You are provided with a battery of e.m.f, E, a standard resistor, R, of resistance 2 \( \Omega \), a key, K, an ammeter, A, a jockey, J, a potentiometer, UV...
You are provided with a battery of e.m.f,E, a standard resistor, R, of resistance 2 \( \Omega \), a key, K, an ammeter, A, a jockey, J, a potentiometer, UV, and some connecting wires.
(i) Measure and record the emf, E, of the battery.
(ii) Set up the circuit as shown in the diagram above with the key open.
(iii) Place the jockey at the point, U, of the potentiometer wire. Close the key and record the reading, i, of the ammeter.
(iv) Place the jockey at a point T on the potentiometer wire UV such that d = UT = 30.0 cm.
(v) Close the circuit, read and record the current, I, on the ammeter,
(vi) Evaluate \(I^1\).
(vi) Repeat the experiment for four other values of d = 40.0 cm, 50.0 cm, 60.0 cm and 70.0 cm. In each case, record I and evaluate \(I^1\).
(vii) Tabulate the results
(ix) Plot a graph with d on the vertical axis and I on the horizontal axis stalling both axes from the origin (0,0).
(x) Determine the slope, s, of the graph.
(xi) From the graph determine the value \(I_1\), of I when d = 0. (ci) Given that=s, calculate 8.
(xii) State two precautions taken to ensure accurate results.
(xii) Given that \( \frac{E}{\delta} = s \), calculate \( \delta \).
(b)(i) Write down the equation that connects the resistance, R, of a wire and the factors on which it depends. State the meaning of each of the symbols.
(ii) An electric fan draws a current of0.75 A in a 240 V circuit. Calculate the cost of using, the fan for 10 hours if the utility rate is $ 0.50 per kWh.
(a) Potentiometer Experiment
The e.m.f. of the battery was measured as:
\(E=3.0\text{ V}\)
With the jockey at \(U\), \(d=0\) and the ammeter reading was:
\(i=1.50\text{ A}\)
The readings obtained are shown below. The reciprocal current was evaluated from \(I^{-1}=1/I\).
Distance, \(d\) (cm)
Current, \(I\) (A)
\(I^{-1}\) (A−1)
30.0
1.154
0.867
40.0
1.111
0.900
50.0
1.071
0.933
60.0
1.034
0.967
70.0
1.000
1.000
A graph of \(d\) against \(I^{-1}\), with both axes beginning at the origin, is plotted below.
Plot of d against I⁻¹. The straight line of best fit has a slope of approximately 3.00 × 10² cm A and cuts d = 0 at I⁻¹ = 0.667 A⁻¹.
Using two widely separated points on the straight line, \((0.867\text{ A}^{-1},30.0\text{ cm})\) and \((1.000\text{ A}^{-1},70.0\text{ cm})\):
\[
s=\frac{70.0-30.0}{1.000-0.867}
=\frac{40.0}{0.133}
\approx 3.00\times10^2\text{ cm A}.
\]
Hence, the slope of the graph is \(3.00\times10^2\text{ cm A}\).
Extrapolating the straight line to \(d=0\),
\(I^{-1}=0.667\text{ A}^{-1}\).
Therefore, \(I=1/0.667=1.50\text{ A}\), which agrees with the current when the jockey is at \(U\).
