Interactive textbook: Cells and circuits

Making a (potential) difference

There is no potential difference between the two spheres below. They have exactly the same charge distributed over the same surface area (much like there being no gravitational potential differences on flat ground).

It would require exactly the same amount of work to take a small positive test charge from infinity to the surface of the left sphere as it would to the right sphere:

To make a potential difference we need to move charges from one sphere to the other:

When charges are moved, work is done. The electric potential energy of the charges increases. If a conductor was connected between the spheres, current would flow from the high potential sphere to the low potential sphere:

Note that positive charges are shown flowing through the conductor; this shows the direction of conventional current. In reality electrons flow the other way. Charge flows until the potential of each sphere is the same. The electric potential energy is converted to thermal energy in the conductor.

This is all well and good but we can't move charges by hand every time we want a potential difference. A cell or battery does it for us and this PhET simulation shows how.

Fairies!

Actually it's not fairies - it's chemistry. But this is a physics course so, as far as we are concerned, it might as well be. 

The important thing to learn is that rearranging charges requires a transfer of energy. Maybe it's worth taking a short detour into the world of chemistry to see how that might be done...

The electrical energy is NOT equal to:

The charges have very little KE because the lattice gets in the way.

Chemical energy

Chemicals such as hydrogen chloride (containing two or more elements) are compounds made of molecules. A molecule is two or more atoms held together by the electric force - hydrogen and chlorine in this case. Both atoms are neutral, so how can there be an electric force?

Well, atoms may be neutral but the electrons within are not evenly spread. This means there can be a force between the atoms.

Making a molecule from atoms requires energy. This energy is transferred to the potential energy of the atoms; we call it chemical energy. If the atoms come apart this energy is released.

How this all happens is impossible to explain in terms of Coulomb's law. Chemists use different models that rely on quantum theory but, for now, we just need a simple model that explains where the energy is coming from.

 

Which arrangement of charges has the most electrical PE?

Most work is done placing the charges as they are in D

 

 

Cells

An electric cell is a single unit of a battery, typically supplying \(1\) or \(2\text{ V}\). For example, a \(12\text{ V}\) battery could be made of six \(2\text{ V}\) cells. There are many types of cell but all use some chemical reaction to convert chemical to electrical energy. One such chemical reaction is between a metal and an acid.

When a metal reacts with an acid it becomes negative. Different metals react at different rates so, in a given time, the fast-reacting metal will become more negative. Zinc reacts faster than copper so if they are both put into acid the zinc becomes more negative than the copper. There will therefore be a potential difference between the metals.

If the metals were connected by a conductor, conventional current would flow from the copper to the zinc.

It is worth acknowledging here that both metals are negative. There isn't a positive and negative side as we have labelled on a shop-bought battery. In this case the copper would be 'positive' and the zinc 'negative', because these labels are really referring to high and low potential, as is made clear by the symbol used for a cell.

The long side is the higher potential.

The positive terminal of a 12 V battery has a potential of 18V relative to the Earth.

The potential of the negative terminal is:

The difference = 12 V

EMF

EMF (\(\varepsilon\)) stands for electromotive force but these words are rarely used as they convey an incorrect meaning. EMF can be defined in two ways:

  1. The work done per unit charge in taking a small positive test charge from the low potential side of a battery to the high potential side for a battery when not connected to a circuit (i.e. no current flows).
  2. The amount of chemical energy converted to electrical potential energy per unit charge flowing through the battery.

The unit for EMF is the volt (\(\text{V}\)).

You might recognise a similarity between these definitions and that for potential difference. In fact, they are the same quantity, it's just that EMF refers only to sources of electrical energy (e.g. batteries) whereas potential difference is more general (e.g. batteries or resistors).

Internal resistance

The internal resistance of a cell is the resistance of the inside of the cell. This is due to the metal parts and chemicals. It is represented on a circuit diagram as a resistance, \(r\), in series with the cell.

The internal resistance of a \(1.5\text{ V}\) AA battery is typically around \(1\text{ Ω}\).

Terminal potential difference

The terminal potential difference is just that - the potential difference across the terminals of the cell. The EMF of a cell is constant but the terminal potential difference depends on the amount of current flowing from the cell:

  • When no current flows, the terminal potential difference is equal to the EMF.
  • When current flows, the terminal potential difference is lower than the EMF.

  • If the cell is short-circuited i.e. a conductor of zero resistance is connected across the terminals, the terminal potential difference would be zero.

