Page 71 - Physics_Form_2

Current electricity

Why does the length increase with Therefore,
an increase of e.m.f.? E
l  2 2  l 1

Theoretically, the e.m.f E of cell E 1
1
P balances the p.d. V on the wire  1.07 V  85 cm
1
FOR ONLINE READING ONLY
because no current flows in G at 1.63 V
the balance point N. As a result, 55.79 cm=
E = V 1 . Similarly, if cell E is the
1
2
e.m.f of cell Q and V is the p.d. The new balance length is 55.79 cm.
2
between A and M, E = V 2 . As a
2
E V Exercise 2.2
result, 1 = 1 .
E V
2 2 1. A wire of length 1.2 m and diameter
The p.d, V across any length, ,l is 0.64 mm has a resistance of 2.4 Ω.
directly proportional to l if the wire Calculate the resistance of a wire of
carries a constant current. As a result, length 0.80 m and diameter 0.32 mm
E 1 = l 1 . Thus, the e.m.f of the two of the same material.
E 2 l 2 2. A wire of length 2 m and a cross
2
cells can be compared. sectional area of 0.5 mm , has a
resistance of 2.2 Ω. Calculate the
resistivity of the material making
Example 2.12 up the wire.
3. A battery of e.m.f 12 V and internal
The balance length of a potentiometer
wire for a cell of e.m.f, E = 1.63 V resistance 1.5 Ω is connected to a
4 Ω resistor. Calculate the:
1
is 85 cm. If the cell is replaced by
another one of e.m.f, E = 1.07 V, (a) total resistance of the circuit
2
calculate the new balance length. (b) current through the battery
(c) p.d across the cell terminals
Solution 4. The p.d across the terminals of a
Given cell is 1.1 V when a current of 0.20 A
E = 1.63 V, l = 1 85 cm, E = 2 1.07 V is being drawn from the cell. If the
1
p.d across the cell is 1.3 V when a
Then, the new length l is obtained from current of 0.10 A is being drawn,
2
the formula: determine the:

E 1 = l 1 (a) internal resistance of the cell.
E 2 l 2 (b) cell’s e.m.f.

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Physics Form 2 Final.indd 65 25/10/2025 10:26