• ## 01 Batteries and voltageEl03PNnugget01 Exposition

### Batteries come in many shapes and sizes, often of different voltage

The batteries commonly used in school science practical work are torch batteries rated at 1.5 volt. If two of these batteries are connected into a circuit one after the other (in series), the total rating is 3.0 volt. If three batteries are used we have 4.5 volt and so on. A single 9 volt battery might be used to supply a radio, whilst a 12 volt battery is used in cars.

A key specification for any battery is its voltage, and this is what we need to know when buying a battery in a shop. But what exactly does the voltage tell us?

### Different voltage

There are two ways of thinking about voltage. These are, in fact, two ways of telling the same story and relate directly to ideas introduced in earlier episodes.

### Voltage as an electrical push

Firstly you can think of voltage as being a measure of the size of the force or push provided by a battery on the charged particles in a circuit. The idea of a battery providing a push to set charged particles in motion was first introduced in episode 01. Furthermore, we saw in episode 02 that adding an extra battery to a circuit (or using a battery with a higher voltage) provides a bigger push, moving charged particles around more quickly and increasing the electric current.

Sometimes the battery voltage is referred to as the battery e.m.f., which stands for electromotive force. This term captures the idea of the battery providing a force or push on all of the charged particles in the circuit.

Thinking about voltage as a push links up closely with the rope loop model, in modelling a larger battery is effected by pulling the rope around with a greater force.

### Voltage in terms of power: energy shifted in each second

Whilst it makes sense to refer to voltage as being a measure of the push of a battery, it makes little sense at all to talk about voltage as being a measure of the push of a bulb. However the impeding caused by the resistance does provide an opposing effect to the push of the battery (again, you'd model this by gripping the rope loop).

Here we need to turn to the alternative view of voltage as a measure of the power dissipated by each current in different parts of the circuit.

More precisely, it is a measure of:

• The amount of power per unit current by the battery (energy shifted from the chemical store in a battery as charge flows through the battery).
• The amount of power per unit current to the surroundings in any circuit resistance (energy shifted to thermal store in surroundings as charge flows through resistance).

If the circuit is to conserve energy (it must!), then the voltage across the battery and across the bulb must be equal, as the power input and output must be equal.

• ## 02 Relationship between current and voltage and resistanceEl03PNnugget02 Expansion – lead me deeper

### Relating voltage, current and resistance

Setting up a simple circuit consists of selecting two components – the battery and the bulb, and linking them with wires. The two choices you make set the push: the voltage of the battery, and how much that push is impeded: the resistance of the bulb. The current in the loop is the result of these two choices. Much more on this in the SPT: Electricity and energy topic.

The relationship between the push of the battery and the size of electric current is summarised by:

current = voltageresistance

You can write this out in symbols:

I = VR

Where V is voltage, measured in volt; I is current, measured in ampere; R is resistance, measured in ohm.

According to this relationship, if the resistance does not change, then increasing the voltage results in an increase in current.

Choosing a 3 volt battery and a 10 ohm resistor results in a current of 0.3 ampere in the resistor (and in the battery, and in the connecting wires).

I = VR

Put in the values to get:

current in loop = 3 volt10 ohm

and work out the current as 0.3 ampere. If the battery voltage is doubled to 6 volt, the new current is still given by I = VR, but you need to insert the new values current in loop now = 6 volt10 ohm and calulate the current as 0.6 ampere.

There is a 0.6 ampere current in each part of the circuit loop: in the lamp; in the battery; in both the connecting wires.

Double the battery voltage gives double the current.

• ## 03 The story developedEl03PNnugget03 Expansion – tell me more

### More about the action of the battery on the charged particles

Why do we say that these ideas about voltage are two ways of telling the same story? The answer to this question takes us back to the detailed explanation, introduced in episode 01, of how electric circuits work.

When a second battery is added to a circuit, increasing the voltage, the effect is to strengthen the electric field around the circuit, between the positive and negative terminals of the battery. There is a greater force (or push) on each and every charge in the circuit due to the stronger field in the wire (as the positive terminal becomes more positively charged with respect to the negative terminal). This stronger field produces a greater acceleration of charged particles between interactions with the ions in the filament of the bulb, and so more energy is shifted from charge to ion during each interaction. So, at one and the same time, increasing the voltage results in a bigger push on each charge and more power being switched by the current.

