# Transformers And Impedance Matching Quiz

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To test and expand your knowledge, take this transformers and impedance matching quiz that we have designed for you. Impedance-matching transformers are used for the purpose of matching the impedance of a source and of its load to get the most efficient transfer of energy. Isolation transformers are mostly applied for reasons of safety to separate a piece of equipment from the power source. So, learn more with these questions, and we wish you good luck.

• 1.

### In a step-up transformer,

• A.

The primary impedance is greater than the secondary impedance.

• B.

The secondary winding is right on top of the primary.

• C.

The primary voltage is less than the secondary voltage.

• D.

All of the above are true.

C. The primary voltage is less than the secondary voltage.
Explanation
In a step-up transformer, the primary voltage is less than the secondary voltage. This is because a step-up transformer is designed to increase the voltage from the primary side to the secondary side. This is achieved by having a greater number of turns in the secondary winding compared to the primary winding, which results in a higher voltage output. Therefore, the statement that the primary voltage is less than the secondary voltage is correct.

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• 2.

### A transformer steps a voltage down from 117 V to 6.00 V. What is its primary-to-secondaryturns ratio?

• A.

1:380

• B.

380:1

• C.

1:19.5

• D.

19.5:1

D. 19.5:1
Explanation
The primary-to-secondary turns ratio of a transformer determines the voltage transformation between the primary and secondary coils. In this case, the voltage is stepped down from 117 V to 6.00 V. The turns ratio can be calculated by dividing the primary voltage by the secondary voltage. Therefore, the turns ratio is 117 V divided by 6.00 V, which equals 19.5:1.

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• 3.

### A step-up transformer has a primary-to-secondary turns ratio of 1:5.00. If 117 V rms appearsat the primary, what is the ac rms voltage across the secondary?

• A.

23.4 V rms

• B.

585 V rms

• C.

117 V rms

• D.

2.93 kV rms

B. 585 V rms
Explanation
The primary-to-secondary turns ratio of 1:5.00 means that for every 1 turn in the primary coil, there are 5 turns in the secondary coil. Since the voltage is directly proportional to the number of turns, the voltage across the secondary coil will be 5 times the voltage across the primary coil. Therefore, if 117 V rms appears at the primary, the ac rms voltage across the secondary will be 117 V rms multiplied by 5, which equals 585 V rms.

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• 4.

### A transformer has a secondary-to-primary turns ratio of 0.167. This transformer is

• A.

A step-up unit.

• B.

A step-down unit.

• C.

Neither a step-up unit nor a step-down unit.

• D.

A reversible unit.

B. A step-down unit.
Explanation
The given turns ratio of 0.167 indicates that the number of turns in the secondary coil is less than the number of turns in the primary coil. This means that the voltage in the secondary coil will be lower than the voltage in the primary coil. Therefore, this transformer is a step-down unit, as it decreases the voltage from the primary side to the secondary side.

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• 5.

### Eddy currents cause

• A.

An increase in efficiency.

• B.

An increase in coupling between windings.

• C.

An increase in core loss.

• D.

An increase in usable frequency range.

C. An increase in core loss.
Explanation
Eddy currents are circulating currents induced in conductive materials when subjected to a changing magnetic field. These currents flow in closed loops and create localized heating in the material. This heating leads to energy losses known as core losses. Therefore, the correct answer is that eddy currents cause an increase in core loss.

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• 6.

### The shell method of transformer winding

• A.

Provides maximum coupling.

• B.

Minimizes capacitance between windings.

• C.

Withstands more voltage than other winding methods.

• D.

Has windings far apart but along a common axis.

A. Provides maximum coupling.
Explanation
The shell method of transformer winding provides maximum coupling. This means that it allows for efficient transfer of energy between the primary and secondary windings of the transformer. By arranging the windings in a way that they are closer together and share a common axis, the magnetic field generated by the primary winding can induce a stronger voltage in the secondary winding. This results in a higher level of coupling between the two windings, leading to improved transformer performance.

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• 7.

### Which of these core types is best if you need a winding inductance of 1.5 H?

• A.

Air core

• B.

Ferromagnetic solenoid core

• C.

Ferromagnetic toroid core

• D.

Ferromagnetic pot core

D. Ferromagnetic pot core
Explanation
A ferromagnetic pot core is the best core type if you need a winding inductance of 1.5 H. This is because ferromagnetic pot cores have a high permeability, which allows for a higher inductance value to be achieved with fewer turns of wire. The pot shape also helps to reduce magnetic leakage and improve the efficiency of the inductor.

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• 8.

### An advantage of a toroid core over a solenoid core is the fact that

• A.

The toroid works at higher frequencies.

• B.

The toroid confines the magnetic flux.

• C.

The toroid can work for dc as well as for ac.

• D.

It is easier to wind the turns on a toroid.

B. The toroid confines the magnetic flux.
Explanation
The advantage of a toroid core over a solenoid core is that the toroid confines the magnetic flux. This means that the magnetic field lines produced by the current in the toroid are concentrated within the core, resulting in a stronger and more efficient magnetic field. In contrast, a solenoid core allows the magnetic field to spread out more, leading to a weaker and less focused magnetic field.

