AS Physics Unit 1 5 Direct Current - animatedscience.co.uk · working out the current flow we have...

AS Physics Unit 1 5 Direct Current

Mr D Powell

Mr Powell 2009 Index

Chapter Map


5.1 Circuit Rules

Specification link-up 3.1.3: Current electricity: Circuits What are the rules for series and parallel circuits? What are the principles behind these rules? How do we use the rules in circuits?


The “current law” states that at a junction all the currents should add up.

I3 = I1 + I2 or I1 + I2 - I3 = 0

• Current towards a point is designated as positive.

• Current away from a point is negative. • In other words the sum of all currents

entering a junction must equal the sum of those leaving it.

• Imagine it like water in a system of canals!

Kirchoffs Law I

Examples;

If I1 = 0.1A, I2 = 0.2A, I3 = 0.3A If I1 = -0.1A, I2 = -0.2A, I3 = -0.3A If I1 = 2A, I2 = 3A, I3 = 5A

Kirchoffs Law I

There are some important multipliers for current:

1 microamp (1 A) = 1 x 10-6 A 1 milliamp (mA) = 1 x 10-3 A

Also remember to make sure you work out current in Amps and time in seconds in your final answers!


Work out the currents and directions missing on these two junctions?

Kirchoffs Law I – Questions?

7A 3A


Simple e.m.f example

1) If I = 100 mA what is that in amps?

100 mA = 0.1 A

2) What is the current in each resistor?

0.1 A

3) Work out the voltage across each resistor.

V = IR so

V1 = 0.1A x 30 = 3 V

V2 = 0.1A x 40 = 4 V ;

V3 = 0.1A x 50 = 5V

4) What is the total resistance?

RT = 30 + 40+ 50 = 120

5) What is the battery voltage?

V = 0.1A x 120 = 12V


PD Rules - Series


This follows from the law of conservation of energy: The total energy per coulomb produced = the total energy per coulomb delivered

Kirchhoff's Voltage Law

For any complete loop of a circuit, the sum of the emfs equal the sum of potential drops round the loop.


battery pd = 12V

This means:

each coulomb of charge leaves the battery

with 12J of electrical energy

If the variable resistor is adjusted so that there

is a pd of 4V across it

This means:

each coulomb of charge uses 4J of energy passing through it

PD Rules - Parallel


The “voltage law” states that the sum of e.m.f’s around a circuit or loop is equal to zero. i.e. all energy is transferred before you return back to the cell.

In other words the sum of all voltage sources must equal the sum of all voltages dropped across resistances in the circuit, or part of circuit.

Think of it like walking around a series of hills and returning back to point of origin – you are then at the same height!

For more complex examples we must note the following rules; There is a potential rise whenever we go through a source of e.m.f from the – to

the + side.

There is a potential fall whenever we go through a resistance in the same direction as the flow of conventional current. i.e. + to -

NB. Both laws become obvious when you start applying them to problems. Just use

these sheets as a reference point.

Kirchoffs Law II


Kirchoff Laws – More Complex Questions

Given the following circuit can you pick out how the current might behave using kirchoffs current laws. We are looking to find the current in the main branch or 0.75 resistor and also through the 1 resistor

You may want to redraw the circuit then apply simple ideas of additive resistance and ratios to find out the currents?

Kirchoff Law I - Answers

In detail this means can rewrite the circuit and fill in the values for V and I as such…

Energy (p.d.) is shared simply according to resistance.

Answers......

This example is more simple than it looks. In fact you have one resistor on its own (0.75)

Then the other three are in parallel with each other. With the 1 on one branch and the two 1.5 resistors on the other.

You can then simply use resistance ratios to determine the current flow.

The ratio for the parallel part is 1:3 so we say that the least current flows through the most resistive part. Hence: current through 1 resistor must be 0.75A.

The current through main branch must be the sum of these i.e. 1A

Complex Examples

P

E1 E2

R1

R2

I2

R3

I3

I1

If we apply Kirchhoffs laws (previous slide) about current we can say that;

I1 = I2 + I3 or 0 = I2 + I3 - I1

If we apply Kirchhoffs laws (previous slide) about pd = 0 in closed loop we can say the following two things

Starting at Point “P” and going clockwise around the left-hand loop;

-I3R2 + E1 - I1R1 = 0 Starting at Point “P” and going

clockwise around the right-hand loop;

-E2 - I2R3 + I3R2 = 0

Hint1: This is complex and you must try and be consistent in your calculations in direction and which way the current is flowing or p.d. is lost!