• ## 04 The voltEl03PNnugget04 Exposition

### Defining a volt, and seeing what that means in practice

The volt is defined in terms of energy shifting from store to store by flowing charged particles, here moving around a circuit.

1 volt = 1 watt1 ampere

You can understanding the implications of this by thinking about a simple circuit.

Set up a situation where current in the resistance (such as a bulb) is one ampere and the voltage across the bulb is also one ampere. As a result one joule of energy is being shifted per second. (This is electrical working; see SPT: Energy topic).

Set up a situation where the current in the resistance is smaller and the energy being shifted will also be reduced.

### Accumulating energy in a store

So, for a fixed voltage, the energy accumulates steadily in the thermal store of the surroundings. The energy shifting from the chemical store also depletes that store by the same amount each second. Change the flow of charge, and you change the energy shifting: the power.

### Changing the voltage

Arrange for the voltage across the bulb to be 3 volt. As a consequence the accumulated energy shifted to the surroundings as each coulomb of charge passes is 3 joule.

The voltage across the bulb (or any other resistor) determines how much energy is shifted to the surroundings as each coulomb of charge passes through. The voltage is the energy shifted per unit charge.

The voltage across the battery sets the amount of energy which is shifted from the battery per coulomb of charge that has passed. You can rewrite the definition above for a duration of one second:

1 volt = 1 joule1 coulomb

This focuses on the accumulated effect of the flow of electrical charge, rather than the process as it happens. As the brightness of bulbs depends on the power, rather than the accumulated energy, this may make better connections with what is seen and easily felt to be important.

• ## 05 PowerEl03PNnugget05 Exposition

### Current and voltage leading to power

To predict the brightness of bulbs you need to think about the power in the electrical pathway. The brightness is the result of two electrical factors:
• The flow – that is the current
• The push – that is the voltage

To make this concrete, let's start with a circuit, in which a 12 volt battery is connected to a single bulb:

This circuit is similar to that suggested for the big circuit in episode 01, where a car headlamp bulb was connected to a 12 volt supply. Assume that an ammeter connected into the circuit shows a current of 2 ampere and start by thinking about the bulb:

Current in the bulb is 2 ampere.

So 2 coulomb of charge pass through the bulb each second.

Voltage across bulb is 12 volt (assuming that each coulomb of charge shifts all 12 joule of energy in the bulb).

12 joule of energy are shifted to the surroundings for every coulomb of charge passing through the bulb.

If these current and voltage figures are taken together, we can see that in the bulb there's continuous activity:

• 2 coulomb of charge pass through every second.
• Each coulomb of charge shifts 12 joule of energy.

So 24 joule of energy are shifted by the bulb each second.

In other words, as the filament of the bulb warms up and radiates, 24 joule of energy are shifted to the surroundings each second. The power of the bulb is 24 watt.

Energy is shifted (or transferred) by the circuit: from the chemical store of the battery to the thermal store of the surroundings; via heating and lighting pathways (see SPT: Energy topic).

What you have calculated here is the energy shifted every second by the electrical working pathway – that is the power input.

All of this ends up heating or lighting.

The power input is equal to the power output: the bulb switches power from one pathway to another.

• ## 06 Electrical powerEl03PNnugget06 Exposition

### Defining electrical power, and calculating the amount

This quantity, the amount of energy shifted by the bulb each second, is called the electrical power of the bulb.

The power of any device (whether electrical or mechanical) is defined as power = energy transferred per second. You can write this as:

power = energyduration

In other words, a device has a high power output if it fills a selected store of energy very quickly. This idea can be applied to electrical devices, mechanical machines and even to people. For example, a trained athlete is able to ride a mountain bike up a steep slope quicker than a person of average fitness (so filling a gravitational store rapidly). The same amount of energy is needed for either person (provided they're about the same mass) to ride the bike up the hill. It's the athlete's ability to perform the task in a shorter period of time that shows that their power output is greater. We don't know, from these measurements, who has emptied their chemical store most rapidly. That would be a measure of the input power.

