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...

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 \(= 5.71\ \Omega\).