A-LevelPhysicsAlternating currents9702/42 · Oct/Nov 2021 · Q1010 Marks

Fig. 10.1 shows a simple laminated iron-cored transformer consisting of a primary coil of 25000 turns and a secondary coil of 625 turns. The output potential difference (p.d.) VOUT is applied to a load resistor of resistance 640Ω. [Figure 10.1] [Figure 10.2]

(a)(i)State the function of the iron core.[1]
(a)(ii)Explain why the iron core is laminated.[2]
(b)(i)The input p.d. VIN is a sinusoidal alternating voltage of peak value 12kV and period 40ms. Calculate the maximum value of VOUT.[1]
(b)(ii)Calculate the root-mean-square (r.m.s.) current in the load resistor.[1]
(b)(iii)On Fig. 10.2, sketch the variation with time t of the power P dissipated in the load resistor for time t = 0 to t = 40 ms. Assume that P = 0 when t = 0.[3]
(c)Explain, with reference to Fig.10.2, why the mean power in the load resistor is 70W.[2]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.

to increase the magnetic flux linkage (between the coils)
to reduce energy losses
by reducing induced currents
maximum VOUT = 12 000 × (625/25000)
= 300 V
r.m.s. current = 300 / (640 × √2)
= 0.33 A
sketch: sinusoidal shape in positive half of the graph, sitting with ‘minima' resting on the time-axis (at P = 0)
each 'cycle' shown repeating every 20 ms
maximum P shown as 140 W
power curve is symmetrical about the midpoint (on the power axis)
mean power is half the peak power

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