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.
(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.
| 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
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