Connect the circuit as shown above. Set the value of R= 30\(\Omega\). Close the key and obtain a balance at point T on the potentiometer wire PQ. Read and record the length TQ = L Evaluate L\(^{-1}\) and R\(^{-1}\), Repeat the experiment for R= 20, 10.5, 3 and 1\(\Omega\) respectively. In each case, determine and record the corresponding values of LL\(^{-1}\) and R\(^{-1}\). Remove the resistance box from the circuit and then determine the length \(L_o\) corresponding to R= 0. Tabulate your readings. Plot a graph at \(R_1\) on the vertical axis, and L\(^{-1}\) on the horizontal axis, starting both axes from the origin (0,0). Determine the slope, s of the graph and its intercept, I on the vertical axis.
Evaluate: (i) k = 1\(^{-1}\)
(ii) \(\frac{Lo}{S}\)
State two precautions taken to ensure accurate result
(b)i. Using your graph, determine the value of L for which R =15\(\Omega\).
ii. if the intercept I = 0.5+ y\(^{-1}\), use your graph to determine the value of y.
iii. Explain what is meant by the e.m.f.of a cell.
Verified Correct Answer: .
Physical Principles & Numerical Solution
Table of values/observation
S/N
R\(\Omega\)
L(cm)
R\(^{-1}\)(\(\Omega^{-1}\))
L\(^{-1}\)(cm\(^{-1}\))
1
30.00
88.80
0.033
0.01126
2
20.00
84.60
0.050
0.01182
3
10.00
79.80
0.100
0.01253
4
5.00
71.50
0.200
0.01399
5
3.00
69.50
0.333
0.01439
6
1.00
42.40
1.000
0.02358
Slope (s) = \(\frac{\bigtriangleup {R^{-1}(\Omega)}}{\bigtriangleup {L^{-1}(cm^{-1})}} = \frac{1.08-(-0.96)}{0.002175-0.00175}\)
= \(\frac{1.08-(-0.96)}{0.002175-0.00175} = \frac{2.04}{0.00425}\) = 4800cm\(\Omega ^{-1}\)
I = -1.14(\(\Omega ^{-1}\)
(i) K I\(^{-1}\) \(\frac{1}{I} = \frac{-1}{1.14}\) = 0.877\(\Omega\)
(ii) C = \(\frac{L_{o}}{S} = \frac{78.50cm}{4800cm \Omega ^{-1}}\) = 0.01635\(\Omega\)
Precautions:
- The key was opened when readings were not taken.
- I ensured tight connections.
- I avoided parallax error when reading potentiometre.
(b)i. When R = 15\(\Omega\), R\(^{-1}\) = \(\frac{1}{R} = \frac{1}{15}\) = 0.0666\(\Omega ^{-1}\)
From the graph, L\(^{-1}\) = 0.001175, \(\frac{1}{L}\) 0.01175, L = \(\frac{1}{0.01175}\)
= 85.11cm
I = 0.5+y\(^{-1}\), -1.14 = 0.5 + y\(^{-1}\), y\(^{-1}\) = 1.14 - 0.5
= -0.64, \(\frac{1}{y}\) = -0.64
\(\therefore\) y = \(\frac{-1}{0.64}\) = -1.6
(iii) e.m.f. is the p.d between the terminals of a cell when an open circuit or the work done in moving unit charge round a complete circuit.
Source: Myschool Past Questions Archive (Record #60145)