(a)(i) Describe, with the aid of a circuit diagram, an experiment to measure the resistance of a wire given an ammeter of low resistance, a battery, a key, a rheostat, va high-resistance voltmeter and some connecting wires.
(ii) State two precautions necessary to obtain an accurate result.
(b) Using the experimental result and any necessary measurements, explain how the resistivity of the wire may be determined.
(c) Two cells, each of e.m.f. 2V and internal resistance 0.552, are connected in series. They are made to supply current to a combination of three resistors, one of resistance 20 connected in series to a parallel combination of two other resistors each of resistance 3Q. Draw the circuit diagram and calculate the:
(i) current in the circuit
(ii) potential difference across the parallel combination of the resistors
(iii) lost volts of the battery.
Explanation
(a)
Description: The circuit is set up as shown above. The key (k) is closed. The rheostat (R) is adjusted so that lowest current is flowing. Ammeter and voltmeter readings (I and V) are recorded. Adjust rheostat to increase current. Ammeter and voltmeter readings are recorded for other positions of the rheostat. Values of (are plotted against values of I. Slope, which is the value of the resistance is determine.
Precautions;
i) Kcy should be removed when readings are not taken
ii) Connections should be tight
(iii) Parallax' error when reading ammeter and voltmeter should be avoided
(iv) Only low current should be used
v) Zero error of the meters should be avoided. Note: Any correct two) Resistance, R = P'/A where = length of wire, A = cross sectional area, p = resistivity )f wire. But slope, S of graph = R. Hence S = P'/A
Measure I with a metre rule. Measure diameter with a micrometer screw gauge and determine A. Substitute p = \(\frac{SA}{I}\)
c) Diagram
i) Total e.m.f. = 4V
Total resistance = 1 + 2 + \(\frac{3 \times 3}{ 2 \times 3}\)
= 3 + 1\(\frac{1}{2}\)
= \(\frac{1}{2} \Omega\)
E = IR\(_T\),
= E/R
:.Current, I = 4/9 x 2/1
= 8/9;
A = 0.9A
(ii) P.d = IR = 8/9 x 3/2 = 4/3 = 1.3V
(iii) Lost volts = Ir = \(\frac{8}{9}\) x 1 = \(\frac{8}{9} = 0.9V\)