• 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. Single Phase Transformer (Experiment) Report.

    - V2(loaded) [x 100%] = V2(no load) 115-110 [x100%] = 4.35% 115 Efficiency = output power = V2 I2 [x 100%] = 110 x 5.5 [x 100%] Input power Pin 637.5 Efficiency = 94.9% Calculation of the Voltage regulation and Efficiency for the load test using the Equivalent circuit parameters.

  2. 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.

  1. Investigating the E.m.f and Internal Resistance of 2 cells on different circuit Structures.

    the equation of the line is found then so can the e.m.f. If gradient can't be found by using the graph then a value of I can be substituted ( x axis) into the equation and then the equation can be solved to find specific component, r or E.

  2. The potato - a source of EMF

    From the data I have been given I can now calculate the internal resistance and the EMF of the potato cell. I will then compare the results from the three different distances between the electrodes and draw conclusions. I predict that the internal resistance will decrease as the electrodes are moved closer.

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

    Moreover, the sand paper may make the wooden surface become smoother, and hence the static and kinetic friction may be different from the original one. So the numbers of pulling process should be minimized. On top of the above precautions, the spring balance is not connected to the wooden block firmly; it may be disconnected during the pulling process.

  2. Choosing a light source

    (which represents the wavelength of the light in meters) and again this is sometimes given or otherwise calculate it. This is the formula used in general M ?=d. sin ? (1) To calculate for ? (wavelength) (4) To calculate for sin ? (angle) You do ?= d. sin ?

  1. Find The Internal Resistance Of A Power Supply

    From Kirchhoff's second law we know that: E = IR + Ir (1) Where E is the Electromotive Force (EMF) of the power supply, maximum energy per unit charge that the power supply can deliver. R is the external (load) resistance. And I is the current flowing through the circuit.

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

    Along the same reasoning, producing heat from convection has almost the same implications. Convection is when heat moves with the material. When the temperature gradient is higher than the material below it, the material would expand and become less dense.

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