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Chapter 20: Problem 31

A step-down transformer has 4000 turns on the primary and 200 turns on the secondary. If the primary voltage amplitude is \(2.2 \mathrm{kV},\) what is the secondary voltage amplitude?

Short Answer

Expert verified
Answer: The secondary voltage amplitude is 0.11 kV or 110 V.

Step by step solution

01

Identify the given information

In this exercise, we are given the following information: - Primary coil turns (N1) = 4000 turns - Secondary coil turns (N2) = 200 turns - Primary voltage amplitude (V1) = 2.2 kV Our goal is to find the secondary voltage amplitude (V2).
02

Find the turns ratio

The turns ratio is the ratio between the number of turns in the primary coil to the number of turns in the secondary coil: Turns Ratio, \(k = \frac{N1}{N2}\) Plug the given values of N1 and N2 into the formula: \(k = \frac{4000}{200} = 20\)
03

Calculate the secondary voltage amplitude

Now that we have the turns ratio, we can calculate the secondary voltage amplitude V2. For transformers, the voltage ratio is equal to the turns ratio: \(k = \frac{V1}{V2}\) Rearrange the formula to find V2: \(V2 = \frac{V1}{k}\) Plug in the given primary voltage amplitude (V1 = 2.2 kV) and the turns ratio (k = 20) into the formula: \(V2 = \frac{2.2\, \text{kV}}{20}\) Calculate the secondary voltage amplitude: \(V2 = 0.11\, \text{kV}\) or \(110\, \text{V}\)
04

Write the final answer

The secondary voltage amplitude of the step-down transformer is 0.11 kV or 110 V.

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Most popular questions from this chapter

A doorbell uses a transformer to deliver an amplitude of \(8.5 \mathrm{V}\) when it is connected to a \(170-\mathrm{V}\) amplitude line. If there are 50 turns on the secondary, (a) what is the turns ratio? (b) How many turns does the primary have?
(a) For a particle moving in simple harmonic motion, the position can be written \(x(t)=x_{\mathrm{m}}\) cos \(\omega t .\) What is the velocity \(v_{x}(t)\) as a function of time for this particle? (b) Using the small-angle approximation for the sine function, find the slope of the graph of \(\Phi(t)=\Phi_{0} \sin \omega t\) at \(t=0 .\) Does your result agree with the value of \(\Delta \Phi / \Delta t=\omega \Phi_{0} \cos \omega t\) at \(t=0 ?\)
A \(0.30-\mathrm{H}\) inductor and a \(200.0-\Omega\) resistor are connected in series to a \(9.0-\mathrm{V}\) battery. (a) What is the maximum current that flows in the circuit? (b) How long after connecting the battery does the current reach half its maximum value? (c) When the current is half its maximum value, find the energy stored in the inductor, the rate at which energy is being stored in the inductor, and the rate at which energy is dissipated in the resistor. (d) Redo parts (a) and (b) if, instead of being negligibly small, the internal resistances of the inductor and battery are \(75 \Omega\) and \(20.0 \Omega\) respectively.
A 2 -m-long copper pipe is held vertically. When a marble is dropped down the pipe, it falls through in about 0.7 s. A magnet of similar size and shape takes much longer to fall through the pipe. (a) As the magnet is falling through the pipe with its north pole below its south pole, what direction do currents flow around the pipe above the magnet? Below the magnet (CW or CCW as viewed from the top)? (b) Sketch a graph of the speed of the magnet as a function of time. [Hint: What would the graph look like for a marble falling through honey?]
A 100 -turn coil with a radius of \(10.0 \mathrm{cm}\) is mounted so the coil's axis can be oriented in any horizontal direction. Initially the axis is oriented so the magnetic flux from Earth's field is maximized. If the coil's axis is rotated through \(90.0^{\circ}\) in \(0.080 \mathrm{s},\) an emf of $0.687 \mathrm{mV}$ is induced in the coil. (a) What is the magnitude of the horizontal component of Earth's magnetic field at this location? (b) If the coil is moved to another place on Earth and the measurement is repeated, will the result be the same?
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