Determine Resistivity by Plotting a Graph

Aim

To determine resistivity of two wires by plotting a graph for potential difference versus current.

Apparatus and Material Required

• DC Power supply (battery eliminator)
• Ammeter (0 – 500 mA) (to measure current)
• Voltmeter (0-5 V) (to measure potential difference)
• Rheostat (for adjusting current)
• Connecting wires
• Plug key (to switch on or off the circuit)
• a piece of sandpaper
• Two sample wires (with known lengths and cross-sectional areas) but unknown resistance (0Ω to 20Ω)
• A scale to measure length of the wires
• A screw gauge to measure diameter of wires

Principle

According to Ohm’s Law, the current passing through a conductor is directly proportional to the voltage applied across it, as long as the conductor’s temperature and other physical properties remain constant.

If I be the current (expressed in Ampere) flowing through the conductor and V be the potential difference (expressed in Volt) across its ends, then according to Ohm’s law \[V \propto I\] and \[ V = RI\] where R is the constant of proportionality and is called as the electrical resistance of the conductor. R is expressed in ohms.

The resistance R depends upon the material, dimensions and temperature of the conductor. At any given temperature, the resistance of a wire of same length and same cross-section varies with the material of the wire. At a given temperature, \[R = \rho \frac{l}{A}\]
where \(\rho\) = resistivity of the material,
\(l\) =length of the wire and
\(A\) = cross-sectional area of the wire.

The graph of potential difference (\(V\)) versus electric current (\(I\)) is a straight line. The slope of the graph is the resistance, \(R\). The resistivity \(\rho\) is thus given by \[\rho = R \frac {A}{l}\]

Procedure

1. Set up:
   • Clean the ends of the connecting wires with the help of sandpaper in order to remove any insulating coating on them.
   • Connect one of the resistance wires with other components (rheostat, battery, key, voltmeter and ammeter). The connection is shown in the diagram
2. Adjust Current and Record Readings:
   • Slide the rheostat contact to one of its extreme ends, so that current passing through the resistance wire is minimum.
   • Insert the key and note down the readings in voltmeter and ammeter.
   • Remove the key and slide the rheostat contact to increase the current passing through the wire. Insert the key again and note down the readings in voltmeter and ammeter.
   • Repeat the step 5 for four different settings of the rheostat. Record the observations in tabular form.
   • Repeat the steps 2 to 6 for another wire resistance.
3. Plot the Graph:
   • For each wire, plot a graph of potential difference (V) on the y-axis versus current (I) on the x-axis.

Observations

1. Range of ammeter = 0 mA to 500 mA
2. Least count of ammeter = 10 mA
3. Range of voltmeter = 0 V to 5 V
4. Least count of voltmeter = 0.1 V
5. Least count of metre scale = 1 mm
6. Length of wire 01, \(l_1\) = 50 mm
7. Length of wire 02, \(l_2\) = 60 mm
8. Diameter of the wire 01, \(d_1\) = 0.7 mm
9. Diameter of the wire 02, \(d_2\) = 0.5 mm.

Wire 01

S.N.	Applied potential difference [voltmeter reading V (V)]	Current flowing through the wire [milliammeter reading I (mA)]
1.	0.4	90
2.	0.6	130
3.	0.8	180
4.	1.0	220

Wire 02

S.N.	Applied potential difference [voltmeter reading V (V)]	Current flowing through the wire [milliammeter reading I (mA)]
1.	1.4	90
2.	2.2	140
3.	3.0	200
4.	4.3	280

Calculations

1. Calculate the slope (Resistance):
   • Determine the slope of the graph. The resistance of the given wire is equal to this slope.
   • Slope \(R\) = \(\frac {\Delta V}{\Delta I}\)
2. Calculate Resistivity\(\rho\):
   • Make sure to convert all measurements to SI units.
   • Wire 01: \(l=50mm\), \(d = 0.7mm\), \(A = \frac {\pi d^2}{4} = 3.85 \times 10^{-7}\) \(\frac{A}{l} = 7.7 \times 10^{-6} \)
   • Wire 02: \(l=60mm\), \(d = 0.5mm\), \(A = \frac {\pi d^2}{4} = 1.96 \times 10^{-7}\) \(\frac{A}{l} = 2.8 \times 10^{-6} \)
   • Use the formula \(\rho = R \frac {A}{l}\) to calculate resistivity
   • Wire 01: \(\rho_1 = R_1 \times \frac{A_1}{l_1}\) = \(4.6 \times 7.7 \times 10^{-6} \Omega – m = 3.54 \times 10^{-5} \Omega – m\)
   • Wire 02: \(\rho_2 = R_2 \times \frac{A_2}{l_2}\) = \(15.3 \times 2.8 \times 10^{-6} \Omega – m= 4.28 \times 10^{-5} \Omega – m\)

Results

1. Resistivity of the first wire is \(3.54 \times 10^{-5} \Omega – m\)
2. Resistivity of the second wire is \(4.28 \times 10^{-5} \Omega – m\).
3. The potential difference across the given wire varies linearly with the current.

Precautions

1. Ensure all connections are tight to avoid errors in readings.
2. Keep the current low to prevent heating in the wire, which could affect resistivity.
3. Take multiple readings for accuracy.
4. Only insert the key when taking observations to avoid the heating of the wire, that can impact its resistivity.
5. Zero error in measuring instruments (voltmeter, ammeter, metre scale) must be taken due care of.

Sources of error

1. The wire used may not be of uniform area of cross-section.
2. Wire length is the length between one terminal of voltmeter to another. The length of wire wound around terminal should not be included in wire length, otherwise it will give error.

Conclusion

The experiment confirms Ohm’s law as evident from the linear V-I graph for each wire. The slope of the graph and known dimensions of the wires help us calculate the resistivity of each wire material.

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Reference: NCERT Lab Manual