Hint2: Solve this using Simultaneous Eq method. You will not be asked to do anything so complex in the real exam!

EXTENSION WORK

Complex Examples

P

E1 E2

R1

R2

I2

R3

I3

I1

Starting at Point “P” and going clockwise around the left-hand loop; -I3R2 + E1 - I1R1 = 0

Starting at Point “P” and going clockwise around the right-hand loop;

- E2 - I2R3 +I3R2 = 0

p.d. down with flow I3 left loop

p.d. is increased with flow I1

p.d. is lowered with flow I1

PD against flow of current

PD against flow of current

p.d. is up with flow I3 right loop

Hint: consider only E direction not I’s

EXTENSION WORK

Complex question….

P

E1 E2

R1

R2

I2

R3

I3

I1

Use the theory from the previous slide to answer the question below working out the current flow we have called I3.

Hint use the technique shown on the previous slide. But try and reason it out yourself with the rules you have been given. This may take some time!

Your answer should include the following;

1. Reference to the rules of current flow & p.d 2. Explanation as to why each contribution is + or – 3. Equation for each loop 4. Answer for I3

E1 = 6V E2 = 2V R1=10 R2=10 R3=2

Answer: I3 = 0.4A

EXTENSION WORK

Answers to question.... P

E1 E2

R1

R2

I2

R3

I3

I1

E1 = 6V E2 = 2V R1=10 R2=10 R3=2

Answer: I3 = 0.4A

Starting at Point “P” and going clockwise around the left-hand loop;

-I3R2 + E1 - I1R1 = 0 Eq 1 -I310 + 6V - I110 = 0 I3+I2 = -6V/10 I3+I2 = -0.6A - Eq 2 or I3 = (0.6A-I2) Starting at Point “P” and going clockwise around the

right-hand loop; -E2 - I2R3 +I3R2 = 0 Eq 3 -2V - I210 +I310 = 0 Sub Eq 2 into 2 to eliminate I3

-2V - I210 + (0.6A-I2) 10 = 0 4V/20 = I2 = 0.2A Hence – feed back into Eq2 to yield I3 = 0.4A

Kirchoffs 1st Law; I1 = I2 + I3

Kirchoffs 2nd law; Sum PD Loop AEDBA 30V = 20I3 + 5I1 - Eqn 1 Sum PD Loop FEDCF 10V = 20I3 - 10I2

10V = 20I3 - 10(I1 - I3) 10V = 30I3 - 10I1 - Eqn 2 Add 2 x Eqn 1 + Eqn 2

70V = 70 I3

I3 = 1 A Substitute this in Eqn 1

I1 = 2A so I2 = 1 A

Complex example..

This worked example relies on two equations found from two loops. Each defined for a separate power source.

Solve simultaneously to find I’s

EXTENSION WORK


Answer….

Plenary Question….

A battery of emf 24 V and negligible internal resistance is connected to a resistor network as shown in the circuit diagram in the diagram below.

Show that the resistance of the single equivalent resistor that could replace the four resistors between the points A and B is 50Ω.

BASIC

A B

30 40

60 120

24 VR1

Extension… if current = 0.16A what is R1



5.2 More about Resistance

Specification link-up 3.1.3: Current electricity: circuits How can we calculate resistances in series or in parallel – In depth Theory? What is resistance heating? How can we calculate the current and pds for each component in a circuit?


Circuit Rules

For resistance calculations at AS you can use the premise of the following rules but they are not sorted can you do this…..


Series Resistance


Resistors in series can be replaced by one single resistor:

I I

R1 R2 R3

V1 V2 V3

RT

VT

I I

VT = V3

But V = IR

I RT = I R1 + I R2 + I R3

V1 + V2 +

I RT = I R1 + I R2 + I R3

RT = R1 + R2 + R3

Resistors in series:


Parallel


Resistors in series can be replaced by one single resistor:

R1

R2

R3

I1

I2

I3

RT

V

I

IT = I3 But V = IR and I = V R

V = V + V +

I1 + I2 +

V

RT R1 R2 R3

V

IT

RT R1 R2 R3

V = V + V + V RT R1 R2 R3

1 1 1 1 = + +

Resistors in Parallel


Generalised Formulae

So all this leads us to a generalised formulae for any number of resistance. In Series resistance is simple but for Parallel resistances we must use the reciprocal.

