National Grid electricity supply: Part 6.6 Examples of transformer calculation practice questions and worked out answers

Doc Brown's Physics exam study revision notes

ANSWERS to the transformer questions

Q1 A transformer has 200 turns on the primary coil and 10 turns on the secondary coil.

If the output p.d. required is 12.0 V from the secondary coil, what p.d. must be put across the primary coil?

V_p / V_s = n_p / n_s

V_pn_s = V_sn_p

V_p = V_sn_p / n_s

V_p = 12 x 200/10 = 240 V

Q2 The p.d. across the primary coil of a transformer was 12000 V and the current flowing through it was 20 A.

If the current flowing through the secondary coil was 1000 A, what is the p.d. across the secondary coil?

V_sI_s = V_pI_p  (assuming 100 % efficiency)

V_s = V_pI_p / I_s  = 12000 x 20/1000 = 240 V

Q3 A power station step-up transformer has 500 turns of wire on the primary coil and 8000 turns of wire on the secondary coil.

(a) If the generator output is 25 000 V, what is the output p.d. in kV for the transmission lines of the National Grid?

V_p / V_s = n_p / n_s

V_pn_s = V_sn_p

p.d. output = V_s = V_pn_s / n_p = (25 000 x 8000)/500 = 400 000 V = 400 kV

(b) What is the power input, in MW, from the generator to the transformer, if the input current to the primary coil is 20 A? (assume 100% efficiency)

P_input  = I_pV_p = 20 x 25 000 = 500 000 W = 500 kW = 0.50 MW

(Note: In reality, a power station will usually have several generators and step-up transformers running at the same time.)

(c)  If the power output from the transformer to the transmission line is 0.48 MW, what is the % efficiency of the energy transfer and what has become of the lost energy?

% efficiency = 100 x useful power output / total power input = 100 x 0.48/0.50 = 96%

There are always thermal energy losses from the transformer to the surrounding thermal energy store.

(d) How much energy is wasted from the transformer every minute?

The power wastage is equal to 0.50 - 0.48 = 0.02 MW

0.02 MW = 20 kW = 20 000 W

This equals a heat loss of 20 000 J/s

Therefore in one minute 20 000 x 60 = 1 200 000 J is lost (1.2 MJ/min)

 

Q4 The p.d. across the primary coil of a transformer is 240 V and carries a current of 5.0 A.

If the p.d. across the secondary coil is 12.0 V, what current is flowing in secondary coil?

V_sI_s = V_pI_p  (assuming 100 % efficiency)

I_s = V_pI_p / V_s  = 240 x 5.0 / 12.0 = 100 A

Q5 A transformer has 20 turns of wire on the secondary coil and the p.d. across it is 3.0 V.

How many turns must there be on the primary coil if the primary coil p.d. is 230 V?

V_p / V_s = n_p / n_s

V_pn_s = V_sn_p

n_p = V_pn_s / V_s = 230 x 20/3.0 = ~1533 turns of wire.

Q6 The output p.d. across the secondary coil of a transformer is 6.0 V and a current flowing through it of 0.50 A.

What p.d. must be applied across the primary coil to give a current of 0.0125 A to flow through it?

V_sI_s = V_pI_p  (assuming 100 % efficiency)

V_p = V_sI_s / I_p = 6.0 x 0.50/0.0125 = 240 V

Q7 The secondary coil of a transformer has a p.d. output of 120 V.

If the primary coil has 300 turns, how many turns must there be on the secondary coil if the primary coil input p.d. is 3.0 V?

V_p / V_s = n_p / n_s

V_pn_s = V_sn_p

n_s = V_sn_p / V_p = 120 x 300/6.0 = 6000 turns

Q8 The p.d. across the secondary coil of a transformer is 12.0 V and 0.30 A flows through it.

If the p.d. across the primary coil is 240 V, what current is flowing through the primary coil?

V_sI_s = V_pI_p  (assuming 100 % efficiency)

I_p = V_sI_s / V_p = 12 x 0.3 / 240 = 0.015 A

Q9 A laptop charger works off the 230 V AC mains supply.

Inside the adapter, the transformer produces an a d.c. supply current of 3.0 A at a p.d. of 19 V.

(a) What current did the charger draw from the mains supply?

power input (mains) = power output (adapter)

P = IV, power output = 3.0 x 19 = 57 W

power input = 57 W, therefore since current drawn I = P / V, I = 57 / 230 = 0.25 A (2 sf)

(b) What assumptions has been made? and how can you tell from your everyday experience that the assumption is incorrect?

The calculation assumes there is no energy loss in the transformer process.

The adapter feels 'warm' due to electrical energy ==> thermal energy by the resistances of the wire causing a rise in temperature - this loss increases the thermal energy store of the surrounding air.