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Chapter 20: Problem 35

When the emf for the primary of a transformer is of amplitude $5.00 \mathrm{V},\( the secondary emf is \)10.0 \mathrm{V}$ in amplitude. What is the transformer turns ratio \(\left(N_{2} / N_{1}\right) ?\)

Short Answer

Expert verified
Answer: The transformer turns ratio (N2/N1) is 2.

Step by step solution

01

Understand the Transformer Turns Ratio Formula

The formula to relate the primary and secondary voltages of a transformer with its turns ratio is given by: \[ \frac{V_{2}}{V_{1}} = \frac{N_{2}}{N_{1}} \] where V1 is the primary voltage, V2 is the secondary voltage, N1 is the number of turns in the primary coil, and N2 is the number of turns in the secondary coil.
02

Plug in the given values

We are given V1 = 5.00 V and V2 = 10.0 V in the problem. We will plug these values into the formula: \[ \frac{10.0}{5.00} = \frac{N_{2}}{N_{1}} \]
03

Solve for N2/N1

Now, we will solve for the transformer turns ratio (N2/N1): \[ \frac{N_{2}}{N_{1}} = \frac{10.0}{5.00} = 2 \] So the transformer turns ratio N2/N1 is 2.

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Most popular questions from this chapter

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?
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?
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 \(U=\frac{1}{2} L I^{2}\)
Two solenoids, of \(N_{1}\) and \(N_{2}\) turns respectively, are wound on the same form. They have the same length \(L\) and radius \(r\) (a) What is the mutual inductance of these two solenoids? (b) If an ac current $$I_{1}(t)=I_{\mathrm{m}} \sin \omega t$$ flows in solenoid $$1\left(N_{1} \text { turns }\right)$$ write an expression for the total flux through solenoid \(2 .\) (c) What is the maximum induced emf in solenoid \(2 ?[\text { Hint: Refer to Eq. }(20-7) .]\)
A de motor has coils with a resistance of \(16 \Omega\) and is connected to an emf of \(120.0 \mathrm{V}\). When the motor operates at full speed, the back emf is \(72 \mathrm{V}\). (a) What is the current in the motor when it first starts up? (b) What is the current when the motor is at full speed? (c) If the current is \(4.0 \mathrm{A}\) with the motor operating at less than full speed, what is the back emf at that time?
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