In this latter case all of the potential created by the cell is dropped across the internal resistance. This means that all the energy would be dissipated inside the battery, so it gets hot. You shouldn't try this in the lab (!) but we can see what happens in a simulation.

An 1.5 V AA battery can deliver 2A for one hour. (2 Ah)

How much charge is there stored in the battery?

1 A = 1C s-1

How much energy is stored in an AA battery?

Energy = qV = 1.5 x 7500

If an AA battery is short circuited what current flows? (r = 1.5 Ω)

I = V/R = 1.5/1.5

If an AA battery is short circuited for how long will the current flow?

I = q/t

t = q/I = 7500/1

The heat capacity of the battery is 2 J K-1

Wht is the temperature change of the battery per second when short circuited?

In 1s 1C flows the energy dissipated = 1.5 J

Q = CΔT

ΔT = Q/C = 1.5/2 = 0.75 K

Will the battery burn?

In one hour the temperature change would be 2700 K however heat is lost to the surroundings so the final temperature will be much lower than this. maybe the simulation was a bit over dramatic.

 

Primary and secondary cells

 

If a cell contains electric potential energy it must have gained that energy from somewhere.

The energy could have been added when the molecules that break down in the cell were put together. This could in turn have come from heat, increasing the kinetic energy of the atoms enabling them to come together. Alternatively the energy may have emerged from the chemical energy of other molecules used in the production of the chemicals in the cell. Cells of this type are called primary cells; once the chemicals have been used, the cell can not produce electrical energy.

Alternatively electrical energy can be used to rebuild the molecules. In this case the cell can be recharged by connecting it to a second source of electrical energy. These cells are called secondary cells.

Discharging

In the example of the two charged balls used previously the potential difference between the balls starts to decrease as soon as the charges start to flow.

This would not be very practical if this system was used to power a light. The light would start to dim from the moment it was switched on! In a cell the chemical reaction replaces the charges so the potential difference remains constant until the chemicals are used up.

You might be wondering why the chemicals don't get used up even when the cell is not delivering current. Maybe this analogy might help. Chemical energy is like the energy stored in a compressed spring between two balls, if the spring is released it can give the balls kinetic energy enabling them to move to separate ends of the container.

As the balls collect at the ends they form hills, as the hills get bigger there comes a point when the balls no longer have enough energy to climb the hill.

The balls can no longer come apart unless balls start to flow off the hills. The gravitational potential energy of the balls at the top of the hill is equal to the elastic potential energy of the spring.

In the same way the electrical potential energy of the charges equals the chemical energy of the molecules. A typical cell has an EMF of about \(1\text{ V}\) so we can deduce that the potential energy of the atoms in a molecule is about \(1 \text{ J C}^{-1}\). This gives an electron energy of \(1.6 \times 10^{-19}\text{ J}\) (which you will see later is about right).

Circuits

A simple circuit consists of a cell connected to an external resistor, \(R\).

Applying the law of conservation of energy to this circuit we can say that when a unit of charge flows:

  • The energy converted from chemical to electrical in the cell...
  • ...is equal to the energy converted to heat due to internal resistance + the energy converted to heat due to external resistance.

Converting these statements to symbols:

  • The energy converted from chemical → electrical per unit charge is \(\varepsilon\)
  • The energy dissipated per unit charge as heat due to internal resistance is the potential difference \(Ir\) (from Ohm's law)
  • The energy dissipated per unit charge as heat due to external resistance is the potential difference \(V=IR\) (from Ohm's law)

\(\Rightarrow \varepsilon=Ir+IR\)

It may not be obvious why the heat dissipated per unit charge in \(R\) is \(V\) but remember that \(V\), the potential difference across \(R\), is equal to the work done per unit charge moving a positive test charge from B to A. If a unit positive charge moves from B to A through \(R\) it must lose this amount of energy. The charge does this by giving kinetic energy to the lattice which is lost to the surroundings as heat.

We can investigate this by measuring the potential around the circuit, here shown with American symbols.

Notice how the potential rises to \(9\text{ V}\) across the cell, drops \(1\text{ V}\) across the internal resistance and then drops the remaining \(8\text{ V}\) across the \(8\text{ Ω}\) resistor.

\(\varepsilon=Ir+IR\)

\(9 = 1 + 8\)

The current (\(I\)) is \(1\text{ A}\).

Now to some laws that apply to all circuits (no matter how simple or tricky).

