th
Discover Physics for Normal (A) 3N/4N (4 Edition): Full Solutions to Textbook Questions Chapter 13
Chapter 13 Current Electricity
Test Yourself 13.1 (page 218)
1. The two types of electric charges are positive charges and negative charges; coulomb (C).
2. Q
I=
t
where I = electric current;
Q = amount of electric charge flowing through the given cross-section;
t = time taken.
3. Ampere
4. Q = It
= 1.5 A × 20 s
= 30 C
Test Yourself 13.2 (page 222)
1. Electromotive force is the work done by a source in driving a unit charge around a complete
circuit, while potential difference is the work done in driving a unit charge through a component.
2. (a) Q
Using I = ,
t
Q=I×t
= 0.40 A × (2 × 60 s)
= 48 C
(b) W
Using V = ,
Q
W=V×Q
= 3.0 V × 48 C
= 144 J
Test Yourself 13.3 (page 226)
(a) V = IR
= 0.90 A × 5.0 Ω
= 4.5 V
(b) W =V×Q
=V×I×t
= 4.5 V × 0.90 A × (1.0 × 60) s
= 243 J
Test Yourself 13.4 (page 228)
1. RA
ρ=
l
where ρ = resistivity;
R = resistance;
A = cross-sectional area;
l = length
© 2013 Marshall Cavendish International (Singapore) Private Limited
13.1
th
Discover Physics for Normal (A) 3N/4N (4 Edition): Full Solutions to Textbook Questions Chapter 13
2. • The resistance of a wire is directly proportional to its length i.e. the longer the wire, the greater
the resistance.
• The resistance of a wire is inversely proportional to its cross-sectional area i.e. the larger the
cross-sectional area, the lower the resistance.
3. Resistance R of one kilometre of copper cable
l
=ρ
A
–8 1000 m
= (1.7 × 10 Ω m)( –4 2 )
1.0 × 10 m
= 0.17 Ω
Get It Right (page 229)
(a) False
There are two types of electric charges — negative charges and positive charges.
The SI unit of electric charge is the coulomb (C).
(b) True
(c) True
(d) False
The e.m.f. and p.d. are both measured in volts. However, while e.m.f. is defined as the work done
to drive a unit charge around a complete circuit, p.d. is defined as the work done to drive a unit
charge through a component.
(e) True
(f) False
A thicker wire has a lower resistance than a thinner wire of equal length and identical material.
Let’s Review (page 230)
Section A: Multiple-Choice Questions
1. A
2. D
Current is the rate of flow of electric charge.
3. B
Q
I=
t
60 C
=
(4 × 60) s
= 0.25 A
4. B
In order to measure the p.d. across the resistor R and the current flowing through it, the voltmeter
must be connected in parallel to R and the ammeter must be connected in series with R. Option
(B) cannot be used to measure the resistance of R because the voltmeter is connected in series
with R.
5. D
V = IR
= 2 A × 10 Ω
= 20 V
© 2013 Marshall Cavendish International (Singapore) Private Limited
13.2
th
Discover Physics for Normal (A) 3N/4N (4 Edition): Full Solutions to Textbook Questions Chapter 13
*6. B
RA l
Using ρ = , resistance R ∝ .
l A
Therefore, for half the resistance,
R 1 l
∝ ×
2 2 A
R 1 1.0
∝ ×
2 2 0.5
∝1
l
The ratio of must be 1 for half the resistance. Only option (B) gives this ratio.
A
Section B: Structured Questions
1. (a) P: Ampere (A) S: Ohm (Ω)
Q: Volt (V) T: Joule (J)
R: Volt (V) U: Coulomb (C)
(b) (i) electrical energy E
Current I =
resistance R × charge Q
(ii) Potential difference V = resistance R × current I
(iii) potential difference V
Current I =
resistance R
2. Potential
Component Current/A Resistance/Ω
difference/V
A 3 0.5 6
B 1.5 0.25 6
C 6 3 2
20 −19
3. (a) Q = 2.0 × 10 electrons × 1.6 × 10 C per electron
= 32 C
(b) Q
I=
t
32 C
=
60 s
= 0.53 A
(c) V
R=
I
12 V
=
32
( )A
60
= 22.5 Ω
© 2013 Marshall Cavendish International (Singapore) Private Limited
13.3
