Transformer Calculations (Oxford AQA IGCSE Physics): Revision Note

A transformer has 20 turns on the primary coil and 800 turns on the secondary coil. The input potential difference across the primary coil is 500 V.

(a) Calculate the output potential difference

(b) State what type of transformer this is

Answer:

Part a)

Step 1: List the known quantities

• Number of turns in the primary coil, n_p = 20

• Number of turns in the secondary coil, n_s = 800

• Potential difference across the primary coil, V_p = 500 V

Step 2: Write the transformer equation from the equation sheet

• The transformer equation on the equation sheet is given as

$\frac{V_p}{V_s}=\frac{n_p}{n_s}$

• Flipping this equation with the secondary coil's variables on the top will make rearranging this equation much easier

$\frac{V_s}{V_p}=\frac{n_s}{n_p}$

Step 3: Rearrange for V_s

• The equation with secondary potential difference as the subject is:

$V_s=V_p\times \frac{n_s}{n_p}$

Step 4: Substitute the known quantities

$V_s=500\times \frac{800}{20}$

$V_s=20000\mathrm{V}$

Part b)

Step 1: Determine if the potential difference has increased or decreased

• The potential difference across the primary coil is 500 V

• The potential difference in the secondary coil is 20 000 V

• Therefore, the potential difference has increased from primary to secondary

Step 2: State whether the transformer is step-up or step-down

• The potential difference has increased

• So this is a step-up transformer

When you are using the transformer equation, there will be less rearranging to do in a calculation if the variable which you are trying to find is on the numerator (top line) of the fraction.

A transformer in a travel adapter steps up a 115 V ac mains electricity supply to the 230 V needed for the hair dryer. A current of 5 A flows through the hairdryer.

Assuming that the transformer is 100% efficient, calculate the current drawn from the mains supply. 

Answer:

Step 1: List the known quantities

• Voltage in primary coil, V_p = 115 V

• Voltage in secondary coil, V_s = 230 V

• Current in secondary coil, I_s = 5 A

Step 2: Write the equation linking the known values to the current from the supply, I_p

$I_p\times V_p=I_s\times V_s$

Step 3: Rearrange the equation to make I_p the subject

$I_p=\frac{I_s\times V_s}{V_p}$

Step 4: Substitute int he known values to calculate

$I_p=\frac{5\times 230}{115}$

Step 5: Calculate a value for I_p and include the correct unit

$I_p=10\mathrm{A}$

A step-down transformer is connected to a supply of 0.16 kV. The primary coil has 32× more turns than the secondary coil.

Calculate the output potential difference.

Answer:

Step 1: List the known quantities

• Potential difference across the primary coil, V_p = 0.16 kV

• Number of turns on primary coil = n_p

• Number of turns on secondary coil = n_s

Step 2: Convert kV into V

• There are 1000 V in 1 kV

$V_p=0.16\times 1000=160\mathrm{V}$

Step 3: Write an expression relating the numbers of turns

• There are 32 times more turns on the primary coil than on the secondary coil

$n_p=32n_s$

Step 4: Write down the transformer equation

• From the equation sheet:

$\frac{V_p}{V_s}=\frac{n_p}{n_s}$

• Flip the equation to find secondary potential difference

$\frac{V_s}{V_p}=\frac{n_s}{n_p}$

Step 5: Substitute in the expression relating number of turns

$\frac{n_p}{n_s}=32$

$\frac{V_p}{V_s}=\frac{n_p}{n_s}=32$

Step 6: Rearrange for potential difference across the secondary coil

$\frac{V_p}{V_s}=32$

$V_s=\frac{V_p}{32}$

Step 7: Substitute in the potential difference across the primary coil

$V_s=\frac{160}{32}$

$V_s=5.0\mathrm{V}$