Exam Style Answers 9 Asal Physics CB

 

 

CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK 

Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded may be different.

 Coursebook answers Chapter 9 8

Exam-style questions 1

B 

[1]

2

D 

[1]

3

a

W = 3.6 A to the right  

[1]

b

X

4.3 2.4 1.9 downwards  

[1]

c

Y

2.1 A to the left  

[1]

d

Z

0 

[1]

=

=

  =

−

=

4.8 2.7

=

−

4.3 4.3 −

=

a

 

5

6.5 2.0 4.5 mA to the right   4.5 4.2 0.3 downwards X 2.2 1.4 0.8 V  

[1] [1]

b

X = 6.3 + 2 .4 − 5 .0 = 3.7 V  

[1]

c

X

[1]

X Y a

d

  6

a

=

=

−

=

−

=

=

=

−

b

6.0 1.4 2.4 −

−

=

2.2 V  

X = 4.3 + 4.7 = 9.0 V 

[1]

Y = 9.0 V 

[1]

current in resistor I   = =

1.8

8.2 mA  

b c

resistance R = V /I  =

  =

−

6.0

0.0082 0082 harge Q = It = 0.0082 × 60    = 0.492 C   n=

Q

=

=

 

19

−

b

p.d. across Y = 0.5 × 6.0 = 3.0 V  3 so, re resi sist stan anceo ceoff Y = = 2 Ω  1.5 p.d. across X 12 3 9.0 V  9.0 so, re resi sist stan anceo ceoff X = = 4.5 Ω   2.0

c

=

−

=

=

[1]

 

2.0 0.5 1.5 A 

9

0.75 A 

 ii  ii

V

 

=

 iii iii

Use the circuit loop including both batteries and the 3 Ω resistor:

 

9.0  = E 2

 

E 2 = 6.0 V 

 iv  iv

I  =

=

IR

[1] =

0.75 75 × 12  

[1]

 9.0 V 

V 

=

[1]

+

(1× 3 ) 

6

R 12 = 0.50 A  

[1] [1] [1]

 

[1]

a

The ammeter goes in the main circuit. It must have have a low resistance so little energy is transferred in it / there is a small p.d. across it.  [1]

 b

i

  [1]

[1]

current in Y

−

4.2 V

i 

 

[1]

[1] [1]

a

=

=

730 Ω 

0.492

e 1. 6 ×10 18 = 3.1 × 10  

1

=

220 p.d. across the lamp V  6.0 1.8

d

7

[1]

=

[1]

The e.m. e.m.f. f. of a battery is the energy transferred per coulomb of charge in the complete circuit.  [1]

  4

The potential difference across the terminals of a battery is the energy transferred per coulomb of charge in the external circuit. 

resista resi stanceof nceof thevoltmete thevoltmeterr and 400 Ω resistor

=

1 1  − (1200 + 400 ) 1

 300 Ω 

[1]

=

urrent in the circuit =

9.0 300

=

0.03 A   [1]

[1]

potentialdropacross

[1]

100 Ω resistor = 0.03 × 10 100 = 3.0 V   therefore, e.m.f. = 9.0 + 3.0 = 12.0 V [1]

[1] [1] [1] [1]

Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside © Cambridge University Press 2020

   

CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK 

 

ii

=

 

 

b

new resistanceof resistanceof thecomb the combinat ination ion (

1 12000

new current =

9.5

387 p.d. ac acro rossth ssthee new

=

+

1 400

)−1

[1] [1]

2

 iii  iii

The voltmeter is in parallel parallel with the main circuit  [1]

 

so it reduces the resistance of any combination it is in, as shown in answers b i and b ii. 

[1]

−1

  1   1  =  +    + 96 = 120 Ω  60 40   

 

resista resi stanceof nceof networ network k  = (

combination = 387 × 0.0246 [1] = 9.53 ≈ 9.5 V  

10 a

= ( 40 + 20 )  = 60 Ω   resista resi stanceof nceof thelowerarm

= 387 Ω

0.0246 0246 A  

resista resi stanceof nceof theupper theupper arm

  =

 

c

1

+

1  

)

[1]

1

−

60 120  40 Ω 

[1]

total potential difference across whole lower arm = 6.0 V  [1] p.d. acrossthe acrossthe parall parallelsecti elsection on

[1]

Resistance is the potential difference across a component divided by the current in it.  [1]

   

=

24 120

×

6.0 = 1.2 V  

[1]

current through 60 Ω resistor = 1.2/60 =  0.02 A  [1]

Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside © Cambridge University Press 2020