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P.d. of a supply delivering current

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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.

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