Outline the features of an ideal step-down transformer.

Real transformers are subject to energy loss. State and explain how two causes of these energy losses may be reduced by suitable features in these transformers.

1.

2.

Calculate the power consumed by the internal circuits when the TV is in “standby” mode.

The efficiency of the transformer is 0.95. Determine the current supplied by the 230 V mains supply.

The TV is on “standby” for 75% of the time. Calculate the energy wasted in one year by not switching off the TV.

\[{\text{1 year}} = 3.2 \times {10^7}{\text{ s}}\]

number of secondary coils < primary coils;

coils wound around an iron core;

ideally no flux leakage/zero resistive heating;

solution;

with linked explanation;

another solution;

with linked explanation;

eg:

laminations in core;

prevents eddy currents in core;

thick/low resistance wires for primary and secondary turns;

reduces heating losses in wires;

4.1 W;

power required at primary \( = \frac{{4.1}}{{0.95}} = 4.3{\text{ W}}\); (allow ECF from (h)(i))

current \(\frac{{4.3}}{{230}} = 1.9 \times {10^{ - 2}}{\text{ A}}\);

approximately \(1.0 \times {10^8}{\text{ J}}\);

Allow use of either power value.

Many candidates described the purpose of a step-down transformer. The question asked for features, such as an iron core or more primary than secondary turns.

Further features such as a laminated core or low resistance coils were not often given, with explanations, as ways of increasing the efficiency of a transformer.

A very easy mark, 4.1W.

To determine the primary current the efficiency needed to be applied correctly to the power in the two circuits. There were many correct answers, with a minority placing the efficiency on the wrong side of the equation.

TV power consumption in standby mode was an easy calculation for 1 mark.