The standard unit of power is the watt, and the symbol used to represetn this is W.

power/watt = energy/jouleduration/second

So: 1 watt is 1 joule of energy transferred per second and 100 watt is 100 joule of energy transferred per second.

• ## 07 Calculating electrical powerEl03PNnugget07 Exposition

### Example calculations

Going back to the circuit we started with, you can see that the electrical power is calculated by multiplying together the current in the bulb and the voltage across it. In general:

powerwatt = voltagevolt × currentampere

Now, the voltage across bulb is 12 volt and the current in bulb is 2 ampere, so

power of bulb = 12 volt × 2 ampere

and the power of the bulb is calculated as 24 watt.

Note that the power calculated here includes the total energy shifted by the bulb through lighting and through heating. The energy shifted by the electrical working pathway does not all end up being shifted by the lighting pathway. The bulb is not as efficient as we'd like. More on this can be found in the SPT: Energy topic.

### An example to explore

Here is a simple loop, where you can choose the battery and choose the bulb, so setting the resistance of that bulb. These choices have consequences for the current in the loop and the power dissipated by the bulb and switched by the battery.

You may well find the code intelligible – the physics is only two lines:

voltageresistance = current

and

power = current × voltage

The rest is all concerned with display.

• ## 08 Shifting energy from a batteryEl03PNnugget08 Expansion – tell me more

### A chemical store, depleted

Find out more about describing the energy shifted from the chemical store.

We said: Energy is shifted by the charged particles as they pass through the battery. What does this mean? To answer the question it is helpful to return to the detailed explanation of how electric circuits work introduced in episode 01.

All of the charged particles (imagine positively charged particles) in the circuit experience a force to move them round from the positive to the negative terminal of the battery. When each charge arrives at the negative terminal, energy must be shifted as it moves across the battery from the negative plate to the positive plate (against the repelling force of the positive plate). The energy required comes from the chemical store of the battery, which is emptied by the electrical working pathway.

An example. For a 12 volt battery with a current of 0.25 ampere, the power switched is 3 watt The result of this accumulated action is that, over one second:

• 0.25 coulomb passes through the bulb (0.25 ampere  ×  1 second.)
• 3 joule is shifted (3 watt  ×  1 second.)

So 3 joule of energy is shifted from the chemical store for each coulomb of charge making the trip from negative to positive plate.

What is in the rest of the circuit determines the stores of energy that are filled as a result of the chemical store being emptied. This ability to shift energy is why electric circuits are so useful.

• ## 09 Measuring voltagesEl03PNnugget09 Exposition

### The instrument used to measure voltages is the voltmeter

Voltmeters are always connected across circuit components, whether a battery, bulb or some other device.

Voltmeters are connected differently to ammeters (see episode 01). With the ammeter, the instrument is connected into the circuit so that the flow of charge through the circuit can be measured.

Why are voltmeters connected across components?

Why are the voltmeter readings sometimes positive and sometimes negative?

To provide an answers to these questions, we need to be clear about what it is that the voltmeter is measuring.

In the circuit here, the battery is rated at 3 volt and therefore shifts 3 joule of energy as each coulomb of charge passes. What do these measurements mean?

Let's keep things simple: assume all of the energy which is shifted from the battery's store is shifted to the stores of the surroundings by the bulb. Therefore no energy is shifted to the surroundings by the action of the connecting wires. We're assuming that the connecting wires do not heat up at all as a result of the current in them: there is no electrical working in the wires. (In fact, the connecting wires must heat up a little, but the energy involved is negligible compared with that in the bulb).

There is electrical working in the battery and in the bulb. In the battery the action is to empty the chemical store, so the voltage is positive. In the bulb the action is to fill the surrounding thermal stores, so the voltage is negative. You can get the correct readings on the voltmeter by ensuring that the negative terminal is always upstream of the positive terminal, as defined by the current.

Any electrical circuit is a system linking the filling of a store in one place with emptying of a second store in another place. For example, as an electric train pulls away from a station a kinetic store is being filled, whilst (say) a coal-fired power station is burning fuel elsewhere and depleting a chemical store). (This is quite unlike a steam train where the depleted and filled stores are associated with objects at same location).

