• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

P.d. of a supply delivering current

Extracts from this document...

Introduction

Physics Experiment Full Report

P.d. of a supply delivering current


Aim : To show how the potential different (p.d.) across a power supply is affected

      by current and demonstrate energy loss due to internal resistance.

Apparatus :


For set up 1

1. 2 Dry Batteries (1.5V each)

2. 3 Light bulbs (2.5V 0.3A)

3. 3 switches

4. Voltmeter (0 – 5 V)

5. Wires


For set up 2

1. 2 Dry Batteries (1.5V each)

2. Ammeter (0-1A)

3 Resistance box (0-100Ω)

4. Switch

5. Wires

For set up 3

1. Extra high tension power supply

2. Milliammeter (0-1mA)

3. Resistance box (0-10MΩ)

4. Wires

Set up 1
(Figure)

image00.png
Set up 2
(Figure)
image01.png

Set up 3
(Figure)
image02.png

Theory

A power supply is a source of electrical energy. When it is connected to a circuit,

it generates a current. In this experiment we are going to measure the p.d. across

the terminals of a power supply as current is drawn from it. We shall see whether

the p.d. is equal to the electromotive force(e.m.f) of the supply.

Points about internal resistance we should know.

ⅰ)The internal resistance of a power supply can be considered as a

resistor connected in series with the power supply.
ⅱ)

...read more.

Middle

) terminal voltage = 0 V
(
) the power supply is in short circuit
(
) the maximum possible current , imax = E / r
(
) the power supply’s entire energy output is being wasted internally as heat .
image06.png
 


We also can develop E = v + i r  into
   
iE      =     iV     +     i²r
power supplied           power delivered to          power dissipated
   by a cell               external circuit             inside battery




Procedure
(set up 1)
(1) Using wires to connect all the apparatus
(2) Connect 3 Light bulbs & a voltmeter in parallel

(3) Each circuit should contain one light bulb and one switch
(4) Switch on the switch one by one
(5) Notice the different shows on the voltmeter
(6) drop down the reading.


Results and Measurements (set up 1)
Tabulate the results as follows
e.m.f of the system (ε) = 2.50 ± 0.05 V


Experimental Data (set up 1)

1st switch

2nd switch

3rd switch

R/Ω

1.1+0.1

10.9+0.1

11.5+0.1

V/V

1.60±0.05

1.40±0.05

1.30±0.05


Calculation (set up 1)
by V = I R , we have

R / Ω

1.1+0.1

10.9+0.1

11.5+0.1

V / V

1.60±0.03

1.40±0.05

1.30±0.05

I / A

1.46

0.128

0.113



Procedure

(set up 2)

(1) Using wires to connect all the apparatus in series
(2) Set the resistance to 20.0
Ω(initial value)
(3) Switch it on
(4) drop down the reading shows on the ammeter
(5) Turn to a larger resistance
(6)

...read more.

Conclusion



Q2 : We have assumed the resistance of the ammeter in the experiment to be negligiblty small. Is such an assumption justified ?
A: The resistance of the ammeter is negligible compared to the internal resistance of the EHT. However,it is not negligible compared to the internal resistance of the dry battery. It is because the resistance of the ammeter is near to the internal resistance of the dry battery.


Conclusion
From the above experiments , we can see that there is energy loss due to the internal resistance of the power supply (dry batteries) . This condition suits the equation
 
 E = V + i r 
( E = e.m.f. of the system,V = terminal voltage, i = current , r = internal resistance of the power supply )


Reference
Further Physics , Book 3 ; Third Edition ; Peter Fung , Peter Sun , Kenneth Young ; Longman Publishing Limited
Current Electricity , Dr.Ken Chan , P.43 – 44 , Wood Fire Publishing Co.


...read more.

This student written piece of work is one of many that can be found in our AS and A Level Electrical & Thermal Physics section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related AS and A Level Electrical & Thermal Physics essays

  1. In this experiment, we will measure the e.m.f. and the internal resistance of a ...

    of the dry cell: = (0.04/1.29) x 100% = 3.1 % Maximum percentage error in the internal resistance of the dry cell: = � (0.3 / 0.7) x 100% = 42.9 % Discussion Accuracy & improvements In this experiment, we had made several experimental errors but we can improve the experiment by the following improvements.

  2. Single Phase Transformer (Experiment) Report.

    Input Power /W Power loss /W 13 12.99999 Short-circuit test. R1 = Psc = 30 = 3.40? I12 2.972 X1 = (Vsc)2 - R12 = 13.86 V ( I1 ) Load Test Voltage regulation = V2(no load) - V2(loaded) [x 100%] = V2(no load)

  1. The potato - a source of EMF

    Knowing that the gradient is negative I am informed that all of the results were recorded in a way that if a result fell short of the expected reading it was tested again. This means there should not be too many anomalies on the graphs.

  2. The aim of the experiment is to verify the maximum power theorem and investigate ...

    1.2 1.2 1.2 1.0 1.2 In the following calculations, we have made several assumptions: 1. Air resistance is neglect 2. The friction is evenly distributed on any surface of the sand paper (i.e. the friction should be the same over the whole sand paper.)

  1. Choosing a light source

    Distance (cm) Light meter (Lux) LDR (light dependent resistance) (ohms) In period 1: I will also try to work out the power output for each light bulb, and this is the kind of table I will produce. Light bulbs (Coloured)

  2. Find The Internal Resistance Of A Power Supply

    And I is the current flowing through the circuit. We know that: V = IR This equation can be substituted into equation (1) to give us the equation: EMF = terminal potential difference + lost volts E = V - Ir (2) Using this equation, we know that the internal resistance of the power supply is constant.

  1. Coursework To Find The Internal Resistance Of A PowerSupply

    take the voltage and ammeter reading and then switching it off again to let it cool. The load resistance will be increased uniformly by moving the contact in the potentiometer across 2cm seven times and therefore current readings will decrease.

  2. Geothermal energy is not easily accessible with our current technology. Our main focus was ...

    By combining all of these ideas we can come up with one new innovative system that could revolutionize geothermal energy production. It starts out with a thin rectangular pipe. This pipe is then pressurized and heated by the magma. Then a surfactant is added to the water before it is placed in the heating chambers.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work