save the faint-hearted from reciprocals!

It is also worth pointing out that when n resistors of the same value (R) are connected in parallel the result is an effective resistance Rn. Link this with an n-fold multiplication of current.

Beware that, even at this level, some students will argue that current always takes the path of least resistance. (Be prepared to discuss what happens in the extreme case when a component is shorted out by a piece of wire in parallel. Point out that a small fraction of the overall current still continues to flow through the component.

This case will re-emerge later in potential dividers in

Episode 118

You may need to mention conductance at this point. Conductance G = 1R and it is measured in siemens, S. In a parallel combination, conductances add.

Worked examples: Adding resistors

Work through examples for two resistors with convenient resistances, e.g. 40  Ω  and 60  Ω . These give 100  Ω  in series and 24  Ω  in parallel. Allow students to perform the calculation for themselves and help out any students who have problems using the reciprocal key on their calculators.

Emphasize that, for resistors in series, the total is always greater than the individual resistances; in parallel, it is less than any of the individual resistances.

Student experiment: Predicting and measuring resistance

Students can measure resistances of different combinations of resistors using an ohm-meter, and compare their answers with calculated values.

You may wish to review the use of the ohm-meter prior to this activity. The range on an ohm-meter can be confusing. Obviously this varies from device to device – make sure they are familiar and competent with the meters you intend them to use; otherwise the points about series and parallel circuits will be lost.

Episode 114-2: Resistors in series and in parallel (Word, 36 KB)

Student questions: Practice with the formulae

Two sets of questions using these formulae.

kiloohms and megaohms will catch some out, as will mA.

Episode 114-3: Circuit resistance (Word, 51 KB)

Episode 114-4: Combining resistances (Word, 31 KB)

In this episode, students find that the energy transferred by an electrical component depends on the potential difference across it, the current through it and the time for which it operates.

Lesson Summary

• Demonstration and discussion: Lamp and joulemeter (10 minutes)
• Student experiment: Power of a lamp (30 minutes)
• Student questions: Electrical power (20 minutes)

Discussion and demonstrations: Lamp and joulemeter

Start by reviewing the effect of increasing the voltage across a filament lamp. The current through the lamp increases and the lamp gets brighter.

Episode 115-1: Energy transferred by a lamp (Word, 28 KB)

Remind your students that the pd is defined as the energy transferred per coulomb and that the current is the number of coulombs per second. If the lamp is connected to a joulemeter you can show that the number of joules transferred increases with time and that the rate increases when voltage and current increase. This should lead to the word equation:

Energy transferred = number of joules per coulomb × number of coulombs

Energy transferred = pd across lamp  ×  current through lamp  ×  time

ending up with the equation:

energy transferred = V  ×  I  ×  t

Now remind them that power is rate of transfer of energy leading to P = I × V (they may also need to be reminded of the SI units, watts).

This exercise can be rounded off by calculating power and showing that (at the rated voltage) it is equal to the quoted power of the lamp.

Student experiment: Power of a lamp

At this stage, students should investigate the way power depends on pd across the lamp. Ask them to predict the effect of doubling the voltage. Some will expect the power to double. Some students may suggest that it quadruples (arguing that both I and V double). In fact neither answer is correct in practice. At higher currents the filament is hotter so the resistance is higher and the current does not double when the voltage doubles.

The results of the experiment will lead to useful discussion about energy, power, voltage, current and resistance.

Episode 115-2: relationship between power & applied potential difference (Word, 35 KB)

Student questions: Electrical power

Applying the above ideas.

Episode 115-3: Types of light bulb (Word, 50 KB)

In this episode, students develop their competence in using the equations for power and energy, in an electrical context.

Lesson Summary

• Discussion: Equations for power (15 minutes)
• Student questions: Using equations (30 minutes)
• Discussion: Paying for electricity (10 minutes)
• Student activity: Calculating an electricity bill (10 minutes)

Discussion

It is worth spending 5 minutes deriving different versions of the equations, i.e. start from P = I × V and V = I × R to deduce P = I² × R and P = V²R.

Now would be a good time to summarise all of these in the remaining time. Work towards completing a summary sheet like the one below.