NB: when multiplying out the second formula make sure you treat all terms in the same way!


Test it out….

If all the bulbs are of equal resistance answer the following;

1) Find all the meter readings

2) What is the total resistance of the

bottom branch

3) If I swapped all three bulbs for one bulb what resistance should it be so that the current flow in the main branch is still 3A (i.e. combine all resistances using formulae)


Answer…. Plenary Question….

A battery of e.m.f 12 V and negligible internal resistance is connected to a resistor network as shown in the circuit diagram. 1. Calculate the total resistance

of the circuit. (3)

2. Calculate the current through the 30Ω resistor. (1)

Basic

1) (three parallel resistors) (1) R = 9.23Ω (1) R = 9.2Ω 9.2 Ω and 30 Ω in series gives 39Ω (1) (allow e.c.f. from value of R)

2) (V = IR gives) 12 = I × 39 I = 0.31 A (1) (allow e.c.f. from (a))

40 50

20

40

12 V

30Ω

30Ω

40

1

20

1

30

11

R


Charge carriers transfer kinetic energy to positive ions through repeated collisions. The pd across the material then provides an accelerating force to the charge carrier which then collides with another positive ion. V = I R

P = I V

P = I2 R

P = V2

R Energy per second transferred

to the component P:

In the steady state this equals the heat transfer to the surroundings

Resistance Heating...


Answer….


Two resistors, A and B, have different resistances but otherwise have identical physical properties. E is a cell of negligible internal resistance. When the resistors are connected in the circuit shown in figure 1, A reaches a higher temperature than B. When connected in the circuit shown in figure 2, B reaches a higher temperature than A. Explain these observations fully, stating which resistance is greater. (6 marks)

iSlice

A B

E

A

B

E

power determines heat produced (1) in series, current is same (1) I2 RA must be > I2 RB (1)

in parallel, p.d. is same (1)

<



What is the link.....


5.3 EMF and Internal Resistance

Specification link-up 3.1.3: Current electricity: electromotive force and internal resistance Modelling Electrical Circuits

Why is the pd of a battery (cell) in real use less

than its emf (electromotive force)?

How can we measure the internal resistance of a battery?

How much power is wasted in a battery?

AO3: HSW – record and the interpret readings from a graph

AO2: Apply Theories to the graphs

IrV

22

rR

RVIP

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dcex6.html

http://www.buchmann.ca/chap9-page1.asp









Modelling Electricity

One idea to help us explain electricity is to think of a electricity like gravitational potential energy. When you are up high you have lots of it. We can even use diagrams to help us to understand what is going on...

transfer to heat energy in resistor

Stored chemical energy

transfer to heat energy

in wires


Activity – Modelling Circuits...

Instructions... Work in a team of 3. Your team will be given some

information about an electricity model.

It is your job to study this, make a “Large Model” with diagram and then explain it to the rest of the class.

Each team is making a different model.

Group Work – try and assign roles… 1. Planner: organise team to complete work in allotted

time.

2. Presenter: writes some notes to explain model to the class or draws a diagram or table of items to help.

3. Maker(s): lead producing the model


What is Internal Resistance....

Let us picture a circuit with a real chemical power source; We expect 6V or 6JC-1 from a cell or the e.m.f. (electromotive force) but because our power source is real some of this energy will be lost inside the cell. Through resistance “r” The energy is converted into heat when a current is actually drawn from the cell through a circuit. We can use our R1+R2 = RT formula to work out the emf . For example if the internal resistance r = 4 and the load resistance R = 8. By using our formula;

= I(R + r) 6V = I(4 +8) I = 0.5A We also know that; Vload = - Ir Vload = 6V - 0.5A x 4 Vload = 4V So 4 Volts is measured across the terminals on the cell.


Practical…

Vload = (-r)I +

y = (m)x + c

Vload = y

-r = gradient

I = x

= c

There are various methods for exploring this idea but they all end up in creating a graph.

You can use a variable resistor method or the method of adding bulbs in parallel.

Both will give similar readings.

Try the adding bulbs one today.

