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.
Explanation
(a) OBSERVATIONS
(i) Value of E correctly read and recorded to at least 1 d.p in volts.
(ii) Value of i correctly read and recorded to at least 1 d.pin amperes.
(iii) Five values of d correctly determined and recorded to at least 1 d.p in cm.
(iv) Five values of I correctly read and recorded to at least 1 d.p in amperes and in trend.
Trend; As d increases, I decreases
(v) Five values of I\(^{1}\) correctly evaluated to at least 3 s.f.
(vi) Composite table showing at least d, I, and I\(^{-1}\).
(ii) P = IV
Power P = \(\frac{240 \times 0.75}{100}\)
0.18kw
Energy consumed in 10hr = 0.18 x 10
= 1.8kWh
Cost of energy = 1.8 x 0.5 = $0.9
OR
Cost of energy = \(\frac{\text {P x cost x time}}{100}\)
\(\frac{0.7 \times 240 \times 10}{1000}\)
= $0.9