Practice Questions

1.A transformer has a primary potential difference of 100 V and a primary current of 0.3 A.The secondary potential difference is 20 V.

What is the secondary current, assuming the transformer is 100% efficient?

Is = (Vp x Ip)/Vs

Is = (100 x 0.3)/20

Is = 1.5A

2.A transformer supplies a secondary coil with a current of 8.0 A at 50 V. The primary coil is supplied with 230 V and carries a current of 2.0 A.

(a) Calculate the input power and output power of the transformer.

Input power = 230 x 2 = 460W

Output power = 50 x 8 = 400W

(b) Determine the efficiency of the transformer.

Efficiency = (Useful power output/total power input) x 100

400W/460W x 100 = 87%

(c) Suggest one reason why the efficiency is less than 100%.

Some energy is lost to surroundings as heat due to the resistance of the wires in the coils.

3.A phone charger contains a transformer that steps the mains voltage down to 5 V for charging. A current of 0.5A runs from the transformer to the phone.

(a) What is the current supplied by the primary coil if the secondary current is 2.0 A?

Vp x Ip = Vs x Is

Ip = (Vs x Is)/Vp

Assuming mains input Vp = 230V

Ip = (5 x 0.5)/230 = 0.011A

(b) Explain why the current is different on each side of the transformer.

A transformer changes the potential difference between the primary and secondary coils. To conserve energy (assuming it’s ideal and there are no losses), the electrical power on the primary side must equal the power on the secondary side:

Power in=Power out

 Vp x Ip = Vs x Is

If the secondary voltage is much lower than the primary voltage, then the secondary current must be much higher to keep the power the same. Conversely, if the secondary voltage is higher than the primary voltage, then the secondary current is lower.