(b)i. State two advantages of a lead-acid accumulator over a Leclanche cell.
ii. A parallel combination of 3\(\Omega\) and 4\(\Omega\) resistors is connected in series with a resistor of 4\(\Omega\) and a battery of negligible internal resistance. Calculate the effective resistance in the circuit.
The e.m.f of the accumulator = 1.5v
Table of value/observations
| R(\(\Omega\)) | 1\(_{amp}\) | 1\(^{-1}_{amp}\) |
| 0 | 0.78 | 1.28 |
| 1 | 0.50 | 2.00 |
| 2 | 0.38 | 2.63 |
| 3 | 0.30 | 3.33 |
| 4 | 0.25 | 4.00 |
| 5 | 0.22 | 4.55 |
Slope (s) = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{4-1}{4-2} = \frac{3}{2}\)
The intercept on the vertical axis is =-1.85\(\Omega\)
Precautions:
(b)i.
- The lead-acid accumulator can be recharged while the Leclanche cell cannot.
- The lead-acid accumulator has the capacity to supply large currents for a long time while the leclanche cell does not.
ii. The two resistors 3\(\Omega\) and 4\(\Omega\) are connected in parallel.
\(\therefore\) The equivalent resistance (\(\Omega\) = \(\frac{1}{R} = \frac{1}{3} = \frac{1}{4} = \frac{4+3}{12}\)
= \(\frac{7}{12}\) = 1.71
The effective resistance in the circuit = 1.71 + 4 = 5.71\(\Omega\)
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