• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3

# To find out the internal resistance and EMF of a given power supply

Extracts from this document...

Introduction

Internal Resistance and EMF (ELC)

Physics Portfolio By Clement Ng 12.6

Aim: To find out the internal resistance and EMF of a given power supply.

Method:

• Arrange apparatus as shown in the below circuit diagram
• Start off by recording the corresponding voltage (terminal pd) and current by figuring out the values available in the voltmeter and ammeter.
• Repeat previous procedures while the external load is varied, so that you obtain a set of voltage and current readings.
• Calculate averages and plot a graph of current against voltage. Use the graph to figure out the internal resistance and the EMF of the power supply.

Middle

0.998

0.09

10.0

0.502

0.40

1.0

1.094

0.07

15.0

0.584

0.35

1.5

1.207

0.01

100.0

1.223

0.01

150.0

 Voltage (v) (± 0.01V) Current (A) (± 0.01A) Resistance (Ω) (Provided on resistor) 1.061 0.10 10.0 0.460 0.45 1.0 1.089 0.07 15.0 0.580 0.38 1.5 1.193 0.01 100.0 1.215 0.01 150.0

Observations: As different external loads, or resistors were plugged into the circuit, several voltage and current readings were recorded. It was also noticed that the readings on both meters kept on jumping, therefore uncertainties of ± 0.01 was deduced for both the current and voltage values.

Skill 4 Data Processing

Arranged in increasing voltage and resistance values:

Conclusion

Improvement 3:

To improve on graph plotting skills, we could use a computer to help us. Many computer software’s nowadays can help us plot the results and calculate the gradients directly. After the experimental results are obtained, simply copy them into the program and plot graphs of V/I. These graphs would be much more accurate then hand plotted ones, and internal resistance values would be calculated to the highest degree of accuracy.

Unfortunately, allowing the computer to do the job for you does not show any skills in data processing. Thus another improvement can be done by using calculus. Calculus is a good tool in mathematics to calculate the gradient of a known equation. The final results maybe even more accurate then the computer values. However, this would be inappropriate in this experiment, since knowing the equation would already give as the internal resistance and EMF values. This improvement maybe effective on other investigations.

By Clement Ng 12.6

Thursday, July 4 2002 05:12AM

--

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

# 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 ...

/ (0.24 - 0.00) = 6.38 ?m = [(8.21-7) + (6.38-7)] /2 = 0.30 The e.m.f. of the dry cell: 1.24 V � 0.04 V Total internal resistance of the dry cell: 0.7? � 0.3 ? The e.m.f. of one cell = 1.29 V� 0.04V Since the resistance of the resistor was assumed to be 5.6 ?

2. ## The potato - a source of EMF

From the table I can see that, as I predicted, the internal resistance decreases as the electrodes are moved closer together. However, the EMF is not exactly what I expected as it increases somewhat from electrode distance 4cm - 2cm.

1. ## Measuring the e.m.f. And Internal Resistance of a Cell

This is called a parallax. * The ammeter also only read to the nearest 0.02 of an Amp, which is an error of � 0.01A. * The digital voltmeter only read to the nearest 0.01 of a Volt and so had an error of � 0.005V.

2. ## Sources of e.m.f. &amp;amp;#150; internal resistance.

Consider a source of e.m.f. E with an internal resistance r, delivering a current I into a load resistance R. The current I is given by: I=E/R + r The power P dissipated in the load is given by: P=I2R=(E/R + r)2R For a given internal resistance r, there will

1. ## Aim: To find out the internal resistance and EMF of a given power supply.

I (Current) V (Volts) 1 2 1.97 2.2 1 2.41 3.9 0.7 2.63 4.7 0.5 2.68 10 0.3 2.85 15 0.2 2.91 22 0.1 2.96 33 0.1 3.01 47 0.05 3.04 100 0 3.1 150 0 3.11 . Skill 5 Conclusions and Evaluations Conclusions The internal resistances are given from the slopes of the graphs.

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

Therefore, we should pay attention to the connection point between the spring balance and the wooden block during the experiment. In addition, beam balance is necessary in the experiment for measuring the masses, which have to be hung at the hanger.

1. ## Investigating Internal Resistance

The exception being if it were, for example, 0.20A it would simply be recorded as 0.2A. Another problem is that with the voltage the values are constantly changing. To overcome and improve the situation the highest value seen on the voltmeter added to the lowest value seen all divided by two should give the average voltage for that particular repeat.

2. ## Investigating the Smoothing Effect of a Capacitor on a Resistive Load

The exponential decay graph of the smoothed output voltage against time can be represented by a general equation of the form x=x0 e- ?/RC. The relevant equation of this form for my experiment is: V=V0 e- ?/RC where V= the voltage after a time, ?

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