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Chapter 22: Problem 63

The resistances of the primary and secondary coils of a transformer are 56 and \(14 \Omega,\) respectively. Both coils are made from lengths of the same copper wire. The circular turns of each coil have the same diameter. Find the turns ratio \(N_{s} / N_{\mathrm{p}}\)

Short Answer

Expert verified
The turns ratio \(N_{s}/N_{p}\) is 1/4.

Step by step solution

01

Understand the Relationship of Resistances to Number of Turns

The resistances of the primary and secondary coils are given as 56 惟 and 14 惟, respectively. Because both coils are made from the same copper wire and have the same diameter, their resistances are proportional to their number of turns. Thus, we can write:\[ \frac{R_{p}}{R_{s}} = \frac{N_{p}}{N_{s}} \] where \(R_{p}\) is the resistance of the primary coil, \(R_{s}\) is the resistance of the secondary coil, \(N_{p}\) is the number of turns in the primary coil, and \(N_{s}\) is the number of turns in the secondary coil.
02

Substitute Known Values Into the Equation

Substitute the given resistance values into the equation from Step 1:\[ \frac{56}{14} = \frac{N_{p}}{N_{s}} \]
03

Simplify the Equation

Simplify the fraction on the left side of the equation:\[ 4 = \frac{N_{p}}{N_{s}} \] This simplifies further to show that the turns ratio \(N_{s}/N_{p}\) is the reciprocal of 4.
04

Solve for the Turns Ratio

Taking the reciprocal of both sides, we get:\[ \frac{N_{s}}{N_{p}} = \frac{1}{4} \] Thus, the turns ratio is 1/4.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Turns Ratio
In transformers, the turns ratio is a crucial concept that helps determine how voltage is transformed from the primary coil to the secondary coil. It's defined as the ratio of the number of turns in the secondary coil ( _{s} ) to the number of turns in the primary coil ( _{p} ). When you know this ratio, you can predict how the voltage will change as it passes through the transformer.

If you have more turns on the secondary coil compared to the primary coil, the output voltage will be higher, which is called step-up transformation. Conversely, if the primary coil has more turns, the voltage is reduced, known as step-down transformation.
  • The equation for the turns ratio is given by ( _{s} / _{p} ).
  • This ratio is helpful in designing transformers to meet specific voltage requirements.
In our exercise, because the primary coil's resistance is 56 惟 and the secondary's is 14 惟 , their turns ratio is ( _{s} / _{p} = 1/4 ). This means the transformer reduces voltage to a quarter of the original value.
The Role of Electrical Resistance
Electrical resistance plays a significant role in determining the efficiency and effectiveness of a transformer. It is defined as the opposition to current flow within an electrical circuit. In the context of transformers, both the primary and secondary coils have intrinsic resistances, which impact how current moves through them.

Resistance ( R ) is associated with the material and dimensions of the wire:
  • Since both coils in the exercise are made from the same copper wire, they share similar resistivity factors.
  • Resistance can be calculated with the formula ( R = 蟻 ( l/A ), where 蟻 is the resistivity, l is length, and A is the cross-sectional area.
The resistance of the coil directly influences the number of turns because longer wire (more turns) means a higher resistance, assuming cross-sectional characteristics remain constant.

In this problem, we see a direct proportionality between resistance and turns, hence the inverse relation in the formula for turns ratio.
Primary and Secondary Coils of a Transformer
Transformers consist of two essential components: the primary and secondary coils. These coils are intricately wound wires that play distinct roles in the transformer鈥檚 operation.

- **Primary Coil:** This is the coil connected to the input voltage source. It's where electrical energy is initially supplied. - **Secondary Coil:** This coil is connected to the load or where the transformed voltage is delivered.

The interaction between these coils allows transformers to modify voltage levels based on the turns ratio. When alternating current passes through the primary coil, it creates a magnetic field, and this field induces a current in the secondary coil through electromagnetic induction.
  • The efficiency of voltage transformation depends significantly on the construction and properties of these coils.
  • For instance, having the same wire diameter, as in the exercise, ensures uniform energy transfer characteristics.
Understanding these transformations and the design of coils helps in building efficient transformers for varied applications.

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Most popular questions from this chapter

A square loop of wire consisting of a single turn is perpendicular to a uniform magnetic field. The square loop is then re-formed into a circular loop, which also consists of a single turn and is also perpendicular to the same magnetic field. The magnetic flux that passes through the square loop is \(7.0 \times 10^{-3}\) Wb. What is the flux that passes through the circular loop?

The magnetic flux that passes through one turn of a 12 -turn coil of wire changes to 4.0 from \(9.0 \mathrm{Wb}\) in a time of \(0.050 \mathrm{s}\). The average induced current in the coil is 230 A. What is the resistance of the wire?

A circular coil of radius 0.11 \(\mathrm{m}\) contains a single turn and is located in a constant magnetic field of magnitude \(0.27 \mathrm{T}\). The magnetic field has the same direction as the normal to the plane of the coil. The radius increases to \(0.30 \mathrm{m}\) in a time of \(0.080 \mathrm{s}\). Concepts: (i) Why is there an emf induced in the coil? (ii) Does the magnitude of the induced emf depend on whether the area is increasing or decreasing? Explain. (iii) What determines the amount of current induced in the coil? (iv) If the coil is cut so it is no longer one continuous piece, are there an induced emf and an induced current? Explain. Calculations: (a) Determine the magnitude of the emf induced in the coil. (b) The coil has a resistance of \(0.70 \Omega .\) Find the magnitude of the induced current.

The drawing shows a coil of copper wire that consists of two semicircles joined by straight sections of wire. In part \(a\) the coil is lying flat on a horizontal surface. The dashed line also lies in the plane of the horizontal surface. Starting from the orientation in part \(a\), the smaller semicircle rotates at an angular frequency \(\omega\) about the dashed line, until its plane becomes perpendicular to the horizontal surface, as shown in part \(b\). A uniform magnetic field is constant in time and is directed upward, perpendicular to the horizontal surface. The field completely fills the region occupied by the coil in either part of the drawing. The magnitude of the magnetic field is 0.35 T. The resistance of the coil is \(0.025 \Omega\), and the smaller semicircle has a radius of \(0.20 \mathrm{m} .\) The angular frequency at which the small semicircle rotates is \(1.5 \mathrm{rad} / \mathrm{s} .\) Determine the average current \(I,\) if any, induced in the coil as the coil changes shape from that in part \(a\) of the drawing to that in part \(b .\) Be sure to include an explicit plus or minus sign along with your answer.

A constant current of \(I=15 \mathrm{A}\) exists in a solenoid whose in- ductance is \(L=3.1 \mathrm{H}\). The current is then reduced to zero in a certain amount of time. (a) If the current goes from 15 to 0 A in a time of 75 ms, what is the emf induced in the solenoid? \(?\) (b) How much electrical energy is stored in the solenoid? (c) At what rate must the electrical energy be removed from the solenoid when the current is reduced to \(0 \mathrm{A}\) in a time of \(75 \mathrm{ms} ?\) Note that the rate at which energy is removed is the power.
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