We’ll show you how you can determine the voltage supplied by different arrangements of cells in an...

We’ll show you how you can determine the voltage supplied by different arrangements of cells in an electrical circuit.

Cells in Series and

Parallel

We’ll start out by looking at cells arranged in series

total 1 2

6.0V

V V V

Cells in Series

The rule is, the total voltage supplied by a battery of cells in series is the sum of the voltages supplied by each cell.

The total voltage supplied by cells in series is the sum of the voltages supplied by each cell.

6.0 V

9.0 V

total 1 2

9.0V 6.0V

15.0V

V V V

Vtotal = V1 + V2 + …

V ? V

The equation we can use is V total equals V1 + V2 etc. The voltage of each cell is added up.


6.0 V

9.0 V

total 1 2

9.0V 6.0V

15.0V

V V V


V ? V

Here are two cells in series, a 9 volt cell and a 6 volt cell.

6.0 V

9.0 V

total 1 2

9.0V 6.0V

15.0V

V V V


Two cells in series


Cells in series are in the same loop or pathway. Each electron has to pass through both cells.

6.0 V

9.0 V

total 1 2

9.0V 6.0V

15.0V

V V V

Two cells in series

The total voltage supplied by cells in series is the sum of the voltages supplied by each cell.Vtotal = V1 + V2 + …

As an electron passes through the 6 volt cell, its electrical potential increases by 6 volts.

6.0 V

9.0 V

total 1 2

9.0V 6.0V

15.0V

V V V

Two cells in series

e–


Then as it goes through the 9 volt cell, its electrical potential increases by another 9 volts.

6.0 V

9.0 V

total 1 2

9.0V 6.0V

15.0V

V V V

Two cells in series

e–


So passing through both cells, its electrical potential goes up by a total of 6 + 9, or15 volts.


6.0 V

9.0 V

total 1 2

9.0V 6.0V

15.0V

V V V

Two cells in series

e–


V

A voltmeter measures the difference in electrical potential of electrons, or what is called potential difference between different points in a circuit.


6.0 V

9.0 V

total 1 2

9.0V 6.0V

15.0V

V V V


In this location, the leads from the voltmeter touch the circuit at point A and point B, so it measures the potential difference between point A and point B


6.0 V

9.0 V

total 1 2

9.0V 6.0V

15.0V

V V V

VA

B


The reading on the voltmeter is 6.0 volts.

6.0 V

9.0 V

total 1 2

9.0V 6.0V

15.0V

V V V

6.0 V

A

B


When the leads from the voltmeter touch point B and point C, it measures the potential difference between points B and C

6.0 V

9.0 V

total 1 2

9.0V 6.0V

15.0V

V V V

V

A

B

C


So it reads 9.0 volts.


6.0 V

9.0 V

total 1 2

9.0V 6.0V

15.0V

V V V

9.0 V

A

B

C


So what do you think the voltmeter will read if its leads touch point A and point C?

6.0 V

9.0 V

total 1 2

9.0V 6.0V

15.0V

V V V

? V

A

B

C


The total potential difference is 6 + 9, or 15 volts, so the voltmeter reads 15 volts.

6.0 V

9.0 V

total 1 2

9.0V 6.0V

15.0V

V V V

15.0 V

A

B

C


To find the total voltage supplied by cells in series, the given equation can also be used.

6.0 V

9.0 V

total 1 2

9.0V 6.0V

15.0V

V V V

V Vtotal

= ?


And because we have only 2 cells, we can say that the total voltage, V total is equal to V1 + V2

6.0 V

9.0 V

total 1 2

9.0V 6.0V

15.0V

V V V

V Vtotal

= ?


We’ll call the voltage of the 6 volt cell, V1

6.0 V

9.0 V

1total 2

6. 9.0V 0V

15.0V

V VV

V Vtotal

= ?


And the voltage of the 9 volt cell, V2

6.0 V

9.0 V

total 1 2

6. 9.0

.0V

0V

15

V

VV V

V Vtotal

= ?


So V total equals V1 + V2, or 6 + 9

6.0 V

9.0 V

total 1 2

6.0V 9.0V

15.0V

V V V

V Vtotal

= ?


Which is equal to 15 volts.

6.0 V

9.0 V

total 1 2

6.0V 9.0V

15.0V

V V V

V Vtotal

= ?


We can represent cells in series in a more compact way. (click) We bring the cells together…

6.0 V

9.0 V

And represent them with alternating short lines for the negative terminals

6.0 V–

9.0 V–

And alternating longer lines for the positive terminals.

