Connect the circuit as shown in the diagram above. PQ is a potentiometer wire 100 cm long and R is a standard resistor of 5\(\Omega\). With the jockey J not...
Question 1 Report
- Connect the circuit as shown in the diagram above. PQ is a potentiometer wire 100 cm long and R is a standard resistor of 5\(\Omega\).
- With the jockey J not making contact with PQ, close the switch. Read and record the ammeter reading I. Open the switch.
- Use the jockey to make contact with PQ at the 20cm mark such that PJ = I = 20 cm. Close the switch, read and record the value I\(_{i}\) of the ammeter. Evaluate I\(^{-1}\).
- Repeat the procedure for other values of I = 35, 50, 65, and 80 cm. In each case, determine the corresponding values of I\(_{i}\), and I\(^{-1}\). Tabulate your readings.
- Plot a graph of I\(^{i}\) on the vertical axis and I\(_{i}\), on the horizontal axis, starting both axes from the origin (0, 0).
- From your graph deduce the value, of I\(_{o}\) of I\(_{i}\), when I\(^{-1}\)= 0.
- Evaluate I\(_{o}\)e
- State two precautions taken to ensure accurate results.
(b)) Define the e. m.f. of a battery
ii. A cell X e.m.f. 1.00 V is balanced by a length of 40.0 cm on a potentiometer wire. Another cell Y is balanced by a length of 60.0 cm on the same wire. Calculate the e.m.f. of Y.
Answer Details
(a) Potentiometer experiment
With the jockey off the potentiometer wire, the ammeter reading is:
I = 0.20 A
The readings obtained when the jockey makes contact at the stated lengths are tabulated below.
| S/N | Length, L (cm) | Ammeter reading, Ii (A) | L-1 (cm-1) |
|---|---|---|---|
| 1 | 20 | 1.00 | 0.050 |
| 2 | 35 | 0.90 | 0.029 |
| 3 | 50 | 0.85 | 0.020 |
| 4 | 65 | 0.80 | 0.015 |
| 5 | 80 | 0.75 | 0.013 |
The graph of Ii against L-1 is shown below.
From the intercept on the vertical axis, when L-1 = 0,
Io = 0.75 A.
Hence,
\[ \frac{I_o}{I}=\frac{0.75}{0.20}=3.75. \]
Precautions
- The ammeter was read with the eye directly in front of the pointer to avoid parallax error.
- The switch was opened after each reading to prevent heating of the potentiometer wire and cell.
(b)(i) E.m.f. of a battery
The e.m.f. of a battery is the work done, or energy supplied, by the battery in driving one coulomb of charge round the complete circuit, including its internal resistance.
(b)(ii) E.m.f. of cell Y
For the same potentiometer wire, e.m.f. is proportional to balancing length:
\[ \frac{E_Y}{E_X}=\frac{l_Y}{l_X} \] \[ E_Y=\frac{60.0}{40.0}\times1.00=1.50\ \text{V}. \]
E.m.f. of cell Y = 1.50 V.
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