You are provided with a potentiometer XY, a voltmeter, V, a standard resistor R, an accumulator, E a plug key, K, a jockey, and connecting wires.
(b)i. State four factors on which the resistance of a wire depends.
ii. A resistance Wire of length 100cm is connected in a circuit. If the resistance per unit length of the wire is 0.02 \(\Omega\)cm\(^{-1}\), how much heat would be produced in the wire if a voltmeter connected across its ends indicates 1.5V while the current runs for 1 minute?
Tables of value/observation
S/N | L 9cm) | V (volts) | L\(^{-1}\)(cm\(^{-1}\)) | V\(^{-1}\)(volts\(^{-1}\)) | L\(^{-1}\)(cm\(^{-1}\))x10\(^{-3}\) |
1 | 15.0 | 0.60 | 0.067 | 1.667 | 67.0 |
2 | 25.0 | 0.80 | 0.040 | 1.250 | 40.0 |
3 | 35.0 | 1.10 | 0.029 | 0.909 | 29.0 |
4 | 45.0 | 1.30 | 0.022 | 0.909 | 22.0 |
5 | 55.5 | 1.60 | 0.018 | 0.625 | 18.0 |
6 | 65.0 | 1.80 | 0.015 | 0.556 | 15.0 |
(6) The graph of V\(^{-1}\)(volts\(^{-1}\)) vs \(\bigtriangleup\) L\(^{-1}\) (cm\(^{-1}\)
Scale: V\(^{-1}\)(volts\(^{-1}\)) axis: 2cm = 0.15units
\(\bigtriangleup\) L\(^{-1}\) (cm\(^{-1}\)) axis: 2cm = 7units
(7) Slope/gradient = \(\frac{\bigtriangleup {V}^{-1}(volts ^{-1})}{\bigtriangleup {L}^{-1}(cm^{-1})}\)
= \(\frac{1.575-0.405}{63-4.2} = \frac{1.17}{58.8}\)
= 0.0199cm/volts
(8) K = \(\frac{1}{s} = \frac{1}{0.0199}\) = 50.25 volts/cm
precautions are:
(b)i. The factors affecting the resistance of a wire are:
- length of wire (R&L)
- cross-sectional area of the (R \(\propto\))
- Nature of material
- Temperature of the wire
ii. R = 100cm x 0.02\(\Omega\)cm = 2\(\Omega\), v =1.5v
t = 1min = 60s
H =p x t = \(\frac{V^{2}t}{R} = \frac{(1.5)^{2}}{2}\) x 60 = 67.5J
OR
I = \(\frac{V}{R} = \frac{1.5}{2}\) = 0.75A
H = P x t = I\(^{2}\)Rt = (0.7)\(^{2}\) x 60 x 2
= \(\frac{9}{16}\)x120 = 67.5J
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