Measure and record the e.m.f. of the accumulator provided. Connect a circuit as shown in the diagram. S is a standard resistor and R is a resistance box. Wi...
Question 1 Report
- Measure and record the e.m.f. of the accumulator provided.
- Connect a circuit as shown in the diagram. S is a standard resistor and R is a resistance box.
- With R = 0\(\Omega\), close the key K. Read and record the ammeter reading I. Evaluate 1\(^{-1}\).
- Repeat the procedure for R=1,2, 3, 4, and 5\(\Omega\). Tabulate your readings.
- Plot a graph of R on the vertical axis and 1\(^{-1}\) on the horizontal axis, starting both axis from the origin (0,0).
- Determine the slope s of the graph and find the intercept C on the vertical axis.
- State two precautions taken to ensure accurate results.
(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.
Answer Details
(a) E.m.f. of the accumulator and the R against \(I^{-1}\) experiment
E.m.f. of the accumulator, \(E = 1.5\ \text{V}\).
Table of readings
| \(R\ (\Omega)\) | \(I\ (\text{A})\) | \(I^{-1}\ (\text{A}^{-1})\) |
|---|---|---|
| 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 |
Graph of \(R\) against \(I^{-1}\)
Slope of the graph
Taking two points on the line of best fit, \((I^{-1}_1, R_1) = (2.00\ \text{A}^{-1},\ 1.1\ \Omega)\) and \((I^{-1}_2, R_2) = (4.00\ \text{A}^{-1},\ 4.1\ \Omega)\):
\[ s = \frac{R_2 - R_1}{I^{-1}_2 - I^{-1}_1} = \frac{4.1 - 1.1}{4.00 - 2.00} = \frac{3.0}{2.00} = 1.5\ \text{V}. \]Intercept on the vertical axis
Producing the line of best fit back to \(I^{-1} = 0\), it cuts the vertical (\(R\)) axis at
\[ C = -1.9\ \Omega. \]Interpretation. For this circuit the current is \(I = \dfrac{E}{R + S + r}\), which rearranges to \(R = E\,(I^{-1}) - (S + r)\). Comparing with \(R = s\,(I^{-1}) + C\): the slope \(s = E = 1.5\ \text{V}\) (equal to the measured e.m.f.), and the intercept \(C = -(S + r) = -1.9\ \Omega\), giving \(S + r = 1.9\ \Omega\).
Two precautions
- I made all connections clean and tight, and opened the key immediately after taking each reading so as not to run down the accumulator.
- I read the ammeter with my eye directly in front of the pointer to avoid errors due to parallax, and I noted and corrected the zero error of the ammeter.
(b)(i) Two advantages of a lead-acid accumulator over a Leclanche cell
- The lead-acid accumulator can be recharged and used again, whereas the Leclanche cell cannot be recharged.
- The lead-acid accumulator has a very low internal resistance, so it can supply a large current for a long time, which the Leclanche cell cannot.
(b)(ii) Effective resistance of the network
The \(3\ \Omega\) and \(4\ \Omega\) resistors are in parallel:
\[ \frac{1}{R_p} = \frac{1}{3} + \frac{1}{4} = \frac{4 + 3}{12} = \frac{7}{12}, \qquad R_p = \frac{12}{7} = 1.71\ \Omega. \]This parallel section is in series with the \(4\ \Omega\) resistor (the battery has negligible internal resistance):
\[ R_{\text{eff}} = R_p + 4 = 1.71 + 4 = 5.71\ \Omega. \]The effective resistance in the circuit \(= 5.71\ \Omega\).
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