Highworth Physics - Keeping Things Moving - Transformers And Power

Front 1

What equation is used to calculate electrical power?

Back 1

electrical power can be calculated using this equation:

P = V × I

P is the power in watts, W

V is the potential difference in volts, V

I is the current in amperes (amps), A

Front 2

What equation is used to calculate the power in the input and output of the coils in a transformer?

Back 2

Assuming that the transformer is 100% efficient (no energy is lost between its primary coil and secondary coil), the power output from the secondary coil will be the same as the power input to the primary coil.

Vp × Ip = Vs × Is

Vp is the potential difference across the primary coil in volts, V

Ip is the current in the primary coil in amperes (amps), A

Vs is the potential difference across the secondary coil in volts, V

Is is the current in the secondary coil amperes (amps), A

Note that, in reality, the assumption that transformers are 100% efficient is not a valid one. Some energy will be lost to the surroundings as heat from the iron core and the coils.

Front 3

A current of 0.2 A is supplied to the primary coil of a transformer at a potential difference of 230 V.

The secondary coil has a 4.0 A current flowing through it.

Calculate the potential difference across the secondary coil, assuming that the transformer is 100% efficient.

Back 3

P = Vp × Ip P = 230 × 0.2 = 46 W

P= Vs × Is Vs = P ÷ Is

Vs = 46 ÷ 4.0 = 11.5 V

Front 4

A student recorded the currents and the secondary potential differences obtained from a transformer for different primary potential differences. The results are shown in the table.

(a) Plot a graph to show the secondary potential difference (Vs) obtained for each primary potential difference (Vp).

(b) Find the ratio ns ÷ np for the transformer by finding the gradient of the line.

(c) Explain why you obtain a more reliable result by using the gradient of the line instead of comparing one set of potential differences.

Back 4

(a) Suitable graph plotted from data.

(b) Take any two points on the line. If using the 10 V and 0 V points: gradient = 5.98 V ÷ 10 V = 0.6.

(c) The line should be a line of best fit, which should ignore any outliers. Using the line therefore effectively takes into account all the results (except any outliers), rather than just using one result which may include an error.

Front 5

What is a Switch mode transformer?

Back 5

Switch mode transformers are often found in the power supplies of electronic devices such as laptop and mobile phone chargers.

Devices like these need a smaller potential difference than the 230 V from the mains electricity. Therefore, they need a step-down transformer to reduce the potential difference, built into the plug or power supply.

Switch mode transformers achieve this by using complex electronic circuits. These rapidly switch the current on and off, allowing the alternating current to be changed to a higher frequency. This is often between 50 Hz and 200 Hz.

Front 6

What are the benefits of using a switch mode transformer?

Back 6

At these frequencies, a much smaller and lighter transformer than normal is able to reduce the potential difference.

As a result, these transformers are suited for use in power supplies such as mobile phone chargers.

When the device is plugged in and the batteries are recharging, a load is being applied (the transformer is drawing power).

Switch mode transformers use very little power when the plug is left switched on but no load is applied (such as when the device’s batteries are not charging).

Front 7

Look at the photo of the transformers in the computer.

Explain why computers need cooling fans inside them.

Back 7

Some of the energy transferred to a transformer is converted to heat in the coils and core.

This could overheat the other components inside the computer if there were no cooling fan to draw cooler air through the case.

Front 8

A step-down transformer changes a 13 A mains supply to a 12 V supply.

What is the current in the secondary coil?

Back 8

Primary p.d. = 230 V.

Is = (Vp × Ip) ÷ Vs

= (230 V × 13 A) ÷ 12 V = 249 V.

Front 9

Explain why the actual current in the secondary coil would be less than the value you calculated in question 2.

Back 9

Some energy would be transferred as heat energy, so the power transferred by the secondary coil would be less than that transferred by the primary coil.

Front 10

Describe three advantages of switch-mode transformers compared to traditional transformers.

Back 10

They are smaller, lighter, and do not waste power when they are switched on but not connected to a load.

Front 11

Explain why switch-mode transformers are more suitable than traditional transformers for cordless phone chargers.

Back 11

Many people leave their phones in the chargers for longer than is necessary to recharge them.

A traditional transformer would waste more energy while it was switched on and not being used.

Front 12

(a) At what frequencies do switch-mode transformers operate?

(b) How are these frequencies achieved?

Back 12

(a) Between 50 kHz and 200 kHz.

(b) By switching the supply on and off very rapidly.

Front 13

A transformer has 1000 turns on the primary coil and 50 turns on the secondary coil, and is plugged into a 230 V supply.

What is the current in the secondary coil if the current in the primary coil is 1 A.

(Hint: you need to work out the potential difference across the secondary coil first.)

Back 13

A Vs = (Vp × Ns) ÷ Np

= (230 V × 1000) ÷ 50 = 4600 V

Is = (Vp × Ip) ÷ Vs

= (230 V × 1 A) ÷ 4600 V = 0.05 A.

Front 14

Look at the data in the table.

Use the equation for power and current in a transformer to calculate what the current should be in the secondary coil, assuming the transformer is 100 per cent efficient, and calculate the percentage of the power going into the transformer that is wasted.

Back 14

Is = (Vp × Ip) ÷ Vs.

Using the 10 V point:

Power in = 10 V x 0.08 A = 0.8 W.

Power out = 5.98 V x 0.127 A =0.76 W.

Power wasted = 0.04 W ÷ 0.8 W x 100% = 5%.

Front 15

What are the key points to remember?

Back 15

Fora 100% efficient transformer, p.d. across primary coil × current through primary coil = p.d. across secondary coil × current through secondary coil; Vp × Ip = Vs × Is.

No transformers are 100% efficient, as some energy is transferred as heat in the coils and core.

Switch mode transformers contain components that switch the supply on and off very rapidly, increasing the frequency.

A step-up transformer increases the potential difference, a step-down transformer decreases it.

The high frequency means that a much smaller and lighter transformer can then be used to change the voltage.

Switch mode transformers waste far less energy as heat than traditional transformers do when they are switched on but with no load.