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...

1. 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\(\Omega\).
2. With the jockey J not making contact with PQ, close the switch. Read and record the ammeter reading I. Open the switch.
3. 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\(^{-1}\).
4. 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.
5. 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).
6. From your graph deduce the value, of I\(_{o}\) of I\(_{i}\), when I\(^{-1}\)= 0.
7. Evaluate I\(_{o}\)e
8.  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.

(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 I_i against L^-1 is shown below.

From the intercept on the vertical axis, when L^-1 = 0,

I_o = 0.75 A.

Hence,

\[ \frac{I_o}{I}=\frac{0.75}{0.20}=3.75. \]

Precautions

1. The ammeter was read with the eye directly in front of the pointer to avoid parallax error.
2. 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.

Read lesson note on Current Electricity (JAMB)