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Chapter 20: Problem 33

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?

Short Answer

Expert verified
Answer: The turns ratio is approximately 20, and the primary coil has approximately 1000 turns.

Step by step solution

01

Identify the given information

We are given: - \(V_p = 170 \mathrm{V}\) (voltage of the primary coil) - \(V_s = 8.5 \mathrm{V}\) (voltage of the secondary coil) - \(N_s = 50\) (number of turns in the secondary coil)
02

Calculate the turns ratio

Using the formula for the turns ratio: $$\frac{N_p}{N_s}=\frac{V_p}{V_s}$$ Plug in the given values: $$\frac{N_p}{50}=\frac{170}{8.5}$$ Now, we can solve for the turns ratio, which is the ratio of \(N_p\) to \(N_s\): $$\frac{N_p}{50}=\frac{170}{8.5} \Rightarrow \frac{N_p}{50} \approx 20$$ So, the turns ratio is approximately 20.
03

Find the number of turns in the primary coil

Now that we have the turns ratio, we can find how many turns are in the primary coil. We know the following: $$\frac{N_p}{50} \approx 20$$ Now, we can solve for \(N_p\): $$N_p \approx 20 \times 50 = 1000$$ Therefore, the primary coil has approximately 1000 turns.

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Most popular questions from this chapter

A transformer with 1800 turns on the primary and 300 turns on the secondary is used in an electric slot car racing set to reduce the input voltage amplitude of \(170 \mathrm{V}\) from the wall output. The current in the secondary coil is of amplitude \(3.2 \mathrm{A}\). What is the voltage amplitude across the secondary coil and the current amplitude in the primary coil?
A horizontal desk surface measures \(1.3 \mathrm{m} \times 1.0 \mathrm{m} .\) If Earth's magnetic field has magnitude \(0.44 \mathrm{mT}\) and is directed \(65^{\circ}\) below the horizontal, what is the magnetic flux through the desk surface?
Compare the electric energy that can be stored in a capacitor to the magnetic energy that can be stored in an inductor of the same size (that is, the same volume). For the capacitor, assume that air is between the plates; the maximum electric field is then the breakdown strength of air, about $3 \mathrm{MV} / \mathrm{m} .$ The maximum magnetic field attainable in an ordinary solenoid with an air core is on the order of \(10 \mathrm{T}\)
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In Section \(20.9,\) in order to find the energy stored in an inductor, we assumed that the current was increased from zero at a constant rate. In this problem, you will prove that the energy stored in an inductor is \(U_{\mathrm{L}}=\frac{1}{2} L I^{2}-\) that is, it only depends on the current \(I\) and not on the previous time dependence of the current. (a) If the current in the inductor increases from \(i\) to \(i+\Delta i\) in a very short time \(\Delta t,\) show that the energy added to the inductor is $$\Delta U=L i \Delta i$$ [Hint: Start with \(\Delta U=P \Delta t .]\) (b) Show that, on a graph of \(L i\) versus \(i,\) for any small current interval \(\Delta i,\) the energy added to the inductor can be interpreted as the area under the graph for that interval.(c) Now show that the energy stored in the inductor when a current \(I\) flows is \(U=\frac{1}{2} L I^{2}\)
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