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# Investigate the current - voltage relationship for a resistor and filament lamp. To determine whether each obeys ohms law.

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Introduction

Page  of

ELECTRICITY

### Aim:

To investigate the current – voltage relationship for a resistor and filament lamp.  To determine whether each obeys ohms law.

Ohms law states that the current through a conductor is proportional to the potential difference across it provided that there is no change in temperature.  It is given by the formula V = IR

### Method:

• The circuit is set up as above, the power supply needs to be set at about 4V.  The first experiment will be carried out with the filament lamp.
• Electrons in a circuit move from the negative terminal to the positive terminal, during which energy can be lost through components causing a difference in the electrical potential energy between the negative and positive terminals.  This is called the potential difference and is measured with a voltmeter.  The voltmeter is connected in parallel with the component, to measure the voltage going through it.
• The ammeter measures the electrical current and is measured in amperes, or amps.  When the current is 1 amp, the flow of charge is 1 coulomb per second.

Care should be taken to make sure the voltmeter and ammeter are working correctly.

• The variable resistor is moved along slightly and readings from the ammeter and voltmeter are recorded in an appropriate table.  Readings are continued to be recorded at intervals along the variable resistor until it reaches the end or till sufficient results are taken.
• The current is then directed backwards, for negative results to be taken. This is done to see if there is any change in the results, as sometimes it can make a difference. It will help prove if a component is an ohmic conductor.
• Now the fixed resistor is put in place of the filament lamp.
• Again, readings are taken from the voltmeter and ammeter as the variable resistor is moved along.
• Readings are recorded in another table.  When the variable resistor has reached the end or there are sufficient results, the current can be directed backwards again for negative readings.

Middle

-0.031

-4

4.3

0.32

Resistor

Positive                                             Negative

 Pd./V I/A Pd./V I/A 0.05 0.003 -0.0025 -0.02 0.5 0.03 -0.04 -0.5 1 0.06 -0.07 -1 1.5 0.1 -0.1 -1.5 2 0.13 -0.14 -2 2.5 0.16 -0.17 -2.5 3 0.2 -0.2 -3 3.5 0.23 -0.23 -3.5

Analysis of results:

The graph for the filament lamp produced a curved line showing no proportionality and that the lamp is not an ohmic conductor.  A resistance cannot be calculated because it is not constant.

The graph for the fixed resistor produced a straight line.  Which means the current – voltage relationship is proportional and it is an ohmic conductor.  A resistance can be calculated because it is constant.

Gradient = ΔΥ   =   3.5   = 15.2Ω

ΔΧ       0.23

Conclusion:

Ohms law is only applicable when the temperature is kept constant.  The graph for the filament lamp had a curved line.  The filament got hotter as the current increased and therefore the resistance increased.  Because the temperature and current were not constant, the filament lamp is not an ohmic conductor.  The gradient of the curve is very small at the beginning showing that there is little resistance.  However as the current increases, the gradient of the curve increases too.  The resistance is changing due to this change in current and temperature.

Overall it is possible to determine two main things.  Firstly, the filament lamp is not an ohmic conductor, and secondly, the resistance becomes more constant as the current increases.

The fixed resistor produced a straight-line graph.

Conclusion

The potential divider is used to supply a variable p.d.  Therefore a desired level can be obtained.  The potential divider can allow the voltage across a component to be varied continuously between 0v and the battery voltage by use of a sliding contact.  Whereas a rheostat varies the resistance, it is designed to resist current flow.  Because it has a maximum resistance it does not allow the voltage across a component, or the current through it, to be reduced to zero.  A rheostat is also quite bulky and dissipates the surplus energy in the form of heat so it can get very hot.

1a)        V =IR

I = V/R

I = 8/4

I = 2a

b)        Current is constant, therefore I = 2a

1. V = IR

V = 2 x 3

V = 6V

1. V = IR

R(total) = 3 + 4

R(total) = 7Ω

V = 2 x 7

V = 14V

2a)        R(total) = 2 + 3 = 5Ω

1. I = V/R

I = 10/5

## I = 2a

1. V = IR

V =2 x 2

V = 4V

1. V = IR

V = 2 x 3

V = 6V

1. Terminal pd. = E – IR

E = terminal pd. + IR

E = 10 + 2 x 1

## E = 12V

4a)        1/R(total) = 1/R(1) + 1/R(2)

1/R(total) = 1/3 + 1/6

1/R(total) = 3/6

R(total) = 6/3 = 2Ω

1. V = IR

V = 6 x 2

V = 12V

1. 12V
2. I = V/R

I = 12/3

1. I = V/R

I = 12/3

## I = 4a

1.      Current across parallel resistors = 0.9a

V(across 5Ω resistor) = IR

Current is constant in series circuit = 0.9a

V = IR

V = 0.9 x 5

V = 4.5V

V(across parallel resistors) = 12 – 4.5

## V = 7.5V

### I = V/R

I1 = 7.5/30

I1 = 0.25a

Need to find the total resistance of the parallel resistors

V = IR

R = V/I

R = 7.5/0.9

### R = 8.3Ω

Equation for 2 parallel resistors must be rearranged to find r.

1/R(total) = 1/r + 1/r

1/R = 1/30 + 1/r

1/r = 1/R – 1/30

1/r = 8 1/3 – 1/30

1/r = 3/25 – 1/30

1/r = 13/150

r = 11.54Ω

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