Question

Learning Goal: To understand the concepts explaining the operation of transformers.

One of the advantages of alternating current (ac) over direct current (dc) is the ease with which voltage levels can be increased or decreased. Such a need is alwayspresent due to the practical requirements of energy distribution. On the one hand, the voltage supplied to the end users must be reasonably low for safety reasons(depending on the country, that voltage may be 110 volts, 220 volts, or some other value of that order). On the other hand, the voltage used in transmitting electricenergy must be as high as possible to minimize losses in the transmission lines. A device that uses the principle of electromagnetic induction to increase or decreasethe voltage by a certain factor is called a transformer.
The main components of a transformer are two coils (windings) that are electrically insulated from each other. The coils are wrapped around the same core, which istypically made of a material with a very large relative permeability to ensure maximum mutual inductance. One coil, called the primary coil, is connected to a voltagesource; the other, the secondary coil, delivers the power. The alternating current in the primary coil induces the changing magnetic flux in the core that creates theemf in the secondary coil. The magnitude of the emf induced in the secondary coil can be controlled by the design of the transformer. The key factor is the number ofturns in each coil.

Consider an ideal transformer, that is, one in which the coils have no ohmic resistance and the magnetic flux is the same for each turn of both the primary andsecondary coils. If the number of turns in the primary coil is and that in the secondary coil is , then the emfs induced in the coils can be written as
,

and therefore,
.

Since both emfs oscillate with the same frequency as the ac source, the formula above can be applied to the instantaneous amplitude or the rms values of the emfs.Moreover, if the coils have zero resistance (as we assumed), then for each coil the terminal voltage will be equal to the induced emf. Therefore, we can write
.

Note that if , then . This is a case of a step-up transformer. Conversely, if , then . This is a case of a step-down transformer. Without energy losses, the power inthe primary and secondary coils is the same:
.

If the secondary circuit is completed by a resistance , then . Combining this with the two equations above gives

.

Dividing the first and last expressions by and then inverting gives

.

In other words, the current in the primary coil is the same as if it were connected directly to a resistance equal to . In a way, transformers "transform" resistancesas well as voltages and currents. In reality, no transformer is ideal. There are always some energy losses. However, modern transformers have very high efficiencies,usually well exceeding 90%.


In answering the questions below, consider the transformer ideal unless otherwise noted.
Part A
The primary coil of a transformer contains 100 turns; the secondary has 200 turns. The primary coil is connected to a size AA battery that supplies a constant voltageof 1.5 volts. What voltage would be measured across the secondary coil?
zero





Part B
A transformer is intended to decrease the rms value of the alternating voltage from 500 volts to 25 volts. The primary coil contains 200 turns. Find the necessarynumber of turns in the secondary coil.
=



Part C
A transformer is intended to decrease the rms value of the alternating current from 500 amperes to 25 amperes. The primary coil contains 200 turns. Find the necessarynumber of turns in the secondary coil.
=



Part D
In a transformer, the primary coil contains 400 turns, and the secondary coil contains 80 turns. If the primary current is 2.5 amperes, what is the secondary current?
Express your answer numerically in amperes.
=




Part E
The primary coil of a transformer has 200 turns and the secondary coil has 800 turns. The power supplied to the primary coil is 400 watts. What is the power generatedin the secondary coil if it is terminated by a 20-ohm resistor?









Part F
The primary coil of a transformer has 200 turns, and the secondary coil has 800 turns. The transformer is connected to a 120-volt (rms) ac source. What is the (rms)current in the primary coil if the secondary coil is terminated by a 20-ohm resistor?
Express your answer numerically in amperes.
=




Part G
A transformer supplies 60 watts of power to a device that is rated at 20 volts (rms). The primary coil is connected to a 120-volt (rms) ac source. What is the currentin the primary coil?
Express your answer in amperes.
=




Part H
The voltage and the current in the primary coil of a nonideal transformer are 120 volts and 2.0 amperes. The voltage and the current in the secondary coil are 19.4volts and 11.8 amperes. What is the efficiency of the transformer? The efficiency of a transformer is defined as the ratio of the output power to the input power,expressed as a percentage: .

Answer 1

Part A
The primary coil of a transformer contains 100 turns; the secondary has 200 turns. The primary coil is connected to a size AA battery that supplies aconstant voltage of 1.5 volts. What voltage would be measured across the secondary coil?
zero

ANSWER:- voltage would be measured across the secondary coil. Since the input voltage is D.C. for which transformer doesnt works.

Part B
A transformer is intended to decrease the rms value of the alternating voltage from 500 volts to 25 volts. The primary coil contains 200 turns. Findthe necessary number of turns in the secondary coil.

ANSWER:-

therefore Ns = 10 turns.

Part C
A transformer is intended to decrease the rms value of the alternating current from 500 amperes to 25 amperes. The primary coil contains 200 turns.Find the necessary number of turns in the secondary coil.
=

Ns = 4000 turns.

Part D
In a transformer, the primary coil contains 400 turns, and the secondary coil contains 80 turns. If the primary current is 2.5 amperes, what is thesecondary current ?
Express your answer numerically in amperes.
=

secondary currentis 12.5 ampere.

Answer 2

For Part E of this problem:
In case of an ideal transformer, the power in the primary circuit is the same as that in the secondary circuit. Thus the answer to Part E is 400 Watts.

Answer 3

For Part F of this problem:

We found from Mastering Physics response to the answer in Part E that

"In case of an ideal transformer, the power in the primary circuit is the same as that in the secondary circuit."  Thus, we can use the relationship V₂ /V₁ =N₂/N₁= I₂/I₁         Since we're given N₂, N₁, and V₁ solve for V₂. Once you've found V₂ substitute that value into the formula Power= (V₂)²/R₂ = V₁I₁ and solve for I₁ Which is the value of I_primary     

This is your answer.