### Making sense of voltmeter readings using a rope loop

The rope loop model, which emphasises electrical working, (so tying in well with the SPT: Energy topic) is the best way to get a for what is happening for flows to or from stores, as signposted by the voltage.

In the electrical circuit, the voltage sets the energy shifting per ampere.

In the rope loop, the force sets the energy shifting per metre / second of rope.

Both larger voltages and larger forces (grip or pull harder) lead to more energy shifting as the charge or rope flows past.

More grip results in a greater retarding force on the loop and so a larger negative voltage (sometimes called a voltage drop) across the resistor or lamp modelled.

More pull results in a greater driving force and so a models a larger voltage across the cell.

• ## 10 Different terms for the same thingEl03PNnugget10 Expansion – tell me more

### Confused? – try this

Read this if you are confused about the use of many different terms.

Voltage and potential difference and potential drop and p.d.

Very often, the voltage across a circuit component, such as a bulb, is referred to as a potential difference (a p.d.).

Teacher: So, if I just take the reading of the voltmeter, it's 2.5 volt. The p.d. across the bulb is 2.5 volt.

Sometimes you might hear people talk about the voltage across the bulb, at other times about the potential difference across the bulb and sometimes about the potential drop or p.d. On first hearing this, it can be confusing! However, the diagrams we use show that the voltage is a measure of the difference in potential energy per coulomb as charge passes through a circuit component. Hence the use of the terms potential difference or potential drop. At this stage is probably simpler for pupils just to use voltage.

There is more on the reasons to prefer thinking about potential difference later in this episode and much more in the SPT: Electricity and energy topic.

• ## 11 Voltage in series circuitsEl03PNnugget11 Exposition

### Voltage in series circuits: what happens?

How do the ideas about voltage work with a circuit where there is more than one bulb? In this circuit, two identical bulbs are connected in series to a 3 volt battery and a voltmeter is connected across each bulb.

What happens in the circuit?

We have already seen that adding a bulb in series results in the current being reduced (due to the increased resistance) and both bulbs becoming dim.

What happens to the voltage across each bulb?

When the second bulb is added, the same voltage appears across the two bulbs. Both are identical, what else could it be as the situation is symmetrical?

The rope loop provides a good teaching model to reason with.

### Voltage across components in series: a summary

The sum of the voltages across the bulbs/resistors must equal the battery voltage. This follows from the energy description.

The powerin must be equal to the powerout.

The energy shifted from the chemical store of the battery must equal the energy shifted by the charged particles as they pass through the bulbs/resistors.

• ## 12 Voltage in parallel circuitsEl03PNnugget12 Exposition

### Voltage in parallel circuits: what happens?

How do the ideas about voltage apply to parallel circuits? In this circuit, two identical bulbs are connected in parallel to a 3 volt battery and a voltmeter is connected across each bulb.

We've already seen that adding a bulb in parallel results in both bulbs being of equal, normal brightness. What happens to the voltage reading across each bulb?

When the second bulb is added in parallel, there is a voltage equal to the full battery voltage across both. Each voltmeter reads 3 volt.

As the second bulb is added, there is a current in both loops. The power in both bulbs is equal and set by the current in and the voltage across each bulb. Both bulbs are as bright as they would be in a simple loop (one battery, one bulb).

### Making sense of the voltmeter readings using a rope loop

Each loop travels at constant speed, so the driving and counter forces must be equal in size (see the SPT: Forces topic). The product of the speed and the force calculates the power switched.

### Voltage across components in parallel: a summary

When bulbs or resistors are connected in parallel, the full battery voltage is dropped across each.

The powerin must be equal to the powerout.

• ## 13 Components not equalEl03PNnugget13 Expansion – lead me deeper

### Non-equal bulbs and resistors in series

Suppose two bulbs are connected in series and they are not identical. Suppose the resistance of bulb A is greater than the resistance of bulb B. How will the battery voltage be shared between the two bulbs?

The answer to this question is that a higher voltage is dropped across the greater resistance. In other words more voltage is dropped across bulb A. Why should this be?