 

If ε = 12 V, R = 11 Ω and I = 1A

Calculate the terminal PD

Terminal PD = IR = 1 x 11

What is the internal resistance

PD cross r = 1 V

R = VI

r = 1 x 1

 

Kirchhoff's second law

Kirchhoff's second law ("K2") states that around any closed loop the sum of the EMFs is equal to the sum of the potential differences. This is a consequence of the law of conservation of energy.

\(\sum\varepsilon=\sum \text{PD}\)

This is equivalent to saying that if you take a positive test charge around a circuit and arrive back at the same place the sum of the increases in potential (EMFs) must equal the sum of the decreases (PDs).

The gravitational analogy would be taking a walk around any path and arriving back at the same point. The sum of your climbs is equal to the sum of your descents.

 

 

 

Let's apply this to the simple circuit:

One way of applying Kirchhoff's second law is to imagine you are taking a positive test charge around the circuit twice, first summing the EMFs and then the PDs.

Starting from A we can go two ways. Let's choose clockwise:

 \(\sum\varepsilon= ε\)

\(\sum\text{PD}=IR+Ir\)

What if we go anticlockwise? This time we go down across the cell (from the long line to the short one) so the EMF is negative:

 \(\sum\varepsilon= -ε\)

Passing a resistor we go up. We can tell we are going up because we go against the current, like walking along a river against the flow. PDs are supposed to represent drops so to make the equation balance a rise in potential is a negative drop:

\(\sum\text{PD}=-IR+(-Ir)\)

Either way?

\(ε = Ir + IR\)

You must remember this sign convention when dealing with more complex circuits. Let's consider some more complex circuits.

\(\sum\varepsilon= ε_2 - ε_1\)

\(\sum\text{PD}=IR\)

\(\Rightarrow ε_2 - ε_1 = IR\)


Starting from A we have two ways round the circuit. Let's take the one passing through \(R_1\):

\(\sum\varepsilon= ε_2 - ε_1\)

\(\sum\text{PD}=I_1R_1\)

\(ε_2 - ε_1 = I_1R_1\)

And now for the \(R_2\) route:

\(\sum\varepsilon= ε_2 - ε_1\)

\(\sum\text{PD}=I_2R_2\)

\(ε_2 - ε_1 = I_2R_2\)


Taking the loop through \(R_1\):

\(\sum\varepsilon= -\varepsilon_3+ε_2 - ε_1\)

\(\sum\text{PD}=-I_1R_1+Ir\)

\(-ε_3 + ε_2 - ε_1 = -I_1R_1 + Ir\)

How did we know that \(I_1\) was in this direction? We didn't, it was a guess. The direction of \(I_1\) depends on the size of \(ε_3\). If you guess wrong, the value calculated will turn out negative.

Kirchhoff's first law

Kirchhoff's first law ("K1") states that the sum of currents into a junction is equal to the sum of the currents out. This is a consequence of the law of conservation of charge.

Applying this to the last circuit (above) gives:

\(I = I_1 + I_2\)

Combinations of resistors

There are many ways to connect resistors but all are based on two simple ways that will help us to solve circuit problems.

In series circuits the resistors are in the same loop. The current through series resistors is constant.

Since energy is conserved the sum of the energies converted to heat per unit charge in \(R_1\) and \(R_2\) are equal to the total energy converted per unit charge in the combination.

\(V = V_1 + V_2\)

Applying Ohm's law:

\(V = IR_1 + IR_2\)

We could replace this combination with a single resistor \(R\) so that the same current flows:

\(IR = IR_1 + IR_2\)

\(\Rightarrow R = R_1 + R_2\)

Two \(4\text{ Ω}\) resistors in series have a total resistance of \(8\text{ Ω}\).


In parallel circuits the resistors are in different loops. The current splits and rejoins at the junctions. 

Applying Kirchhoff's first law:

\(I = I_1 + I_2\)

Applying Ohm's law:

\(I = {V\over R_1} + {V\over R_2}\)

We could replace this combination with a single resistor \(R\) so that the same current flows:

\({V\over R} = {V\over R_1} + {V\over R_2}\)

\({1\over R} = {1\over R_1} + {1\over R_2}\)

Two \(4\text{ Ω}\) resistors in parallel have a total resistance of \(2\text{ Ω}\).

 

A 2 Ω and a 4 Ω resistor are connected in parallel.

The total resistance of the combination is:

1/R = 1/4 + 1/2 = 3/4

R = 4/3 Ω

 

 

A 2 Ω and a 4 Ω resistor are connected in series.

