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Chapter 20: Problem 34

The primary coil of a transformer has 250 turns; the secondary coil has 1000 turns. An alternating current is sent through the primary coil. The emf in the primary is of amplitude \(16 \mathrm{V}\). What is the emf amplitude in the secondary? (tutorial: transformer)

Short Answer

Expert verified
Answer: 64V

Step by step solution

01

Identify the transformer equation.

The transformer equation is given by: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\) where \(V_p\) is the emf amplitude in the primary coil, \(V_s\) is the emf amplitude in 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

Plug in the given values.

We are given \(N_p = 250\) turns, \(N_s = 1000\) turns, and \(V_p = 16 \mathrm{V}\). Substituting these values into the transformer equation, we get: \(\frac{16}{V_s} = \frac{250}{1000}\)
03

Solve for \(V_s\).

To solve for \(V_s\), we can cross-multiply and simplify: \(16 \times 1000 = V_s \times 250\) \(16000 = 250V_s\) Now, divide both sides by 250: \(V_s = \frac{16000}{250}\)
04

Calculate the emf amplitude in the secondary coil.

After the division, we get the emf amplitude in the secondary coil: \(V_s = 64 \mathrm{V}\) Therefore, the emf amplitude in the secondary coil is \(64 \mathrm{V}\).

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Most popular questions from this chapter

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?
A step-down transformer has a turns ratio of \(1 / 100 .\) An ac voltage of amplitude \(170 \mathrm{V}\) is applied to the primary. If the primary current amplitude is \(1.0 \mathrm{mA},\) what is the secondary current amplitude?
The armature of an ac generator is a rectangular coil \(2.0 \mathrm{cm}\) by \(6.0 \mathrm{cm}\) with 80 turns. It is immersed in a uniform magnetic field of magnitude 0.45 T. If the amplitude of the emf in the coil is $17.0 \mathrm{V},$ at what angular speed is the armature rotating?
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 \(\left(N_{2} / N_{1}\right) ?\)
In this problem, you derive the expression for the self inductance of a long solenoid [Eq. \((20-15 a)] .\) The solenoid has \(n\) turns per unit length, length \(\ell,\) and radius \(r\) Assume that the current flowing in the solenoid is \(I\) (a) Write an expression for the magnetic field inside the solenoid in terms of \(n, \ell, r, I,\) and universal constants. (b) Assume that all of the field lines cut through each turn of the solenoid. In other words, assume the field is uniform right out to the ends of the solenoid-a good approximation if the solenoid is tightly wound and sufficiently long. Write an expression for the magnetic flux through one turn. (c) What is the total flux linkage through all turns of the solenoid? (d) Use the definition of self-inductance [Eq. \((20-14)]\) to find the self inductance of the solenoid.
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