You have been provided with an accumulator E, a standard resistor Rx, two resistance boxes RB\(_{1}\) and RB\(_{2}\), two keys K\(_{1}\) and K\(_{2}\) and o...

Answer Details

Measurement of e.m.f. of the accumulator

E.m.f. of the accumulator, \(E = 2.0\ \text{V}\).

Observation and table of readings

For each setting the resistance in \(RB_1\) was made equal to the resistance in \(RB_2\) (\(R\) in \(RB_1 = R\) in \(RB_2 = R\)). With \(K_1\) and \(K_2\) closed the potential difference \(V_1\) across the standard resistor \(R_x\) was read, and \(V_1^{-1}\) evaluated.

\(R\ (\Omega)\)	\(V_1\ (\text{V})\)	\(V_1^{-1}\ (\text{V}^{-1})\)
1	1.50	0.667
2	1.30	0.769
3	1.15	0.870
4	1.10	0.909
5	0.90	1.111

Graph of \(V_1^{-1}\) against \(R\)

The graph of \(V_1^{-1}\) (vertical axis) against \(R\) (horizontal axis), with both axes starting from the origin, is a rising straight line.

Slope and intercept

Taking two widely spaced points on the line of best fit, \((0.3,\ 0.60)\) and \((6.5,\ 1.22)\):

\[ s = \frac{\Delta (V_1^{-1})}{\Delta R} = \frac{1.22 - 0.60}{6.5 - 0.3} = \frac{0.62}{6.2} = 0.10\ \text{V}^{-1}\,\Omega^{-1} \]

The intercept on the vertical axis (value of \(V_1^{-1}\) where the line cuts \(R = 0\)) is

\[ c = 0.57\ \text{V}^{-1}. \]

Evaluation of \(y = \dfrac{1}{s}\)

\[ y = \frac{1}{s} = \frac{1}{0.10} = 10\ \text{V}\,\Omega. \]

Two precautions

1. I opened the keys when readings were not being taken, to avoid heating of the resistors and drift in the cell.
2. I made clean, tight connections at all terminals and read the voltmeter with my eye directly in front of the pointer to avoid parallax error (also correcting for any zero error).

(b)(i) Potential difference between two points

The potential difference between two points in an electric circuit is the work done (in joules) in moving one coulomb of positive charge from one point to the other. It is measured in volts.

(b)(ii) Internal resistance of the cell

Given \(E = 3\ \text{V}\), \(R = 4\ \Omega\) and current \(I = 0.5\ \text{A}\):

\[ E = IR + Ir \] \[ 3 = (0.5 \times 4) + 0.5r \] \[ 3 = 2 + 0.5r \] \[ 0.5r = 1 \] \[ r = \frac{1}{0.5} = 2\ \Omega. \]