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• 9.

### High voltage is used in long-distance power transmission because

• A.

It is easier to regulate than low voltage.

• B.

The I^2 R losses are minimized.

• C.

The electromagnetic fields are strong.

• D.

Small transformers can be used.

B. The I^2 R losses are minimized.
Explanation
High voltage is used in long-distance power transmission because it helps minimize the I^2 R losses. When electricity is transmitted over long distances, there is resistance in the transmission lines which leads to energy losses in the form of heat. By using high voltage, the current can be reduced, which in turn reduces the I^2 R losses. This helps to maximize the efficiency of power transmission and minimize the amount of energy wasted as heat.

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• 10.

### In a household circuit, 234-V rms electricity usually has

• A.

One pHase.

• B.

Two pHases.

• C.

Three pHases.

• D.

Four pHases.

C. Three pHases.
Explanation
In a household circuit, 234-V rms electricity usually has three phases. This is because most household circuits operate on a three-phase system, where the electrical power is distributed across three separate phases. Each phase carries a voltage of 234-V rms, resulting in a total of three phases. This allows for a more efficient and balanced distribution of electricity throughout the household.

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• 11.

### In a transformer, a center tap often exists in

• A.

The primary winding.

• B.

The secondary winding.

• C.

An unbalanced winding.

• D.

A balanced winding.

D. A balanced winding.
Explanation
A center tap often exists in a balanced winding in a transformer. This is because a balanced winding is designed to have equal numbers of turns on each side of the center tap. The center tap allows for the connection of a load to either side of the winding, providing flexibility in the transformer's output voltage. It also helps in achieving a balanced distribution of voltage across the winding, ensuring efficient power transfer between the primary and secondary sides of the transformer.

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• 12.

### An autotransformer

• A.

• B.

Has a center-tapped secondary.

• C.

Consists of a single tapped winding.

• D.

Is useful only for impedance matching.

C. Consists of a single tapped winding.
Explanation
An autotransformer consists of a single tapped winding, which means that it has only one winding with multiple taps along its length. This allows for different voltage ratios to be achieved by connecting the load to different taps on the winding. Unlike a traditional transformer with separate primary and secondary windings, an autotransformer can provide a variable voltage output by adjusting the connection point on the winding. This makes it a versatile and efficient option for voltage regulation in various applications.

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• 13.

### Suppose a transformer has a primary-to-secondary turns ratio of 2.00:1. The inputimpedance is 300 Ω, purely resistive. What is the output impedance?

• A.

75 Ω, purely resistive

• B.

150 Ω, purely resistive

• C.

600 Ω, purely resistive

• D.

1200 Ω, purely resistive

A. 75 Ω, purely resistive
Explanation
The output impedance of a transformer can be calculated using the turns ratio squared. In this case, the turns ratio is 2.00:1, so when squared, it becomes 4. Therefore, the output impedance is 4 times the input impedance. Since the input impedance is 300 Ω, the output impedance would be 4 * 300 Ω = 1200 Ω. However, the question specifies that the output impedance is purely resistive, so we need to divide this by 2 to get the final answer of 600 Ω. Therefore, the correct answer is 600 Ω, purely resistive.

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• 14.

### Suppose a purely resistive input impedance of 50 Ω must be matched to a purely resistiveoutput impedance of 450 Ω. The primary-to-secondary turns ratio of the transformer must bewhich of the following?

• A.

9.00

• B.

3.00

• C.

1/3.00

• D.

1/9.00

C. 1/3.00
Explanation
The input and output impedances of a transformer can be matched by using the turns ratio of the transformer. The turns ratio is calculated by taking the square root of the ratio of the output impedance to the input impedance. In this case, the output impedance is 450 Ω and the input impedance is 50 Ω. Taking the square root of the ratio gives us the turns ratio of 3.00. However, the question asks for the primary-to-secondary turns ratio, so the inverse of 3.00 is taken, resulting in 1/3.00.

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• 15.

### Suppose a quarter-wave matching section has a characteristic impedance of 75.0 Ω. Theinput impedance is 50.0 Ω, purely resistive. What is the output impedance?

• A.

150 Ω, purely resistive

• B.

125 Ω, purely resistive

• C.

100 Ω, purely resistive

• D.

113 Ω, purely resistive

D. 113 Ω, purely resistive
Explanation
A quarter-wave matching section is used to match the impedance of the input to the impedance of the output. In this case, the input impedance is 50.0 Ω and the characteristic impedance of the matching section is 75.0 Ω. To match these impedances, the output impedance of the matching section must be equal to the square of the characteristic impedance divided by the input impedance, which is (75.0^2) / 50.0 = 113 Ω. Since there is no mention of any reactive components, the output impedance is purely resistive. Therefore, the correct answer is 113 Ω, purely resistive.

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• Current Version
• Sep 02, 2023
Quiz Edited by
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• Dec 11, 2010
Quiz Created by
BATANGMAGALING

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