Page 54 - Physics_Form_2

Physics for Secondary Schools

is increased in a parallel connection. 9V
One might wrongly perceive that bulbs  15  = 0.6 A
connected in a parallel connection are
brighter only because of the combined
brightness of all the bulbs. (c) p.d across R 1
FOR ONLINE READING ONLY
V = IR 1
1 R
Example 2.3  0.6 A 4 2.4 V=

Figure 2.16 shows an electric circuit
p.d across R
5
where, V = 9 V, R  1 4  , R  , 2
2
B
6
and R  . V = IR 2
2 R
3
 0.6 A 4 3.0 V=
R
1 A
p.d across R 3
V = IR 3
3 R
V R
B 2  0.6 A 5 3.60 V=

(d) To find the electric potential at point
A, take note that, the potential is
C R B
3 measured with reference to the
Figure 2.16 negative terminal of the source.
This becomes our reference, 0 V.
(a) What is the total resistance of the Therefore, we start at the negative
circuit? terminal of the battery, where the
(b) What is the value of the current flows potential is 0 V. Travelling around
in the circuit?
(c) What is the potential drop across the circuit in the direction of electron
each resistor? flow, we pass through R where there
3
(d) What is the electric potential at point is a voltage drop of 3.6 V, then pass
A? R where the p.d is 3.0 V up to point
2
A. Therefore, the potential at point
Solution A is,

4
(a) R = R + R + R    
6
5
T 1 2 3 V = A V + 1 R V 2 R
 15  3.6 V 3.0 V= +
V
(b) I = 6.6 V=
R
T

48
Student’s Book Form One

Physics Form 2 Final.indd 48 25/10/2025 10:25