Quantity	Symbol	SI unit	Comments	Equations
Potential difference	V	Volt (V)	Also use mV and kV	V = W / Q
Charge	Q	Coulomb (C)		Q = It
Current	I	Amp (A)	Also use mA
Energy	E or W	Joule (J)	Also use kJ, MJ
Power	P	Watt (W)	Also use mW, kW, MW	P = W / t, P = I V = V2 / R = I2 R
Resistance	R	Ohm (W)	Also use kW	R = V / I
Time	t	Second (s)

Student questions

Students can now work on a series of exercises. These could be set as a homework activity but there is something to be said for letting them work through the problems in class in front of you and pausing to go through solutions every now and again. This will identify problems quickly (and deal with them).

Episode 116-1: The algebra of power (Word, 22 KB)

Discussion

Paying for electricity: this should be revision of work covered at a previous stage in your school or college. The main points to get across are that:

• We pay for the electrical work done by our electricity supply (not charge or current or voltage). Usually this is an energy source such as a fuel at a power station but the cost also includes the production costs (infrastructure and maintanance).
• The electricity companies use a non-SI unit, the kWh, to calculate our bills.
• You could start by showing that domestic appliances have a high power rating (often in kilo watts).

(e.g. a 100 W lamp illuminated for 10 hours transfers 3 600 000 J of energy to the surroundings). Define the kilowatt-hour (kWh) as the amount of energy transferred by a 1 kW appliance operating continuously for 1 hour.

Amount of energy in kWh is then just:

energy / kWh = power / kW) × time / h)

so 1 kWh = 1000 W  ×  3600 s

1 kWh = 3 600 000 J

To calculate the cost of electricity, multiply the energy transferred (kWh) by the cost per kWh (p).

Student activity

At this point it would be helpful to show them an electricity meter and a bill and then to get them to calculate the costs of running common appliances. This will emphasize the large power of devices that transfer energy by heating (e.g. immersion heaters, electric cookers, electric showers).

This episode links Kirchhoff's circuit laws to conservation of charge and energy. Students can verify the laws experimentally, and use them to solve simple circuit problems.

Lesson Summary

• Demonstration and discussion: Explaining the laws (15 minutes)
• Student experiment: Verifying the laws (30 minutes)
• Worked example: Focussing on the Second law (10 minutes)

Discussion and demonstrations: Explaining the laws

Remind the class that charge and energy are conserved quantities. This is best done in the context of a demonstration – e.g. an electric motor lifting a load (or any other device that transfers energy electrically – perhaps an electric heater).

Episode 117-1: Energy transfer by an electric motor (Word, 33 KB)

The current is the same before and after the motor. The voltage drop across the motor is a measure of the electrical work done (energy transferred) per coulomb of charge passing a point. It is simple to verify that the pd across the motor and across the supply are the same, leading to the idea that the electrical work done by the supply is equal to the energy transferred out of the circuit by the motor. This can be generalized to the ideas:

• Charge simply flows around a circuit – it is not used up.
• The electrical work done (by cell/power pack/generator etc...) is equal to the energy transferred to the surroundings by the circuit.
• The first of these statements leads to Kirchhoff’s first law, the second to his second law.

Episode 117-2: Kirchhoff's Laws 1 (Word, 31 KB)

Student experiment: Verifying the laws

With a less able group (or simply to provide more opportunities to build and test circuits) you might get them to build a sequence of circuits and to measure currents and voltages. The parallel circuits are particularly good practice and the exercise will reinforce their understanding that ammeters must be connected in series and voltmeters in parallel.

Episode 117-3: Verifying Kirchhoff's Laws (Word, 64 KB)

Worked examples: Focussing on the second law

Students are unlikely to be required to solve complex problems involving circuits with two or more loops. However, they should be able to apply Kirchhoff's laws to simple circuits.

The first law is not difficult; the second law is harder. Teach your students to use a finger to trace round a complete loop in a circuit, starting at a source of emf. The first time round, they add up all the emfs (taking account of their directions). The second time round, they add up the values of IR for each component (again, algebraically, and including contributions for internal resistance). These two quantities are then equal.

Show a worked example on the board.

