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\(_{1}\) = (\(\frac{|_{p}}{|_{s}}\))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\(_{1}\) = (\(\frac{|_{p}}{|_{s}}\))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\(_{1}\) = (\(\frac{|_{p}}{|_{s}}\))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 x10\(^{3}\). A has a resistance of 50\(\Omega\). Determine the resistance required to convert it into a voltmeter reading up to 1.5V.
Explanation
Table of values
| L(cm)\(^{3}\) | Ls(cm) | Lp(cm) | R\(_{1}\) |
| 100 | 5.3 | 94.7 | 17.90 |
| 90 | 6.5 | 93.5 | 14.40 |
| 80 | 7.0 | 93.0 | 13.30 |
| 70 | 8.5 | 91.5 | 10.80 |
| 60 | 9.5 | 90.5 | 9.50 |
| L(cm) | Is\(^{1}\) | Lg | R\(_{2}\) |
| 100 | 21.60 | 78.4 | 3.60 |
| 90 | 22.1 | 77.9 | 3.50 |
| 80 | 23.4 | 76.6 | 3.30 |
| 70 | 24.3 | 75.7 | 3.10 |
| 60 | 25.3 | 74.7 | 2.95 |
| L(cm) | Ls(cm) | Lp(cm) | R\(_{1}\) | Ls\(^{1}\)cm\(^{3}\) | Lq | R\(^{2}\) |
| 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 |
Slope (s) = of the graph = \(\frac{DR_{2}}{DR_{1}} = \frac{1.4}{8.0}\) = 0.175
K = \(\sqrt 5\) = 0.17555 = 0.42
Precautions:
- Key was opened where readings were not being taken.
- Tight connections were ensured
- ParallaX errors were avoided in reading the metre rule.
- Zero error was noted in reading the metre rule.
- The jockey was not rolled on or allowed to scratch the wire.
(b) The resistivity of a material is defined as the resistance of a unit length of the material/wire of a uni cross-sectional area.
OR
P = \(\frac{AR}{L}\) where
A = cross-sectional area
R = eletrical resistance
L = length of the wire
V = Vg + V\(_{R}\) = IRg + IR = I (Rg + R) \(\therefore\) \(\frac{V}{I}\) = Rg+ R
= \(\frac{1.5}{1.5 \times 10^{-3}}\) = 50 + R = 10\(^{3}\) = 50 + R \(\therefore\) R = 950\(\Omega\)