Questions on transformers

Question 1

Explain how a transformer transfers energy from its primary coil to its secondary coil.

Accepted Answer

The alternating current in the primary produces a continually changing magnetic flux in the iron core, which links the secondary coil (1); by Faraday's law the changing flux linkage induces an alternating emf in the secondary, so energy is transferred without an electrical connection (1).

Question 2

A transformer has 1000 turns on its primary coil and 100 turns on its secondary coil. The primary voltage is 230 V. Calculate the secondary voltage.

Accepted Answer

V_s = V_p × N_s/N_p = 230 × 100/1000 (1) V_s = 23 V (1)

Question 3

A transformer has fewer turns on its secondary coil than on its primary coil. State whether it is a step-up or a step-down transformer.

Accepted Answer

Step-down: the secondary voltage is lower than the primary voltage (1).

Question 4

Explain why a transformer produces no output when its primary coil is connected to a steady dc supply.

Accepted Answer

A steady current produces a constant flux in the core (1); with no change in the flux linkage of the secondary, no emf is induced (1).

Question 5

State two causes of energy loss in a real transformer.

Accepted Answer

Any two of the following, one mark each (2): resistance of the windings heating the coils; eddy currents induced in the core; flux leakage, with some flux missing the secondary; energy used in repeatedly magnetising the core each cycle.

Question 6

A phone charger contains a transformer that steps 230 V down to 5.0 V. Calculate the turns ratio N_p:N_s.

Accepted Answer

N_p/N_s = V_p/V_s = 230/5.0 (1) N_p:N_s = 46:1 (1)

Question 7

A transformer steps 240 V down to 12 V. The primary coil has 2000 turns. Calculate the number of turns on the secondary coil.

Accepted Answer

N_s = N_p × V_s/V_p = 2000 × 12/240 (1) N_s = 100 turns (1)

Question 8

An ideal transformer has a primary voltage of 230 V and draws a primary current of 2.0 A. The secondary voltage is 23 V. Calculate the power transferred and the secondary current.

Accepted Answer

P = V_pI_p = 230 × 2.0 = 460 W (1) Ideal, so V_sI_s = V_pI_p (1) I_s = 460/23 = 20 A (1)

Question 9

A transformer has an input power of 500 W and a useful output power of 450 W. Calculate its efficiency.

Accepted Answer

Efficiency = output power/input power = 450/500 (1) Efficiency = 0.90 (90%) (1)

Question 10

An ideal step-up transformer raises 230 V to 1150 V. The secondary current is 0.40 A. Calculate the primary current.

Accepted Answer

I_p = V_sI_s/V_p = (1150 × 0.40)/230 (1) I_p = 2.0 A (1)

Question 11

Explain what eddy currents are, why they waste energy in a transformer, and how the design of the core reduces them.

Accepted Answer

Eddy currents are currents induced in the iron core itself by the changing flux (1); they dissipate energy by heating the core as they flow through its resistance (1); the core is made of thin laminations separated by insulation, which break the current loops into small ones and greatly reduce the loss (1).

Question 12

The alternating flux in a transformer core links every turn of both coils. Explain why the induced emf per turn is the same for both coils, and how this leads to the transformer equation V_s/V_p = N_s/N_p.

Accepted Answer

The core carries the same changing flux through every turn of both coils (1); by Faraday's law each turn therefore has the same rate of change of flux, so the same emf is induced in every turn (1); the total voltage across each coil is the emf per turn multiplied by its number of turns, so the voltages are in the ratio of the turns (1).

Question 13

A student tests a transformer with four meters. The primary takes 0.26 A at 230 V; the secondary delivers 4.6 A at 11.5 V. Calculate the efficiency of the transformer and the power dissipated inside it.

Accepted Answer

Input power = 230 × 0.26 = 59.8 W (1) Output power = 11.5 × 4.6 = 52.9 W (1) Efficiency = 52.9/59.8 = 0.88 (88%) (1) Power dissipated = 59.8 − 52.9 = 6.9 W (1)

Question 14

A power of 100 kW is transmitted along a cable of resistance 5.0 Ω. Calculate the power lost in the cable when the transmission voltage is 10 kV, and when it is 100 kV. Comment on the results.

Accepted Answer

At 10 kV: I = P/V = 100000/10000 = 10 A (1)
Loss = I²R = 10² × 5.0 = 500 W (1)
At 100 kV: I = 1 A, so loss = 1² × 5.0 = 5 W (1)
Raising the voltage by a factor of 10 cuts the current by 10 and the loss by 100 (1).

Question 15

A transformer is 96% efficient. The primary coil takes 5.0 A at 230 V and the secondary output is at 46 V. Calculate the input power, the output power and the secondary current.

Accepted Answer

Input power = 230 × 5.0 = 1150 W (1) Output power = 0.96 × 1150 = 1104 W (1) I_s = 1104/46 = 24 A (1)

Question 16

Explain why electrical energy is transmitted across the National Grid at very high voltage.

Accepted Answer

For a given power, a higher voltage means a smaller current, since P = VI (1); the power wasted heating the cables is I²R (1); reducing the current therefore greatly reduces the losses, making transmission more efficient (1).

Question 17

A garden lighting system needs an rms supply of between 11.5 V and 12.5 V, taken from the 230 V mains. Three transformers are available.
T1: 1000 primary turns, 52 secondary turns
T2: 1000 primary turns, 60 secondary turns
T3: 750 primary turns, 30 secondary turns
Deduce which transformer, if any, is suitable.

Accepted Answer

V_s = V_p × N_s/N_p for each transformer (1)
T1: V_s = 230 × 52/1000 = 12.0 V (1)
T2: V_s = 13.8 V and T3: V_s = 9.2 V (1)
Only T1 gives an output between 11.5 V and 12.5 V, so T1 is the suitable transformer (1)

Question 18

An ideal transformer steps the 230 V mains down with a turns ratio of 10:1. The secondary is connected to a 4.6 Ω resistor. Calculate the secondary voltage, the secondary current and the primary current.

Accepted Answer

V_s = 230/10 = 23 V (1) I_s = V_s/R = 23/4.6 = 5.0 A (1) Ideal, so I_p = V_sI_s/V_p (1) I_p = (23 × 5.0)/230 = 0.50 A (1)