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.
- Measure and record the e.m.f of the accumulator.
- Connect a circuit as shown above.
- Set the resistance R, in the resistance box such that R in RB\(_{1}\) = R in RB\(_{2}\) = 1\(\Omega\).
- With K\(_{1}\) open and K\(_{2}\) closed, measure and record the potential difference V across the standard resistor Rx. (v) Close K, and K,. Read and record the potential difference V\(_{o}\) across Rx.
- Evaluate V\(_{1}^{-1}\)
- Repeat procedure (v) for four other values of R = 2, 3, 4 and 5\(\Omega\) respectively. In each case, ensure that the value of R in RB\(_{1}\) is equal to the value of R in RB\(_{2}\).
- Evaluate V\(_{1}^{-1}\) in each case. Tabulate your readings.
- Plot a graph of V\(_{1}^{-1}\) on the vertical axis against R on the horizontal axis starting both axes from the origin (0,0).
- Determine the slope, s, of the graph and the intercept on the vertical axis.
- Evaluate y = \(\frac{1}{s}\)
- State two precautions taken to ensure accurate results
(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.
Explanation
Table of values (Observation)
| R (\(\Omega\) | V\(_{1}\)(v) | V\(_{1}^{-1}\) |
| 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 |
Slope (s) = \(\frac{\bigtriangleup {V}_{1}^{-1}}{DR} = \frac{1.22 - 0.6}{6.5 - 0.3}\)
= \(\frac{0.62}{6.20}\) = 0.1
Intercept (c) = 0.57
Evaluation: y = \(\frac{1}{S} = \frac{1}{0.1}\) = 10
Precautions:
- I removed the key when not taking readings
- I ensured tight connections/clean terminals
- I avoided parallax error in reading the voltmeter
- I corrected for zero error on the voltmeter
(b)i. The potential difference (pd) (in volts) between two points in a circuit is the work done (in joules) when one unit of positive charge (coulomb) moves from one point to the other
ii. E = IR + Ir, 3 =0.5 x 4 + 0.5r, 3 = 2 +0.5r
r = \(\frac{3 - 2}{0.5}\) = 2\(\Omega\)