You are provided with a potentiometer, an ammeter, a voltmeter, a standard resistor, and other necessary apparatus. Using the circuit diagram above as a guide carry out the following instructions.
Set up a circuit as illustrated in the diagram above.
Close the key, K.
Read and record the ammeter reading \(l_o\) and the voltmeter reading \(V_o\) when jockey J is not making contact with the potentiometer wire OQ.
Using J, make a contact with the potentiometer wire OQ at a point P such that OP = 1Ocm.
Read and record the current and the corresponding value of the voltage V.
Repeat the procedure for other values of OP= 20cm, 30cm, 40cm, 50cm, and 60cm.
Tabulate your readings.
Plot a graph with V on the vertical axis and I on the horizontal axis, starting both axes from the origin (0, 0).
Determine the slope, s, of the graph.
Determine the value of V when I = 0.
State two precautions taken to obtain accurate results.
(b) i. State two advantages of a lead-acid accumulator over a dry Leclanche cell.
ii. A cell of emf 2V and internal resistance of 1\(\Omega\) passes current through an external load of 9\(\Omega\). Calculate the potential drop across the cell.
Verified Correct Answer: .
Physical Principles & Numerical Solution
S/N
OP(\(_{cm}\))
I(\(_{A}\))
V(\(_{v}\))
1
20.0
0.13
1.08
2
30.0
0.14
0.97
3
40.0
0.15
0.85
4
50.0
0.16
0.74
5
60.0
0.17
0.60
Slope, s = \(\frac{y_{2} - y_{1}}{x_{2} - x _{1}} = \frac{0.85 - 0.30}{0.20 - 0.5}\)
\(\frac{0.55}{0.05}\)
= 11
When I = o
V = 3.15v
Precautions; i
- ensured that the key was open whenever the circuit was not in use so as to avoid the cells being run down.
- ensured there was no zero error in the ammeter and the voltmeter by making necessary adjustments before taking my readings.
- ensured that the jockey did not scrap the potentiometer wire.
- ensured that the electrical wires were tightly connected to the terminals in order to avoid current fluctuations.
(b)i.
Lead-acid accumulator
Leclanche cell
(i). It has the capacity to supply large current for a long time
(ii). It can be charged
It has no such capacity
It cannot be recharged
ii. e.m.f = 2V
r = 1\(\Omega\) (internal resistance)
R = 9\(\Omega\) (external resistance)
Calculate the potential drop across the cell.
By formula;
E = I(R +r)
2 = I(9 + 1)
2 = 101
I = \(\frac{2}{10}\) = 0.2A
Therefore, the p.d drop across the cell is Ir.
= 0.2 x 1 = 0.2V
Source: Myschool Past Questions Archive (Record #60004)