Measure and record the e.m.f of the accumulator provided. Connect the circuit as shown in the diagram. With zero resistance in the resistance box, adjust the rheostat to obtain the maximum possible reading on the ammeter. Do not adjust the rheostat again throughout the experiment. Open the key. With R= 1\(\Omega\), close the key, read and record the reading of the ammeter I. Calculate 1\(^{-1}\) Repeat the experiment for R=2,3,4 and 5\(\Omega\) respectively. In each case, read and record the value of \(1_A\) and calculate the Corresponding value of l\(_{A}^{-1}\). Now connect the resistor Q in series into the circuit and without altering the setting of the rheostat, record the new ammeter readings \(l_B\) for R= 1,2,3, 4, and 5\(\Omega\). Calculate the corresponding value of \(l_B\) in each case. Tabulate all your readings. On the same graph and using the same axes and scales, plot the graphs of:
(i) I\(_{A}^{-1}\)' on the vertical axis and R on the horizontal axis;
(ii) l\(_{B}^{-1}\) on the vertical axis and R on the horizontal axis. Calculate the slopes \(S_A\) in (i) and \(S_B\) in (ii) above. Determine the difference D between the intercepts of the graphs on the vertical axis. State two precautions taken to ensure accurate results.
(b) i. Explain why a battery of eight dry Leclanche cells, each of e.m.f. 1.5v is not normally used in place of a motor-car battery of 12V to start a car.
ii. State two sources of e.m.f. other than the chemical cell.
Verified Correct Answer: .
Physical Principles & Numerical Solution
Table of values/observation
S/N
R(\(\Omega\))
\(I_A\)(A)
I\(^{-1}_{A}\)
\(I_B\)(A)
I\(^{-1}_{B}\)(A\(^{-1}\))
1
1.00
1.80
0.556
0.95
1.053
2
2.00
0.95
1.053
0.60
1.667
3
3.00
0.60
1.667
0.50
2.000
4
4.00
0.45
2.222
0.40
2.500
5
5.00
0.40
2.500
0.35
2.857
Slop\(e_A\) (\(s_A\)) = \(\frac{2.5-0.25}{5-0.35} = \frac{2.25}{4.65}\) = 0.4838\(\Omega ^{-1}\) A\(^{-1}\)
Slop\(e_B\) (\(S_B\)) = \(\frac{2.85-0.95}{5-0.55}\) = \(\frac{1.9}{4.45}\) = 0.4270\(\Omega ^{-1} A^{-1}\)
Interpret on I\(_{A}^{-1}\) (A\(^{-1}\)) axis = 0.075A\(^{-1}\)
Interpret on I\(_{B}^{-1}\) (A\(^{-1}\)) axis = 0.725A\(^{-1}\)
= D = \(S_B\) - \(S_A\)
= 0.725 - 0.075 = 0.65A\(^{-1}\)
Precautions:
- Always open the key when readings are not taken.
- Ensure tight connections.
(b)i. The current produced by the cells is less than that of the car battery because of the total internal resistance of the cells. A car requires a large amount of Current to start it. This amount of current cannot be supplied by the cells
ii. Photocell and Dynamo
Source: Myschool Past Questions Archive (Record #60140)