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.
- Connect a circuit as shown in the diagram above.
- Close the key and use the jockey to make contact with the potentiometer with XY at a point N such that l = XN= 15cm.
- Read and record the value of the potential difference V on the voltmeter.
- Evaluate l\(^{-1}\) and V\(^{-1}\).
- Repeat the procedure for five other values of I= 25, 35, 45, 55, and 65cm respectively. Read and record the value of V and evaluate V\(^{-1}\) and l\(^{-1}\) in each case. Tabulate your readings.
- Plot a graph V\(^{-1}\) on the vertical axis against l\(^{-1}\)on the horizontal axis starting both axes from the origin (0,0).
- Determine the slope, s, of the graph.
- Evaluate k = \(\frac{1}{s}\)
- State two precautions taken to ensure accurate re- results.
(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?
Explanation
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:
- I would ensure that the key is opened when the circuit is not in use to avoid the cell being used up.
- I would ensure that the electrical wires are tightly connected
- I would ensure that I take note of the zero error in the pointer instruments to avoid systematic error
- I would ensure that I take pointer readings within the vertical eye level to avoid parallax
- I used short connecting wires to ensure that the required energy to dive the electrical charges was not used up in overcoming the resistance of the wire.
(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