Just work through the sheets and create a quality graphs and write a conclusions….


Panasonic Cell Example Results

y = -1.2914x + 6.5421 R² = 0.9987

4.50

5.00

5.50

6.00

6.50

7.00

0.00 0.50 1.00 1.50

Vo

ltag

e /

V

Current / A

Current /Amps +/-0.01A

Potential Difference /V +/-0.01V

0.27 6.17

0.50 5.90

0.72 5.63

0.90 5.40

1.09 5.15

1.24 4.92

1.37 4.76


D Cell Sample results....

Variable resistor method using 3 chemical cell or 3*r in series.

Current /Amps +/-0.01A

Potential Difference /V +/-0.01V

0.01 4.32

0.03 4.26

0.05 4.17

0.07 4.08

0.09 4.01

0.11 3.94

0.13 3.89

0.15 3.8

0.17 3.74

0.20 3.63


D Cell Results

y = -3.6383x + 4.3515 R² = 0.9974

3

3.2

3.4

3.6

3.8

4

4.2

4.4

4.6

0 0.05 0.1 0.15 0.2 0.25

Vo

ltag

e /

V

Current/ A

Graph to Show Internal Resistance of 3 Cells

Hence; 4.35V = emf -r = -3.638 But this is of 3

cells So r = 1.213



In the circuit shown the battery has emf and internal resistance r.

1) State what is meant by the emf of a battery….

Basic

2) When the switch S is open, the voltmeter, which has infinite resistance, reads 8.0 V. When the switch is closed, the voltmeter reads 6.0 V. a) Determine the current in the circuit when the switch is closed….. b) Show that r = 0.80 Ω.

2.4

V

S

r

• energy changed to electrical energy per unit charge/coulomb passing through

• [or electrical energy produced per coulomb or unit charge]

• [or pd when no current passes through/or open circuit]

iSlice

8.0

5.268

68

)(

r

r

Ir

rRI


Power Transfer Ideas....

We can also think about the situation as a power transfer and then rearrange…..

Pcircuit = Pcell + Pload or

= Ir + IR P= I = I2r + I2R P = = I(R + r)

but we know IR = Vload then

Vload = - Ir So we now have a formula which gives us the POTENTIAL DIFFERENCE across our power supply. Obviously this will be less than the emf , since you drop some voltage over the internal resistor, r. The Ir part is the volts lost in the power supply.


Load Resistance Matching…

This is very useful as then we can derive and plot a graph of Power against load resistance.

This peaks when R=r….

22

2

rR

RVIP

PRI

IrR

rRI

IrIR


Basic Fudge Maths...

RVI

R

RVI

RR

RVI

4

2

2

2

2

2

2

• There are 3 scenarios to consider from the data plotted in this graph about the Power delivery....

• We can think (using fudge maths) about three cases to give us an idea of the value of Power….

22

rR

RVI

RVI

R

RVI

R

RVI

2

2

2

2

2

0

rr

RVI

r

RVI

2

2

2

0

r<<R r=R

r>>R

Extension Work


Neat Calculus... We can also look at the idea of the

maximum of the graph and differentiate the function.

You will never have to do this but you can look at the function as a quotient differential form (AS Maths)

This means you differentiate using the concept of u & v to yield a complex derivative. You cannot do simple differentiation but have to do a complex sum as shown below......

rR

rR

Rr

rR

Rr

dR

dP

rR

RP

rR

RVI

3

2

3

2

2

2

2

2

0

2

2

v

dR

dvu

dR

duv

dR

dy

v

u

rR

Ry

Power delivered Gradient of curve Place at maximum

NB: you only need to know the outcome not the maths for AS

Extension Work


5.4 More Circuit Calcs

Specification link-up 3.1.3: Current electricity How can we calculate currents in circuits with: resistors in series and parallel? more than one cell? diodes in the circuit?


How a Solar Cell Works

Carbon, silicon and germanium have a unique property in their electron structure - each has four electrons in its outer orbital.

In silicon is a near perfect insulator. But if we coat or “dope” the surface with an impurity such as phosphorus which has five electrons in the outer shell we have a spare electron in the lattice as all outer shells have 8 for a perfect octet.

