TEST OF PRACTICAL KNOWLEDGE QUESTION You are provided with two wires marked P and C. a resistor \(R_s\) = \(1\Omega\) and other necessary apparatus. Connect...

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

1. 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.
2. 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.

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(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.