Why an electromotive force (e.m.f.) is not induced at the output when a constant direct voltage is at the input.

Question 3

An ideal transformer is shown in Fig. 9.1.

Fig. 9.1

(a) Explain

(i) why the core is made of iron, [1]

(ii) why an electromotive force (e.m.f.) is not induced at the output when a constant direct voltage is at the input. [2]

(b) An alternating voltage of peak value 150 V is applied across the 1200 turns of the primary coil. The variation with time t of the e.m.f. E induced across the secondary coil is shown in Fig. 9.2.

Fig. 9.2

Use data from Fig. 9.2 to

(i) calculate the number of turns of the secondary coil, [2]

(ii) state one time when the magnetic flux linking the secondary coil is a maximum. [1]

(c) A resistor is connected between the output terminals of the secondary coil. The mean power dissipated in the resistor is 1.2 W. It may be assumed that the varying voltage across the resistor is equal to the varying e.m.f. E shown in Fig. 9.2.

(i) Calculate the resistance of the resistor. [2]

(ii) On Fig. 9.3, sketch the variation with time t of the power P dissipated in the resistor for t = 0 to t = 22.5 ms.

Fig. 9.3

[3]

[Total: 11]

Reference: Past Exam Paper – March 2017 Paper 42 Q9

Solution:

(a)

(i) Iron increases the (magnetic) flux linkage (with the secondary coil) / to reduce flux loss

(ii) An e.m.f. is induced only when the flux (in the core/coil) is changing.

A constant / direct voltage gives a constant flux / field {which is not changing}

(b)

(i)

NS / NP = VS / VP

NS = (52 / 150) × 1200 = 416 turns

(ii)

0 ms or 7.5 ms or 15.0 ms or 22.5 ms

{From Faraday’s law, the gradient of a flux linkage graph gives the induced e.m.f.}

(c)

(i)

EITHER

{Mean power = max power / 2

Max power = V_max² / R

Mean power = V_max² / 2R

From the graph, the maximum voltage V_max = 52 V}

mean power = V² / 2R            and V = 52 (V)

R = 52² / (2 × 1.2) = 1100 (1127) Ω

OR

{Mean power = V_rms² / R

V_rms = V_max / √2           }

mean power = Vrms² / R              and Vrms = 52 / √2 (= 36.8 V)

R = 36.8² / 1.2 = 1100 Ω

(ii)

sinusoidal shape with troughs at zero power

only 3 ‘cycles’

each ‘cycle’ is 2.4 W high and zero power at correct times

{Max (peak) power = 2 × mean power = 2 × 1.2 = 2.4 W

Mean power = max power / 2            or         Max power = V_max² / R

 

The graph of e.m.f. is a sine curve.

The graph of power will be a sine square curve. (Power = V² / R)

 

Power depends on the square on the e.m.f. à Power will always be positive, even when e.m.f. is negative (the square of a negative value is positive).

 

e.m.f. (and thus, power) is zero at 0 ms, 7.5 ms, 15.0 ms, 22.5 ms.

e.m.f. (and thus, power) has max value at halfway between each of the above times.}