2014 WAEC Physics Practical You are provided with a potentiometer AB, a 102\(\Omega\) standard resistor R, a battery of...

WAEC 2014

You are provided with a potentiometer AB, a 102\(\Omega\) standard resistor R, a battery of emf 4.5V, a jockey J, and other necessary materials.

Connect a circuit as shown in the diagram above.
Close key K. Without J making contact with AB, read and record the ammeter reading I. Open the key.
Use the jockey to make contact with AB at the 20cm mark such that AJ = x = 20cm. Close the key, read and record the ammeter reading.
Evaluate x\(^{-1}\).
Repeat the procedure for values of x= 35cm, 45cm, 60cm, and 80cm respectively.
Tabulate your readings.
Plot a graph with x\(^{-1}\) on the vertical axis and l\(_{i}\) on the horizontal axis, starting both axes from the origin (0, 0).
Determine the slope, s, of the graph.
From your graph, determine the value l\(_{o}\) of I\(_{1}\) for which x\(^{-1}\)= 0.
Evaluate \(\frac{I_{o}}{I}\).
State two precautions taken to obtain accurate results.

(b)i. Define the emf of a battery.

ii. A cell X of emf 1.018V is balanced by a length of 50.0cm on a potentiometer wire. Another cell Y is balanced by a length of 75.0cm on the same wire. Calculate the emf of Y.

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Explanation

Electricity
I = 0.36A

\(\frac{x}{cm}\)	\(\frac{I}{A}\)	\(\frac{X^{-1}}{cm^{-1}}\)
20.0	1.14	0.050
35.0	0.90	0.029
45.0	0.84	0.022
60.0	0.78	0.017
80.0	0.66	0.013

Slope (s) = \(\frac{\bigtriangleup {X}^{-1}}{\bigtriangleup {I}}\) = \(\frac{0.08-0.01}{1.68-0.644} = \frac{0.07}{1.036}\)

From the graph,
When X\(^{-1}\) = 0
I\(_{0}\) of I\(_{1}\) = 0.448A

x. Evaluate \(\frac{I_{0}}{I}\)

= \(\frac{0.448}{0.36}\) = 1.244

Precaution;

The key was opened when readings were not taken.
I ensured tight connections of the connecting wire
I avoided parallax errors when reading the potentiometer.

(bi) Define emf of a battery: The e.m.f of a cell is the potential difference across its terminals when it is in an open circuit. i.e. not Supplying current to an external circuit.

ii. A cell x emf is 1.018V.
Length\(_{1}\) on the potentiometer x = 50.0cm
Length\(_{2}\) on the potentiometer, y,= 75.0cm
A cell Y emf is?

Solution
By formula;

\(\frac{E_{1}}{E_{2}}\) = \(\frac{L_{2}}{L_{1}}\)

\(\frac{1.018}{E_{2}} = \frac{75.0}{50.0}\)

75.E\(_{2}\) = 1.018 x 50

E\(_{2}\) = \(\frac{1.018\times 50}{75}\)

E\(_{2}\) = 0.612V.

see graph above

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