You are provided with two wires marked P and C. a resistor \(R_s\) = \(1\Omega\) and other necessary apparatus. Connect \(R_s\) in the left-hand gap of the ...
Question 1 Report
You are provided with two wires marked P and C. a resistor \(R_s\) = \(1\Omega\) and other necessary apparatus.
- Connect \(R_s\) in the left-hand gap of the metre bridge, a length L = 100cm of wire P in the right-hand gap and the other apparatus as shown in the diagram: above
- Determine the balance point B on the bridge wire AC
- Measure and record AB =/s, and BC = /
- Evaluate \(R_1\) = \(\left(\frac{|_{p}}{|_{s}}\right)Rs\)
- Repeat the procedure for four other values of L = 90, 80, 70 and 60cm. In each case obtain and record the value of \(|_{s}\) and \(|_{p}\) and evaluate \(R_1\) = \(\left(\frac{|_{p}}{|_{s}}\right)Rs\)
- Repeat the experiment with the second wire, Q. Obtain the value of \(|_{s}\) and \(|_{Q}\) for equal lengths of wire as used in wire P.
- Evaluate \(R_1\) = \(\left(\frac{|_{p}}{|_{s}}\right)Rs\). In each case, tabulate your readings.
- Plot a graph of \(R_2\) on the Vertical axis against \(R_1\) on the horizontal axis.
- Determine the slope S, of the graph.
- Evaluate the k = \(\sqrt s\).
- State two precautions taken to ensure accurate results.
(b)i. Define the resistivity of the material of a wire.
ii. A galvanometer with a full-scale-deflection of 1.5 x \(10^{3}\). A has a resistance of \(50\Omega\). Determine the resistance required to convert it into a voltmeter reading up to 1.5V.
Answer Details
Metre-bridge determination of wire resistances
The standard resistor \(R_s = 1\,\Omega\) is connected in the left-hand gap and a length \(L\) of the test wire in the right-hand gap. The jockey is moved along the bridge wire AC until the galvanometer shows no deflection at the balance point B. Writing \(l_s = AB\) and \(l_p = BC\), the unknown resistance is
\[ R = \left(\frac{l_p}{l_s}\right)R_s .\]The reading is taken for \(L = 100, 90, 80, 70\) and \(60\ \text{cm}\), first with wire P (giving \(R_1\)) and then with wire Q at the same lengths (giving \(R_2\)).
Table of readings
| \(L\)/cm | \(l_s\)/cm (P) | \(l_p\)/cm | \(R_1\)/\(\Omega\) | \(l_s\)/cm (Q) | \(l_q\)/cm | \(R_2\)/\(\Omega\) |
|---|---|---|---|---|---|---|
| 100 | 5.3 | 94.7 | 17.90 | 21.6 | 78.4 | 3.60 |
| 90 | 6.5 | 93.5 | 14.40 | 22.1 | 77.9 | 3.50 |
| 80 | 7.0 | 93.0 | 13.30 | 23.4 | 76.6 | 3.30 |
| 70 | 8.5 | 91.5 | 10.80 | 24.3 | 75.7 | 3.10 |
| 60 | 9.5 | 90.5 | 9.50 | 25.3 | 74.7 | 2.95 |
Sample evaluation (L = 100 cm, wire P): \(R_1 = \dfrac{l_p}{l_s}R_s = \dfrac{94.7}{5.3}\times 1 = 17.90\,\Omega\); and for wire Q: \(R_2 = \dfrac{74.7\ \text{to}\ 78.4}{21.6}\times 1\), e.g. \(\dfrac{78.4}{21.6}\times 1 = 3.60\,\Omega\).
Graph of \(R_2\) against \(R_1\)
Plotting \(R_2\) (vertical axis) against \(R_1\) (horizontal axis) gives a straight line, showing that \(R_2\) increases uniformly with \(R_1\).
Slope of the graph
Taking two well-separated points on the line of best fit, \((R_1, R_2) = (9.5,\ 3.00)\) and \((17.9,\ 3.67)\):
\[ S = \frac{\Delta R_2}{\Delta R_1} = \frac{3.67 - 3.00}{17.9 - 9.5} = \frac{0.67}{8.4} = 0.080 .\]Hence
\[ k = \sqrt{S} = \sqrt{0.080} = 0.28 .\]Two precautions
- The key was opened when readings were not being taken, so that the wires did not heat up and change their resistance; all connections were kept clean and tight.
- The jockey was tapped gently (not slid or allowed to scratch the bridge wire) and the balance length was read with the eye directly above the wire to avoid parallax error.
(b)(i) Resistivity
The resistivity of the material of a wire is the resistance of a specimen of the material having unit length and unit cross-sectional area. From \(R = \dfrac{\rho L}{A}\),
\[ \rho = \frac{RA}{L},\]where \(A\) is the cross-sectional area, \(R\) the resistance and \(L\) the length of the wire. Its SI unit is the ohm-metre \((\Omega\,\text{m})\).
(b)(ii) Converting the galvanometer to a voltmeter
Full-scale-deflection current \(I_g = 1.5\times10^{-3}\,\text{A}\), coil resistance \(G = 50\,\Omega\), required full-scale reading \(V = 1.5\,\text{V}\). A multiplier resistance \(R\) is connected in series with the galvanometer:
\[ V = I_g(G + R) \;\Rightarrow\; G + R = \frac{V}{I_g} = \frac{1.5}{1.5\times10^{-3}} = 1000\,\Omega .\]\[ R = 1000 - 50 = 950\,\Omega .\]A resistance of \(\mathbf{950\,\Omega}\) must be connected in series with the galvanometer.
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