You are provided with a potentiometer x y; a jockey, J; a standard resistor, R, and other necessary apparatus.
Connect a circuit as shown in the diagram above.
Close the key. Read and record the current 10 when J is not in contact with XY.
Let J make contact with XY at C, such that XC = l= 25 cm. Close the key. Read and record the current l.
Evaluate I\(^{-1}\).
Repeat the procedure for four other values of I = 40, 55, 70, and 85 cm. Tabulate your readings.
Plot a graph of l on the vertical axis against I\(^{-1}\) on the horizontal axis.
From your graph, deduce the value of I when l\(^{-1}\) = 0. Evaluate \(\frac{|O}{1}\)
State two precautions taken to ensure accurate results.
(b)i) Explain what is meant by the potential difference between two points in an electric circuit.
ii. A piece of resistance wire of diameter 0.2m and resistance 7\(\Omega\) has a resistivity of 8.8 x 10\(^{-7}\) \(\Omega\)m. Calculate the length of the wire. [\(\pi\) = \(\frac{22}{7}\)].
Verified Correct Answer: .
Physical Principles & Numerical Solution
Table of values/observation
\(l_o\) = 0.8A
S/N
L(cm)
l(A)
(L\(^{-1}\)cm\(^{-1}\))
1
25.0
1.20
0.040
2
40.0
1.00
0.025
3
55.0
0.95
0.018
4
70.0
0.85
0.014
5
85.0
0.8
0.012
When 1\(^{-1}\) =0, l = 0.66A
\(\frac{|_{o}}{|} = \frac{0.8}{0.66}\) = 1.212
Precautions:
The Key was open when readings were not being taken.
Tight connections were ensured.
Parallax errors were avoided in reading the meter-rule
Zero error was noted in reading metre-rule
The jockey was not rolled on or allowed to scratch the wire
(b) The p.d between two points is the work done (in joules) in moving an electric charge of one coulomb across the two points.
ii. R = \(\frac{PL}{A} = \frac{4pL}{\pi d^{2}}\)
d = 0.2m, \(\pi\) = \(\frac{22}{7}\), R = 7\(\Omega\), p = 8.8 x 10\(^{-7} \pi \), L = ?
\(\therefore\) 7 = \(\frac{4 \times 8.8 \times 10^{-7} \times L}{\frac{22}{7} \times (0.2)^{2}}\)
7 = \(\frac{4 \times 8.8 \times 10^{-7} \times L \times 7}{22 \times 0.04} = \frac{28 \times 8.8 \times 10^{-7} \times L}{0.88}\)
L = \(\frac{7 \times 0.88}{28 \times 8.8 \times 10^{-7}}\)
L = \(\frac{6.16}{246.4}\) x 10\(^{-7}\)
L = 0.025 x 10\(^{7}\)
L = 2.5 x 10\(^{-2}\) x 10\(^{7}\)
L = 2.5 x 10\(^{5}\)
Source: Myschool Past Questions Archive (Record #60107)