The total resistance of the combination is:

1/R = 1/4 + 1/2 = 3/4

R = 4/3 Ω

Kirchhoff's second law is a consequence of:

the stored energy = the energy converted fro electrical to heat.

Measuring current and potential difference

These days you are most likely to use a multimeter to measure both current and potential difference (and resistance!). However, your physics lab may have separate ammeters and voltmeters. These are actually the same instrument and use the deflection of a coil in a magnetic field due to the current flowing through it. The difference is that a voltmeter has a large resistance in series with the coil whereas an ammeter has a low resistance in parallel with it.

An ammeter is placed in series with a component. Its low resistance means that it does not reduce the current that it is supposed to measure.

If this was not the case, the current in the whole circuit would decrease significantly:


A voltmeter is placed in parallel with a component. Its large resistance means it does not draw much current from the component, and therefore does not significantly reduce the potential difference across it.

Without the large resistance?

 

What is the ideal resistance of an Ammeter and a voltmeter?

There should be  no potential dropped across the ammeter and no current flowing through the voltmeter.

Potential divider

A potential divider is a circuit used to provide a different potential difference to the terminal potential difference of the source. Here the new potential difference is that across just one of the resistors:

The \(10\text{ V}\) potential difference of the cell is divided into two steps of \(5\text{ V}\) by the two equal resistors.

In general if we have two resistors \(R_1\) and \(R_2\)...

...total resistance is \(R_1 + R_2\).

Now, current is:

\(I={V_\text{s}\over R_1+R_2}\)

Potential difference across \(R_2\) is therefore:

\(\Rightarrow V=IR_2={V_\text{s}R_2\over R_1+R_2}\)

To create a variable potential difference, a variable resister is used. Overall, this circuit is called a potentiometer:

If the slider is moved to A, \(V = V_\text{s}\). If it is moved to B, \(V = 0\text{ V}\).

Note that if a resistance much bigger than \(R_2\) is connected across \(R_2\) then the potential difference is not changed significantly. However, a comparable resistance will reduce the total resistance increasing the current in the circuit. This increases the potential difference across \(R_1\) and reduces the potential difference across \(R_2\):

R1/R2 = 3 and Vs = 12 V

V =

ratio of V1/V2 = R1/R2

Measuring EMF and internal resistance

In this required practical to measure EMF and internal resistance, a variable resistor is used to draw varying currents from a battery.

Applying Kirchhoff's second law gives:

\(\varepsilon=IR+Ir\)

\(IR=-Ir+\varepsilon\)

Now, \(IR\) is the terminal potential difference measured by the voltmeter so:

\(V=-Ir+\varepsilon\)

This is of the form \(y=mx+c \). If a graph of \(V\) against \(I\) is plotted it will have gradient \(-r\) and \(y\)-intercept \(ε\).

 

 

The internal resistance and EMF for the previous graph are

- gradient and y intercept

 

This graph was for an AA battery which statement is most likely to be true?

There is no PD across r when the current = 0 A

The EMF should be 1.5 V

The random errors are small

Power

As well as mechanical power, we are now able to calculate electrical power.

Consider a charge \(q\) flowing through a resistor from A to B in time \(t\):

The heat dissipated per unit charge is \(V\), so the heat dissipated when charge \(q\) flows is \(Vq\).

Now power is the rate at which energy is transferred:

\(P={E\over t}={Vq\over t}\)

And current is the rate of flow of charge:

\(I={q\over t}\)

\(\Rightarrow P=IV\)

From Ohm's law we know that \(V = IR\) so we can substitute for \(V\) or \(I\) to alternatively get:

\(P=I^2R={V^2\over R}\)

Which equation you use depends on the question, here are two typical examples:

  1. Which resistor dissipates the most power?

Both have the same potential difference across them so if we use \(P ={V^2\over R}\) we can deduce that resistor \(R\) has twice as much power.

  1. How about in this series circuit?

Both have the same current so if we use \(P = I^2R\) we can deduce that resistor \(2R\) has twice as much power.

The power dissipated by a lightbulb determines its brightness - good to know in questions about brightness rather than power!

 

The ratio Power dissipated in R / Power dissipated in each 2R resistor

Resistance of parallel combination = R so power in combination is the same as R, power in each is 1/2 power in R.

or

Current through each 2R = 1/2 current through R

Power in R = I2R

Power in each 2R = (I/2)22R =I2R/2

Total Score:

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