More energy is shifted as charged particles pass through a bigger resistance. Think of the rope loop teaching model. If two pupils act as resistors but one pupil grips the rope more tightly than the other (greater resistance), the hand with the tighter grip warms up more than that with the looser grip.

The same argument is formalised using the relationship:

V = R × I

The current (I) is the same through each bulb as the two bulbs are connected in series. Therefore a higher voltage must be dropped across the greater resistor.

Remember that the sum of the voltages across the two bulbs must still be equal to the battery voltage.

### An example of a circuit with series connections to explore

Here is a loop with series connections, where you can choose the battery and choose the bulbs, so setting the resistance of those bulbs. These choices have consequences for the current in the loop and the power dissipated by the bulbs and switched by the battery.

You may well find the code intelligible – the physics is only a few lines, containing these essential relationships:

resistance of loop = resistance of first bulb + resistance of second bulb

voltageresistance = current

voltage = current × resistance

power = current × voltage

The rest is all concerned with display.

### Non-equal components in parallel

Suppose two bulbs are connected in parallel to a 3 volt battery and they are not identical: the resistance of bulb A is greater than the resistance of bulb B. What happens to the current in, and voltage across, each bulb in this case?

The voltage across both bulbs must be the same. Consider the two loops separately: the current in loop A is small (due to the high resistance), so the power switched by the bulb is small; the current in loop B is large (due to the lower resistance) so the power switched by the bulb is small.

The rope loop model can help to make sense of this; the power switched depends only on the grip and on the speed at which the rope passes through the hand.

Voltage across bulb A is 3 volt

Current in bulb A: relatively small

Power switched by bulb A: relatively small

Voltage across bulb B is 3 volt

Current in bulb B: relatively large

Power switched by bulb B: relatively large

### An example of a circuit with parallel connections to explore

Here are two loops, so a circuit with parallel connections, where you can choose the battery and choose the bulbs, so setting the resistance of those bulbs. These choices have consequences for the current in the loops and the power dissipated by the bulbs and switched by the battery.

You may well find the code intelligible – the physics is only a few lines, containing these essential relationships:

voltageresistance = current

current through battery = current in inner loop + current in outer loop

power = current × voltage

The rest is all concerned with display.

• ## 14 Starting to understand potential differenceEl03PNnugget14 Expansion – lead me deeper

### Making sense of voltmeter readings using energy hills

So, the voltmeter is connected across the circuit component (battery or bulb) in order to measure the energy shifted for each coulomb of charge passing through the component. It measures the energy shifted each time one unit of charge passes through the component.

### Making sense of voltmeter readings with series connections using energy hills

As before, we assume that no energy is shifted to the surroundings through the connecting wires.

So, the sum of the energies shifted by each bulb is equal to the energy shifted from the chemical store of the battery. So we have X + X is equal to 3.0 volt. So X is 1.5 volt, assuming the bulbs have equal resistances.

### Making sense of voltmeter readings with parallel connections using energy hills

As before, we assume that no energy is shifted to the surroundings through the connecting wires.

So, the full battery voltage of 3 volt is dropped across each bulb. Once again, we suggest you keep the description simple, and assume that no energy is shifted to the surroundings by the connecting wires.

• ## 15 Final reviewEl03PNnugget15 Summary

### Bringing together ideas about circuits

The final part of the electric circuits topic brings together all of the ideas introduced in the preceding episodes to think through a range of different circuits.

### Four different circuits to work through

For each circuit you can work through key questions, starting with what actually happens when the circuit is completed:

1. What happens? What do you see happening when the circuit is completed?
2. Can you produce a description in terms of the electric circuit model? Explain what happens in terms of the charged particles moving around the circuit and the which elements shift energy.
3. Can you produce a description in terms of a teaching model? How would you draw on the rope loop teaching model to help explain this circuit to pupils?
4. Can you produce a description in terms of current, voltage? What is the electric current in different parts of this circuit? Explain the value of the electric current in the different parts. What is the voltage drop across each bulb in this circuit? Explain the value of the voltage drop across each bulb.
5. Can you predict the power output from bulb? What is the power output of each bulb in this circuit? Explain the value of the power output from each bulb.

You can find out more in the SPT: Electricity and energy topic.

You'll find the answers on TalkPhysics.org.

•