6.0 V

+

9.0 V

+

So its positive to negative (click)

6.0 V9.0 V

To positive (click)

6.0 V9.0 V

To negative (click)

6.0 V9.0 V

And we can replace the 6 volts and 9 volts (click) by 15 volts, for the total battery.

6.0 V9.0 V15.0 V

A very common voltage for cells is 1.5 volts, so two 1.5 volt cells in series can be depicted like this.

1.5 V1.5 V

And because they are series, the total voltage (click) of this battery is 3 volts

1.5 V1.5 V3.0 V

A group of four 1.5 volt cells in series would have a total of four times 1.5…

1.5 V1.5 V1.5 V1.5 V

Which is equal to 6 volts

1.5 V1.5 V1.5 V1.5 V

6.0 V

Cells in parallel behave differently than cells in series. .

total 1 2

6.0V

V V V

Cells in Parallel

The rule is, the total voltage supplied by cells of equal voltage in parallel is the same as the voltage supplied by each cell.

total 1 2

6.0V

V V V

The total voltage supplied by cells of equal voltage in parallel is the same as the voltage supplied by each cell.

Just a note here, in this course, any cells that are in parallel to each other, will be of equal voltage.

total 1 2

6.0V

V V V


The equation we can use for cells in parallel is V total = V1 = V2 etc.

total 1 2

6.0V

V V V

Vtotal = V1 = V2 = …


Here is an arrangement of two 6 volt cells in parallel

total 1 2

6.0V

V V V


Two cells in paralle

l


6.0 V6.0 V6.0 V

If we attach the leads of a voltmeter across the first cell, the voltage reads 6 volts

total 1 2

6.0V

V V V



6.0 V

6.0 V6.0 V6.0 V

And if we attach the leads across the second cell, it also reads 6 volts.

total 1 2

6.0V

V V V



6.0 V

6.0 V

What do you think the voltmeter will read if we attach it to the whole combination, like this?

total 1 2

6.0V

V V V



? V6.0 V6.0 V6.0 V

We see it also reads 6 volts.

total 1 2

6.0V

V V V



6.0 V

6.0 V6.0 V6.0 V

If we use the given equation, we have just two cells in parallel, so we can say V total = V1 = V2

total 1 2

6.0V

V V V



6.0 V

6.0 V6.0 V6.0 V

V1 and V2 are both 6 volts

total 1 2

6.0V

V V V



6.0 V

6.0 V6.0 V6.0 V

V total equals V1 equals V2, so V total is also 6 volts.

total 1 2

6.0V

V V V



6.0 V

6.0 V6.0 V6.0 V

Now, you may be wondering why one would put more than one cell in parallel when the voltage stays exactly the same?

WHY ?

6.0 V6.0 V6.0 V

The simple reason is, putting cells in parallel will make them last longer than having an individual cell. And the more cells there are in parallel, the longer they will last, given a certain load.

Putting cells in parallel makes

them last longer

6.0 V6.0 V6.0 V

Look at this simulation for a while. We’ve coloured the electrons that go through the left cell green, and the ones that go through the right cell blue. Here, we’re showing electron flow rather than conventional current.

e–

e–

e–

e–

e–

e–

e–

e–

e– e– e– e– e– e– e–

e– e– e– e– e– e– e–

e–

e–

e–

e–

e–

e– e– e– e–

e–

e–

e–

e–e–e–e–

6.0 V6.0 V 6.0 V6.0 V

e–e–

e–e–

All of the electrons pass though the resistor, where some of their potential energy is converted to heat. You can see that half of the electrons that pass through in a given time interval are green and half are blue

e–

e–

e–

e–

e–

e–

e–

e–

e– e– e– e– e– e– e–

e– e– e– e– e– e– e–

e–

e–

e–

e–

e–

e– e– e– e–

e–

e–

e–

e–

e–e–e–e–

resistor6.0 V6.0 V 6.0 V6.0 V

e–e–

e–

We can see that half of the electrons that go through the resistor (the green ones), get potential energy from the cell on the left.

e–

e–

e–

e–

e–

e–

e–

e–

e– e– e– e– e– e– e–

e– e– e– e– e– e– e–

e–

e–

e–

e–

e–

e– e– e– e–

e–

e–

e–

e–

e–e–e–e–

The green electrons get their energy from this

cell.