This means that we can separate charged electrons in our sandwich and produce an electric field. If a light photon through a glass screen falls onto our sandwich it releases the electrons to move through the field and produce an electric current. This is how a solar cell works.

p-type

n-type

Mono-crystaline solar panel providing an output of 4.4V, 90mA


Energy from the Sun

At the outer edge of the Earth’s atmosphere, solar radiation has a power of about 1.4 kilowatts per square metre.

There is more to the Sun’s radiation than just the light we see. The visible region represents only 49% of the energy that arrives from the Sun. Most of the rest of the energy (also about 49%) arrives as infra red radiation; ultraviolet radiation accounts for the remaining 2%. Many different ways have been developed for capturing and using this energy from the Sun. One of which is a solar cell. Light photons in the UV part of the spectrum are able to release electrons in a solar cell to transfer the suns energy to more useful forms….


5.5 Potential Divider

Specification link-up 3.1.3: Current electricity: potential divider What is a potential divider? How can we supply a variable pd from a battery? How can we design sensor circuits?

sVRR

RV

21

22

V1

V2

Vs

R1

R2


Potential Dividers Explained…

• Take a simple series circuit with uniform current flow and two equal resistors. The p.d. drop across each is the same

• Then ‘open-out’ the cell to show as a “rail”

• Then label the supply as Vs and the 0V as ground rail, the resistors and voltmeters as 1 & 2 (you could use a & b)

V

V

V

V

V1

V2

Vs

R1

R2


What is the output voltage……

• We can use this circuit to be able to find the output voltage across R2 so we can see a change in a component such as a thermistor.

• So add the output voltage Vout

• Output voltage is the same as the voltage across R2 i.e. V2 = Vout

V1

V2

Vs

R1

R2 Vout


Calculations

• Since the current is the same through both resistors we can define

2

21

2

2

21

2

21

21

VVRR

RV

RRR

VV

RIV

IRR

V

RRIV

IRV

sout

sout

out

s

s

Vs

R1

R2 Vout

I

V1

V2

NB: Now we can express Vout as ratio of resistance multiplied by Vs


I

Alternative Maths...


Simple Examples to think about...

?

?

?

?

?

?

A B

C

D

?


Temperature Sensors?

sout VRR

RV

21

2

Redraw the circuit shown and label the variables according to the rules.

Work out the Vout voltage for the two temperatures to verify the formulae that we have just derived;


AC Audio input

Variable voltage output to a loud speaker

A simple volume control

AC Audio input


Theory Summary

sout VRR

RV

21

2

V1

V2

Vs

R1

R2 Vout

A potential divider does just what is states. It divides a potential difference Think of a p.d. of 10V across a resistor. The p.d. will drop by 1V for each 10% of the resistor that the current passes through. From this theory two resistors will have a ratio which from the idea that V=IR will relate the output voltage on a resistor to the source voltage as shown. Obviously if resistor 1 and 2 are swapped Vout also swaps. We can replace one of the fixed resistors with; Variable resistor, which could act as a volume control or sensor i.e. Thermistor or LDR.


Summary of Uses….



In the circuit shown, the battery has negligible internal resistance.

Basic Calc

3) The circuit shown in the diagram acts as a potential divider. The circuit is now modified by replacing R1 with a temperature sensor, whose resistance decreases as the temperature increases. Explain whether the reading on the voltmeter increases or decreases as the temperature increases from a low value. (3)

iSlice Explaining

V

I

R

R

9.0 V

1

2

1) If the emf of the battery = 9.0V, R1 = 120 and R2 = 60, calculate the current I flowing in the circuit. (3)

2) Calculate the voltage reading on the voltmeter. (1)

(temperature increases, resistance decreases), total resistance decreases (1) current increases (1) voltage across R2 increases (1) or R2 has increased share of (total) resistance (1) new current is same in both resistors (1) larger share of the 9 V (1) or R1 decreases (1) Vout decreases (1)]

21

2inout

RR

RVV


Summary Questions…




Plenary Question....

1. Give an example of what you might use a potential divider for as well as a Light sensor.

2. What is the output voltage of this potential divider?

4.4V


More Simple examples to try...

0V

12V

VOUT

0V

100

100

0V

1.5V

VOUT

0V

50

45

0V

50V

VOUT

0V

10

75

0V

3V

VOUT

0V

75

25

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AS Physics Unit 1 5 Direct Current - animatedscience.co.uk · working out the current flow we have...

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