You are provided with two wires marked P and C. a resistor R\(_{s}\) = 1\(\Omega\) and other necessary apparatus.
(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.
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:
(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\(\Omega\)
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