6.0 V6.0 V 6.0 V6.0 V

e–e–

e–

And half of the electrons that go through the resistor (the blue ones), get potential energy from the cell on the right.

e–

e–

e–

e–

e–

e–

e–

e–

e– e– e– e– e– e– e–

e– e– e– e– e– e– e–

e–

e–

e–

e–

e–

e– e– e– e–

e–

e–

e–

e–

e–e–e–e–

The blue electrons get their energy

from this cell.

6.0 V6.0 V 6.0 V6.0 V

e–e–

e–

So half of the energy converted to heat in the resistor came from the cell on the left, carried by the green electrons, and half came from the cell on the right, carried by the blue electrons.

e–

e–

e–

e–

e–

e–

e–

e–

e– e– e– e– e– e– e–

e– e– e– e– e– e– e–

e–

e–

e–

e–

e–

e– e– e– e–

e–

e–

e–

e–

e–e–e–e–

Half of the energy converted to heat in

the resistor came from the cell on the left, and half came from the cell on the

right.

6.0 V6.0 V 6.0 V6.0 V

e–e–

e–

So the two cells are sharing the work.

e–

e–

e–

e–

e–

e–

e–

e–

e– e– e– e– e– e– e–

e– e– e– e– e– e– e–

e–

e–

e–

e–

e–

e– e– e– e–

e–

e–

e–

e–

e–e–e–e–

These two cells are sharing the work

6.0 V6.0 V 6.0 V6.0 Ve–

e–

e–

You can see that each cell is supplying energy at half the rate the resistor is using the energy.

e–

e–

e–

e–

e–

e–

e–

e–

e– e– e– e– e– e– e–

e– e– e– e– e– e– e–

e–

e–

e–

e–

e–

e– e– e– e–

e–

e–

e–

e–

e–e–e–e–

Each cell is supplying energy at half the rate the resistor is using the

energy.

6.0 V6.0 V 6.0 V6.0 V

e–e–

e–

Now we’ll cut the wire from the cell on the right, closing off this loop of the circuit.

e–

e–

e–

e–

e–

e–

e–

e–

e– e– e– e– e– e– e–

e– e– e– e– e– e– e–

e–

e–

e–

e–

e–

e–

6.0 V6.0 V

e–

e–

In order for the resistor to convert energy to heat at the same rate as it was before,

e–

e–

e–

e–

e–

e–

e–

e–

e– e– e– e– e– e– e–

e– e– e– e– e– e– e–

e–

e–

e–

e–

e–

e–

In order for the resistor to convert energy to heat at

the same rate as it was before, the green electrons

must extract energy from the cell on the left twice as quickly as it did when both cells were working

6.0 V6.0 V

e–

e–

the green electrons must extract energy from the cell on the left twice as quickly as they did when both cells were working

e–

e–

e–

e–

e–

e–

e–

e–

e– e– e– e– e– e– e–

e– e– e– e– e– e– e–

e–

e–

e–

e–

e–

e–

In order for the resistor to convert energy to heat at

the same rate as it was before, the green electrons

must extract energy from the cell on the left twice as quickly

as they did when both cells were

working

6.0 V6.0 V

e–

e–

The cell on the left is doing all the work now, and losing energy faster, so it will burn out much more quickly than it would if both cells were sharing the work.

6.0 V6.0 V

e–

e–

e–

e–

e–

e–

e–

e–

e– e– e– e– e– e– e–

e– e– e– e– e– e– e–

e–

e–

e–

e–

e–

e–

This cell will burn out much faster than it would if both cells were

working

e–

e–

You can see that when two cells in parallel are both operating, this cell loses its energy more slowly, so it will last longer.

e–

e–

e–

e–

e–

e–

e–

e–

e– e– e– e– e– e– e–

e– e– e– e– e– e– e–

e–

e–

e–

e–

e–

e– e– e– e–

e–

e–

e–

e–

e–e–e–e–

When two cells in parallel are both operating, this cell loses its energy more slowly, so it will last

longer

6.0 V6.0 V 6.0 V6.0 V

e–e–

e–

Having more cells in parallel will share the load even more, making each cell last even longer!

6.0 V6.0 V6.0 V 6.0 V 6.0 V

Having more cells in parallel will share the load even more, making each cell last even

longer!

Now we’ll look at some cases where some cells are in series and some are in parallel.

total 1 2

6.0V

V V V

Cells in Series and

Parallel

In the following examples, we’ll assume that each individual cell, shown by one short line and one long line, has a voltage of 1.5 volts.

