Exam Style Answers 11 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 11 5

Exam-style questions 1

B 

[1]

2

C 

[1]

3

a

 

terminal p.d. = 2.5 × 0.30 = 0.75 V  

E

=

V

+

1.5 = 0.75 + 2.5

 

so, r   = 0.30 Ω   c

i 

×

[1]

r so 2.5r = 0.75 

 

1.8 87 75 ≈ 1.8 88 8W Power wer for 0.5 Ω: total resistance resistance  ii  Po ii R + r = 0.80 Ω 1.5 = 1.875 A     current = 0.8

b

power

   

Power for 0.2 Ω: total resistance Power R + r = 0.50 Ω 1.5 = 3.0 A   current =

 

power

i 

The test cell is the wrong way round

[1]

=

0.5 3.02 × 0.20 = 1.80 W  

 

2.25 34.6

 

r  = 0.40   Ω 

[1]

 ii  ii

Substitution into E  = I (R + r) so E  = 1.50 V 

[1]

Internal resistance is too high 

[1]

Maximum current < 4 A 

[1]

There is no/negligible current through the high resistance voltmeter and, hence, the cell. When the resistor is connected connected in parallel there is a much larger current through that and the cell.  [1]  There is now a potential drop as electrical electr ical work is done against the internal

[1] [1]

 ii  ii

[1] [1] [1]

lost volts = 0.54 V  [1] lost vo volts 0.54 = = 0.77   Ω    r= [1] I  0.7  iii  The resistance of the voltmeter > iii >> > r  or R  [1] a

[1]

  b  

1

0.341( 4 + r )

b

c

7

1.43 4 34 = 0.93 9 33 V  

[1]

The e.m. e.m.f. f. of a cell is the work done per coulomb of charge [1] in the complete circuit.  [1]

[1]

×

I ( R + r )  = 0.625 ( 2 + r ) 

resistance of the cell.  V  8.40 i  I  =   = R 12   0.70 A  

[1]

34.6

so, e .m.f . =

=

a 

[1]

At the balance point, the ammeter reading is zero.  [1] 2.25

 

[1]

 ii  ii

=

 c

[1]

so he must reverse it. 

e.m.f .

E

=

6

 

1.434  

1.8752 × 0.5 = 1.76 W  

 

=

i 

[1]

=

a

b

[1]

power P = I 2 R = 2.52 × 0.30

The resistance due to the work done (or energy transferred) in driving current through the cell  [1] which equals the ‘lost volts’ divided by the current.  [1]

Ir [1]

 

4

 

There is work done inside the cell against the internal resistance. or There is a voltage (lost volts) across the internal resistance.  [1] b

a

In circuit 1, the p. p.d. d. across the bulb varies from 0 to 240 V.   In circuit 2, it never falls to zero.  V 2 2402 =   i  R = 60 P =

 960 Ω 

[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 

Resistance is greater when switched on. or Resistance is lower at room temperature.   [1]

 ii ii

a

  b

c

2

a

[1]

0.6 × (8 + r )  = 1.50 × ( 2 + r ) 

[1]

r = 2.0 Ω 

 

Substitution into either equation gives E  = 6.0 V  [1]

[1]

10 a

A diagram similar to Figure 11.13 

because current through R2 decreases.  [1]

b

Voltmeter reading will decrease  

V out

=

R2 ( R1 + R2 )

V in so, 2.0 =

×

470 470 + R1

×

9  [1] [1]

resistance resis tance of R2 and voltmeter in parallel

   1       1         = 903 Ω   +    1645   2000 )  

 

9

I ( R + r ) 

 

Resistance of a metal increases with increasing temperature (or decreases with decreasing temperature).  [1]

−1

V out

=

=

R1 = 1645 ≈ 1600 Ω 

 

E

 

 

8

b

=

R2 ( R1 + R2 )

i 

V in

×

=

903 (1645 + 903)

=

[1]

5.9V   [1]

Straight line through origin with positive gradient 

i   ii ii

Graph axes labelled V  (x  (x-axis) and  ( y-axis) l  (  y-axis)  [1]

ii 

A: 0 V; B: 2.2 V 

iii 

General diagram diagram (with one or two two cells)  [1]

 

Two cells in correct polarity  

 

Switches, or suitable comment indicating that only one cell is used at a time  [1]

R1 R2

 

=

42.6

=

15.4 =

( 42.6

−

[2] [1]

15.4 )

0.36 

 

0.57  

 iv iv

uncertainty in R1 = ±0.2 cm; percentage uncertainty = 1.3%

 

uncertainty in R2 = ± 0.2 cm; percentage uncertainty = 0.5% 

 

total percentage uncertainty = 10.7 + 0.5 = 11.2 ≈ 11%

 

total uncertainty = ±0.06 

[1]

 

± 0.2 cm  R1 15.4 ( R1 + R2 )

 iii iii

[1]

[1] [1] [1]

[1]

[1]

[1]

[1]

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