You are provided with two resistance wires labeled and B, a1\(\Omega\) standard resistor Rx, and other necessary apparatus. Connect \(R_x\) in the left hand...
Question 1 Report
You are provided with two resistance wires labeled and B, a1\(\Omega\) standard resistor Rx, and other necessary apparatus.
- Connect \(R_x\) in the left hand gap of the metre bridge, a length \(L = 100\text{cm}\) of wire A in the right hand gap, and the other apparatus as shown in the diagram above.
- Determine and record the balance point P on the bridge wire NQ.
- Measure and record \(l_x = \text{NP}\) and \(l_y = \text{PQ}\).
- Evaluate \(R_1 = \frac{I_y}{L_x} = R_x\)
- Repeat the procedure for four other values of \(L = 95, 85, 75\) and \(65\text{cm}\). In each case, determine and record the balance point P and the length \(l_x\) and \(l_y\). Also, evaluate \(R_1\).
- Repeat the experiment with the second wire B. Obtain the balance points P and the values of and \(l_x\) and \(l_y\).
- Evaluate \(R_1 = \frac{I_y}{L_x} = R_x\) in each case. Tabulate your readings.
- Plot a graph of \(R_2\) on the vertical axis against \(R_1\) on the horizontal axis.
- Determine the slope, s, of the graph.
- Evaluate \(k = \sqrt{s}\).
- State two precautions taken to obtain accurate results.
(b)i. State two advantages of using a potentiometer over a voltmeter for measuring the potential difference.
ii. Define the internal resistance of a cell.
Answer Details
Metre-bridge determination of R1 and R2
Principle. The bridge wire NQ is 100 cm long and uniform. At the balance (null) point P no current flows through the galvanometer, so, with the standard resistor \(R_x = 1\,\Omega\) in the left gap and the wire in the right gap,
\[ \frac{R_x}{R} = \frac{l_x}{l_y} \quad\Rightarrow\quad R = R_x\,\frac{l_y}{l_x}, \qquad l_x = NP,\; l_y = PQ. \]For wire A this gives \(R_1 = \dfrac{l_y}{l_x}\,R_x\); repeating with wire B at the same lengths gives the corresponding values \(R_2 = \dfrac{l_y'}{l_x'}\,R_x\).
Table of readings (\(R_x = 1\,\Omega\)):
| L /cm | lx = NP /cm | ly = PQ /cm | R1 = (ly/lx)Rx /Ω | l′x = NP /cm | l′y = PQ /cm | R2 = (l′y/l′x)Rx /Ω |
|---|---|---|---|---|---|---|
| 100 | 5.2 | 94.8 | 18.23 | 21.5 | 78.5 | 3.65 |
| 95 | 6.0 | 94.0 | 15.67 | 22.5 | 77.5 | 3.44 |
| 85 | 6.5 | 93.5 | 14.38 | 23.0 | 77.0 | 3.35 |
| 75 | 7.7 | 92.3 | 11.99 | 23.6 | 76.4 | 3.24 |
| 65 | 8.2 | 91.8 | 11.20 | 24.4 | 75.6 | 3.10 |
Sample evaluation (L = 100 cm row).
\[ R_1 = \frac{l_y}{l_x}R_x = \frac{94.8}{5.2}\times 1 = 18.23\,\Omega, \qquad R_2 = \frac{l_y'}{l_x'}R_x = \frac{78.5}{21.5}\times 1 = 3.65\,\Omega. \]The remaining rows are evaluated in the same way, as tabulated above.
Graph of R2 against R1.
Slope. Reading two widely spaced points on the line of best fit:
\[ (R_1, R_2) = (11.0,\;3.12)\ \text{and}\ (18.5,\;3.66). \] \[ s = \frac{\Delta R_2}{\Delta R_1} = \frac{3.66 - 3.12}{18.5 - 11.0} = \frac{0.54}{7.5} = 0.072. \]Evaluation of k.
\[ k = \sqrt{s} = \sqrt{0.072} = 0.27. \]Two precautions.
- The jockey was tapped gently on the bridge wire (never dragged) so that the wire was not scratched or its cross-section altered.
- All connecting terminals were kept clean and tight, and readings of the balance point were taken with the eye placed vertically above the wire to avoid parallax error.
(b)(i) Two advantages of a potentiometer over a voltmeter
- At the balance point the potentiometer draws no current from the cell, so it measures the true e.m.f./potential difference without the loading error that a voltmeter introduces.
- It gives readings of very high accuracy and its sensitivity (range) can be extended as required, since it depends only on lengths and a standard cell rather than on a moving pointer.
(b)(ii) Internal resistance of a cell
The internal resistance of a cell is the opposition offered by the materials inside the cell (its electrolyte and electrodes) to the flow of current through the cell itself.
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