Each single cell is 1.5 V

Let’s say we’re asked to determine the reading on the voltmeter for this arrangement of cells

V


V = ? V

We see that the group of cells on the red (click) line on the left has three 1.5 volt cells in series

V


V = ? V

Three 1.5 V

cells in series

When cells are in series, their voltages add up, so the total voltage of this (click) group of three cells is 3 times 1.5, which is 4.5 volts

V


V = ? V

3 × 1.5= 4.5 V

This group of 3 cells in series on the second red line, (click) will also have a total of 4.5 volts.

V


V = ? V

3 × 1.5= 4.5 V

4.5 V

As well as (click) this group

V


V = ? V

3 × 1.5= 4.5 V

4.5 V 4.5 V

And (click) this one

V


V = ? V

3 × 1.5= 4.5 V

4.5 V 4.5 V 4.5 V

This is just like having four 4.5 V cells in parallel

V


V = ? V

4.5 V 4.5 V 4.5 V 4.5 V

Four 4.5 V cells in parallel

When cells are in parallel, (click) the total voltage is the same as the voltage of each cell,

V


V = ? V

4.5 V 4.5 V 4.5 V 4.5 V


When cells are in parallel, the total voltage is the same as the voltage of

each cell

So the total voltage of this combination of cells is 4.5 volts

V


V = 4.5 V

4.5 V 4.5 V 4.5 V 4.5 V



each cell

Here’s another example. We’re given this arrangement of 1.5 volt cells and we’re asked to determine the voltage on the voltmeter.

V


V = ? V

Looking at these two cells on the left, (click) we see that they are two 1.5 volt cells in parallel

V


V = ? V

Two 1.5 V

cells in paralle

l

Remember, for a group of cells in parallel, the total voltage of the group is the same as the voltage of each cell in the group.

V


V = ? V

Two 1.5 V

cells in paralle

l


each cell

So as far as voltage is concerned, (click) this combination of cells is like a single 1.5 volt cell

V


V = ? V

1.5 V

Now, we’ll consider this group of cells up here (click) they are three 1.5 volt cells in series.

V


V = ? V

1.5 V

Three 1.5 V cells in series

Remember, when cells are in series, their voltages add up..

V


V = ? V

1.5 V

In series, voltages add

up

So the total voltage of three 1.5 volts in series is 3 times 1.5, which is 4.5 volts

V


V = ? V

1.5 V

3 × 1.5 = 4.5 V

We can consider these two groups of cells as being in series with each other.

V


V = ? V

1.5 V

4.5 V

If we replace the voltmeter with a resistor so current can flow, (click) every electron that goes through a cell on the left, must then go through the cells on the top.


1.5 V

4.5 V

e– e–

Adding up the voltages of these two groups in series,


1.5 V

4.5 V

V V = ? V

Adding up the voltages of these two groups in series,Vtotal = 1.5 V + 4.5 V = 6.0 V

We get V total is equal to…


1.5 V

4.5 V

V V = ? V


1.5 volts


1.5 V

4.5 V

V V = ? V


Plus 4.5 volts


1.5 V

4.5 V

V V = ? V


Which equals 6 volts


1.5 V

4.5 V

V V = ? V


So the voltage on the voltmeter will read 6.0 volts


1.5 V

4.5 V

V V = 6.0 V


Another way we can look at it is, an electron will pass through one of the cells on the left, increasing its potential by 1.5 volts, then through the cells on top, increasing its potential by another 4.5 V, so this electron has increased its electrical potential by a total of 6 volts.


1.5 V

4.5 V

e–

Has increased its

electrical potential by

6.0 volts

A different electron will pass through the other cell on the left, increasing its potential by 1.5 volts, then through the cells on top, increasing its potential by another 4.5 V, so this electron has also increased its electrical potential by a total of 6 volts.


1.5 V

4.5 V

e–

Has increased its

electrical potential by

6.0 volts

So every electron that passes through this arrangement of cells, increases its electrical potential by a total of 6.0 volts


1.5 V

4.5 V

Every electron that goes

through this arrangement of cells, increases

its electrical potential by a total of 6.0 V

So again, the voltage supplied by this arrangement is 6.0 volts.


1.5 V

4.5 V

V V = 6.0 V

We’ll show you how you can determine the voltage supplied by different arrangements of cells in an...

Documents

Transcript of We’ll show you how you can determine the voltage supplied by different arrangements of cells in an...