IBDP Physics 11.2 – Power generation and transmission: IB Style Question Bank HL Paper 2


(a) The primary coil of a transformer is connected to a 110 V alternating current (ac) supply. The secondary coil of the transformer is connected to a 15 V garden lighting system that consists of 8 lamps connected in parallel. Each lamp is rated at 35 W when working at its normal brightness. Root mean square (rms) values are used throughout this question.

    1. The primary coil has 3300 turns. Calculate the number of turns on the secondary coil. [1]

    2. Determine the total resistance of the lamps when they are working normally. [2]

    3. Calculate the current in the primary of the transformer assuming that it is ideal. [2]

    4. Flux leakage is one reason why a transformer may not be ideal. Explain the effect of flux leakage on the transformer. [2]

(b) A pendulum with a metal bob comes to rest after 200 swings. The same pendulum, released from the same position, now swings at 90° to the direction of a strong magnetic field and comes to rest after 20 swings.

Explain why the pendulum comes to rest after a smaller number of swings. [4]



a i « \frac{15}{110}\times 3300  = » 450 «turns» 

a ii

ALTERNATIVE 1  calculates total current = \frac{35}{15}\times 8 « = 18.7 A»  resistance = « \frac{15}{18.7} = » 0.80 «Ω» 

ALTERNATIVE 2 calculates total power = 35×8 « = 280 W»  resistance =« \frac{15^2}{280} =» 0.80 «Ω» 

ALTERNATIVE 3 calculates individual resistance = \frac{15^2}{35}« =  6.43 Ω»  resistance = «\frac{6.43}{8} » = 0.80 «Ω» 

a iii

total power required = 280 «W»

OR uses factor \frac{3300}{450}

OR total current = 18.7 « A» 

current = 2.5 OR 2.6 «A» 

a iv

the secondary coil does not enclose all flux «lines from core»  induced emf in secondary

OR power transferred to the secondary

OR efficiency is less than expected 

Award [0] for references to eddy currents/heating of the core as the reason. Award MP2 if no reason stated.


bob cuts mag field lines

OR there is a change in flux linkage 

induced emf across bob  leading to eddy/induced current in bob 

eddy/induced current produces a magnetic field that opposes «direction of» motion 

force due to the induced magnetic field decelerates bob  damping of pendulum increases/there is additional «magnetic»

damping  MP4 and MP5 can be expressed in terms of energy transfer from kinetic energy of bob to electrical/thermal energy in bob


The diagram shows a sketch of an ideal step-down transformer.

The number of turns in the primary coil is 1800 and that in the secondary coil is 90.


State Faraday’s law of induction.[2]


Explain, using Faraday’s law of induction, how the transformer steps down the voltage.[4]


The input voltage is 240 V. Calculate the output voltage.[2]


Outline how energy losses are reduced in the core of a practical transformer.[2]


Step-up transformers are used in power stations to increase the voltage at which the electricity is transmitted. Explain why this is done.[2]




the size of the induced emf
is proportional/equal to the rate of change of flux linkage

The word ‘induced’ is required here.
Allow correctly defined symbols from a correct equation. ‘Induced’ is required for MP1.


varying voltage/current in primary coil produces a varying magnetic field

this produces a change in flux linkage / change in magnetic field in the secondary coil

a «varying» emf is induced/produced/generated in the secondary coil

voltage is stepped down as there are more turns on the primary than the secondary

Comparison of number of turns is required for MP4.


output voltage \( = \frac{{90 \times 240}}{{1800}}\)

= 12 «V»


laminated core reduces eddy currents

less thermal energy is transferred to the surroundings


for a certain power to be transmitted, large V means low I

less thermal energy loss as P = I2R / joule heating