[Search Doc Brown's science website]

SITEMAP   School-college Physics Notes: Electricity 6.6 Transformer calculations

UK GCSE level age ~14-16 ~US grades 9-10 Scroll down, take time to study content or follow links

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

INDEX for physics notes on National Grid power supply, use of transformers-calculations and environmental issues


This page contains online questions only. Jot down your answers and check them against the worked out answers at the end of the page


6.6 Examples of transformer calculation practice questions and worked out answers

step-up transformer step-down transformer calculations formula

KEY to transformer diagram and transformer formula for calculations

Everything you need to know is on the transformer diagram above and take note of the abbreviations below!

To save repetition in the questions PLEASE note the following abbreviations which I will use for problem solving:

P = power (W) = I (A) x V (V),  P = IV = I2R

Vp = p.d. across the input primary coil

Vs = induced p.d. generated across the output secondary coil

np = number of wire turns on the primary input coil

ns = number of wire turns on the secondary output coil

Ip = input current flowing through the primary coil

Is = induced output current flowing out through the secondary coil

The ratio of the potential differences across the coil equals the ratio of the turns on each coil.

Vp   np
---- = ----
Vs   ns

Vp / Vs = np / ns    The p.d. and coils transformer equation

Vpns = Vsnp

VsIs = VpIp  The transformer power equation

Since P = IV, it means power input = power output,

assuming 100 % efficiency (never this in reality)

Rearrangement: Vs / Vp = Ip / Is

 

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?

ANSWERS

 

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?

ANSWERS

 

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?

(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)

(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?

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

ANSWERS

 

Q4 The p.d. across the primary coil of a transformer is 240 V carrying 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?

ANSWERS

 

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

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

ANSWERS

 

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.

ANSWERS

 

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

ANSWERS

 

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

ANSWERS

 

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?

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

ANSWERS


Keywords, phrases and learning objectives for National Grid electricity supply

Be able to do exam questions based on the formula for transformers e.g. calculations involving ratio of turns of wire in primary coil and secondary coil, input voltage and output voltage and the input and output current.

Appreciate the calculations assume there is no loss of power i.e. power input = power output, even though there is some wasted thermal energy.


WHAT NEXT?

TOP of page

INDEX for physics notes on National Grid power supply, use of transformers-calculations & environmental issues

ALL my electricity and magnetism notes

email doc brown - comments - query?

INDEX of all my PHYSICS NOTES

BIG website and using the [SEARCH BOX] below, maybe quicker than navigating the many sub-indexes


Basic Science Quizzes for UK KS3 science students aged ~12-14, ~US grades 6-8

BiologyChemistryPhysics for UK GCSE level students aged ~14-16, ~US grades 9-10

Advanced Level Chemistry for pre-university age ~16-18 ~US grades 11-12, K12 Honors

Find your GCSE/IGCSE science course for more help links to all science revision notes

Use your mobile phone in 'landscape' mode?

SITEMAP Website content Dr Phil Brown 2000+. All copyrights reserved on Doc Brown's physics revision notes, images, quizzes, worksheets etc. Copying of website material is NOT permitted. Exam revision summaries and references to GCSE science course specifications are unofficial.

INDEX of notes on National Grid power supply & use of transformers

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?

Vp / Vs = np / ns

Vpns = Vsnp

Vp = Vsnp / ns

Vp = 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?

VsIs = VpIp  (assuming 100 % efficiency)

Vs = VpIp / Is  = 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?

Vp / Vs = np / ns

Vpns = Vsnp

p.d. output = Vs = Vpns / np = (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)

Pinput  = IpVp = 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?

VsIs = VpIp  (assuming 100 % efficiency)

Is = VpIp / Vs  = 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?

Vp / Vs = np / ns

Vpns = Vsnp

np = Vpns / Vs = 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?

VsIs = VpIp  (assuming 100 % efficiency)

Vp = VsIs / Ip = 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?

Vp / Vs = np / ns

Vpns = Vsnp

ns = Vsnp / Vp = 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?

VsIs = VpIp  (assuming 100 % efficiency)

Ip = VsIs / Vp = 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.

Using SEARCH some initial results may be ad links you can ignore - look for docbrown

INDEX of notes on National Grid power supply & use of transformers

TOP OF PAGE