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7.6 Transformer Efficiency

7.6 Transformer Efficiency

The transformer efficiency measures the amount of energy it delivers versus the energy needed to deliver it. We know that in both, the primary and the secondary, there are losses because of resistance of coils (Copper losses). It also appears leakage flux (Iron losses) (Fig. 7.21). We define the efficiency of a transformer as:

η = P_util / P_absorbed = P_useful / (P_useful + P_losses) = P₂ / (P₂ + P_Fe + P_Cu) (7.36)

Diagram of an electrical circuit featuring a voltage source labeled U₀, resistors R_η, R₁, R₂, and R_s, inductors L_η, L₁, and L₂, and a transformer. The transformer is marked as "Ideal" with a turns ratio formula U₁/U₂ = N₁/N₂. Current paths I_p and I_s are indicated, along with voltages U₁ and U₂.

Fig. 7.21 Power transformer losses P_fe, and P_cu. Source From wikicommons under free to use license [17]

Suppose we have a transformer where we supply the rated voltage V1N to the primary, and an inductive load that connects to the secondary, which absorbs a certain current, I2.

In this case, the useful output power is:

P₂ = V₂ · I₂ · cos φ

练习题

What does transformer efficiency measure?

A. The ratio of input voltage to output voltage
B. The amount of energy delivered versus the energy needed to deliver it
C. The ratio of primary turns to secondary turns
D. The amount of current in the primary versus secondary winding

A transformer has an output power kW, iron losses W, and copper losses W. What is the efficiency of this transformer?

A. 97.1%
B. 96.2%
C. 98.0%
D. 95.0%

Which of the following are types of losses that occur in a transformer? Select all that apply.

A. Copper losses due to resistance of coils
B. Iron losses due to leakage flux
C. Magnetic saturation losses
D. Dielectric losses in insulation

Copper losses in a transformer occur only in the primary winding due to its resistance.

Transformer efficiency can be calculated as the ratio of useful output power to the sum of useful output power and all losses.

For a transformer with an inductive load, the useful output power is calculated as .

In the transformer efficiency formula , the term represents ___ losses.

When a transformer supplies an inductive load with secondary voltage and current , the useful output power is calculated using the formula ___.

A transformer delivers 80 kW to a load with iron losses of 800 W and copper losses of 1200 W. Calculate the input power and the efficiency of the transformer.

Explain why transformer efficiency is important and describe the two main categories of losses that reduce efficiency. How do these losses relate to heat generation in the transformer?

In the transformer efficiency formula , the iron losses contribute to the heat generated in the transformer. Which of the following correctly identifies the components that make up these iron losses?

A. includes only hysteresis losses in the magnetic circuit
B. includes only eddy current losses in the core
C. includes both hysteresis losses and eddy current losses
D. includes copper losses in the primary and secondary windings

An ideal transformer would achieve 100% efficiency. Which of the following conditions of an ideal transformer directly eliminate the losses that appear in the efficiency formula ?

A. Zero winding resistance eliminates copper losses ()
B. No iron losses condition eliminates
C. Unsaturated magnetic circuit eliminates all transformer losses
D. No stray flux condition eliminates leakage flux losses

The copper losses () in the transformer efficiency formula are also referred to as load losses because they depend on the load current flowing through the transformer windings.

When calculating transformer efficiency using , the iron losses are considered ___ losses because they occur whenever the transformer is energized, regardless of whether it is supplying a load.

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