TEST OF PRACTICAL KNOWLEDGE QUESTION You are provided with a potentiometer, an ammeter, a voltmeter, a standard resistor, and other necessary apparatus. Usi...

TEST OF PRACTICAL KNOWLEDGE QUESTION

You are provided with a potentiometer, an ammeter, a voltmeter, a standard resistor, and other necessary apparatus. Using the circuit diagram above as a guide carry out the following instructions.

1. Set up a circuit as illustrated in the diagram above.
2. Close the key, K.
3. Read and record the ammeter reading l\(_{o}\) and the voltmeter reading V\(_{o}\) when jockey J is not making contact with the potentiometer wire OQ.
4. Using J, make a contact with the potentiometer wire OQ at a point P such that OP = 1Ocm.
5. Read and record the current and the corresponding value of the voltage V.
6. Repeat the procedure for other values of OP= 20cm, 30cm, 40cm, 50cm, and 60cm.
7. Tabulate your readings.
8. Plot a graph with V on the vertical axis and I on the horizontal axis, starting both axes from the origin (0, 0).
9. Determine the slope, s, of the graph.
10. Determine the value of V when I = 0.
11. State two precautions taken to obtain accurate results.

(b) i. State two advantages of a lead-acid accumulator over a dry Leclanche cell.

ii. A cell of emf 2V and internal resistance of 1\(\Omega\) passes current through an external load of 9\(\Omega\). Calculate the potential drop across the cell.

(a) Tabulation, graph and deductions

The key was closed only while a reading was being taken. The readings obtained were as follows.

Position of jockey	OP (cm)	Current, I (A)	Voltmeter reading, V (V)
J not in contact	—	0.11	1.32
P	10.0	0.12	1.20
P	20.0	0.13	1.08
P	30.0	0.14	0.97
P	40.0	0.15	0.85
P	50.0	0.16	0.74
P	60.0	0.17	0.60

The graph of terminal voltage against current is plotted below. A straight line of best fit has been drawn.

Using two widely separated points on the line of best fit, displayed as \[ (I_1,V_1)=(0.13\,\text{A},1.08\,\text{V}),\qquad (I_2,V_2)=(0.17\,\text{A},0.60\,\text{V}). \]
The slope is

\[s=\frac{V_2-V_1}{I_2-I_1}=\frac{0.60-1.08}{0.17-0.13}=\frac{-0.48}{0.04}=-12.0\ \text{V A}^{-1}.\]

Therefore, the slope is \(-12.0\ \text{V A}^{-1}\). The negative sign shows that the terminal voltage decreases as the current increases.

From the intercept of the best-fit line on the voltage axis, when \(I=0\),

\[V\approx 2.63\ \text{V}.\]

Precautions

1. The key was opened when readings were not being taken, to prevent heating and polarization of the cell.
2. All connections were made tight, and the ammeter and voltmeter were checked for zero error before use.
3. The jockey was touched lightly on the potentiometer wire and was not dragged along it.

(b)

(i) Advantages of a lead-acid accumulator over a dry Leclanché cell

1. It is rechargeable and can therefore be used repeatedly.
2. It can supply a relatively large current for a long period.

(ii)

\[E=2.0\ \text{V},\qquad r=1.0\ \Omega,\qquad R=9.0\ \Omega.\]

\[I=\frac{E}{R+r}=\frac{2.0}{9.0+1.0}=0.20\ \text{A}.\]

The potential drop across the cell, that is across its internal resistance, is

\[V_{\text{internal}}=Ir=(0.20)(1.0)=\boxed{0.20\ \text{V}}.\]

Read lesson note on Current Electricity (JAMB)

Read lesson note on Capacitors (JAMB)