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Question 1 Report
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 other necessary apparatus.
(b)i. Explain what is meant by the potential difference between two points in an electric circuit
ii. A cell has an e.m.f. of 3 V. When it is connected across a resistor of resistance \(4\Omega\), a current 0.5A passes through the circuit. Calculate the internal resistance of the cell.
E.m.f. of the accumulator, \(E = 2.0\ \text{V}\).
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 |
The graph of \(V_1^{-1}\) (vertical axis) against \(R\) (horizontal axis), with both axes starting from the origin, is a rising straight line.
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
\[ c = 0.57\ \text{V}^{-1}. \]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.
Given \(E = 3\ \text{V}\), \(R = 4\ \Omega\) and current \(I = 0.5\ \text{A}\):
\[ E = IR + Ir \] \[ 3 = (0.5 \times 4) + 0.5r \] \[ 3 = 2 + 0.5r \] \[ 0.5r = 1 \] \[ r = \frac{1}{0.5} = 2\ \Omega. \]