Episode 117-4: Kirchhoff's Laws 2 (Word, 41 KB)

Episode 117-5: Questions on Kirchhoff's Laws (Word, 43 KB)

This episode introduces the use of a potential divider as a source of variable pd. Students will also learn to use potential dividers to detect temperature or light levels.

The potential divider circuit is a particularly useful arrangement but many students find this difficult to grasp at first. It is worth spending time to make sure they understand what is going on. It may also be worth emphasizing, especially with more able groups, that the output voltage is affected by the load resistance (many, on first meeting the device, do not realize this).

Lesson Summary

• Discussion and demonstration: The potential divider formula (15 minutes)
• Student experiment: Using potential dividers (40 minutes)
• Student questions: Using the formula (20 minutes)
• Discussion: The effect of the load on output (10 minutes)
• Student questions: The effect of the load on output (30 minutes)

Discussion and demonstrations: The potential divider formula

The potential divider is one of the most useful circuits your students will meet so it is particularly important that they understand how it works and how to calculate the voltage across each of its resistors. For an unloaded potential divider the current through each resistor is the same so the voltage is proportional to the resistance. This means that the pd across the pair of resistors is divided in the same ratio as the resistors themselves:

i.e.

V₁V₂ = IR₁IR₂

or

V₁V₂ = R₁R₂

It is worth emphasizing the practical implication of this – if R₁ >> R₂ then V₁ is more or less the supply voltage and if R₁ << R₂ then V₁ is close to 0 V. You could encourage them to see V_source as an input to the potential divider and V₁ as an output. The circuit itself provides a way to tap off a voltage between 0 V and V_source .

This can, of course be done continuously using a rheostat or potentiometer and it is well worth demonstrating a variety of these including the rotary potentiometers used as volume controls in hi-fi systems.

The potential divider equation can be derived by rearranging the ratios above to give:

V₁ = R₁R₂  ×  V₂

Student experiment: Using potential dividers

Start with Parts 1-3 of these experiments.

Once they have experimented with several versions of the potential divider they can apply the ideas to a simple sensor circuit (Part 4). Set them the task of building and testing either an electronic temperature sensor or an electronic light meter (or both).

You may feel that the latter experiment as presented is sufficient to introduce the idea of a temperature or light intensity sensor, but this could be extended. The thermistor can be used as part of an electronic thermometer if it is connected into a suitable potential divider circuit and calibrated. One way to do this is to use a water bath to warm the thermistor (you can use ice with salt to get well below 0  ° C and hot water to get above 80  ° C giving a good range). However, to do this you need water-proofed thermistors (e.g. wrapped in polythene or embedded in an epoxy). If the water reaches the electrical connections then the readings will be unreliable. Beware of safety in the laboratory with both hot water and mercury thermometers (they need these for their calibration). Hot water from a kettle is safer than using a beaker on a tripod heated by a Bunsen burner. A similar approach can be used to make a light intensity meter (this can be calibrated using a lux meter).

Episode 118-1: Potential dividers (Word, 42 KB)

Student questions: Using the formula

Practice in potential divider calculations.

Episode 118-2: Tapping off a potential difference (Word, 44 KB)

Discussion: The effect of the load on output

With strong groups you might discuss the effect of loading a potential divider on its output voltage. The ideas to get across are:

Connecting a load across R₁ reduces the output voltage.

This is because the effective resistance in the lower arm of the potential divider is now a parallel combination of R₁ and R_load (less than R₁ ) so a smaller fraction of the voltage is tapped off.

If R_load >> R₁ then there is no significant effect on the output voltage.

It is worth going back to their experiences as younger pupils tackling simple circuits, and considering what was happening when a lit bulb went out when shorted out by a piece of wire. Pupils will often regress to a current based explanation, particularly under the pressure of examination conditions. We need to encourage them now to think in terms of potential difference, and resistance. In the shorting out case it was not that much that all the current wanted to take the easier parallel route, but that the low resistance of the wire in parallel reduced the combination’s total resistance, compared to the rest of the circuit.

Student questions

Episode 118-3: Loading a potential divider (Word, 36 KB)

Episode 118-4: Brightness of bulbs (Word, 29 KB)