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Chapter 20: Problem 38

A transformer for an answering machine takes an ac voltage of amplitude $170 \mathrm{V}\( as its input and supplies a \)7.8-\mathrm{V}$ amplitude to the answering machine. The primary has 300 turns. (a) How many turns does the secondary have? (b) When idle, the answering machine uses a maximum power of \(5.0 \mathrm{W}\). What is the amplitude of the current drawn from the 170 -V line?

Short Answer

Expert verified
Answer: The number of turns in the secondary coil is approximately 14, and the amplitude of the current drawn from the 170V line is approximately 0.0294 A.

Step by step solution

01

Calculate the turns ratio of the transformer

The turns ratio of a transformer is given by the equation: \( turns \ ratio = \frac{V_{secondary}}{V_{primary}} = \frac{N_{secondary}}{N_{primary}} \) We can solve for the number of turns in the secondary coil: \( N_{secondary} = turns \ ratio \times N_{primary} \)
02

Calculate the number of turns in the secondary coil

We can now plug in the given values to find the number of turns in the secondary coil: \( N_{secondary} = \frac{7.8 \ V}{170 \ V} \times 300 \) \( N_{secondary} \approx 13.71 \) Since the number of turns must be a whole number, we can round it to the nearest integer: \( N_{secondary} \approx 14 \)
03

Calculate the amplitude of the current drawn from the 170V line

We know that the answering machine uses a maximum power of 5.0 W when idle. We can use this power and the primary voltage to find the current amplitude: \( P = V_{primary} \times I_{primary} \) Rearranging the equation to solve for the primary current amplitude, we get: \( I_{primary} = \frac{P}{V_{primary}} \) Now plug in the given values: \( I_{primary} = \frac{5.0 \ W}{170 \ V} \) \( I_{primary} \approx 0.0294 \ A \) Thus, the amplitude of the current drawn from the 170V line is approximately 0.0294 A.

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Most popular questions from this chapter

A flip coil is a device used to measure a magnetic field. A coil of radius $r, N\( turns, and electrical resistance \)R$ is initially perpendicular to a magnetic field of magnitude B. The coil is connected to a special kind of galvanometer that measures the total charge \(Q\) that flows through it. To measure the field, the flip coil is rapidly flipped upside down. (a) What is the change in magnetic flux through the coil in one flip? (b) If the time interval during which the coil is flipped is \(\Delta t,\) what is the average induced emf in the coil? (c) What is the average current that flows through the galvanometer? (d) What is the total charge \(Q\) in terms of \(r, N, R,\) and \(B ?\)
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 \(\left(N_{2} / N_{1}\right) ?\)
The time constant \(\tau\) for an \(L R\) circuit must be some combination of $L, R,\( and \)\mathscr{E} .$ (a) Write the units of each of these three quantities in terms of \(\mathrm{V}, \mathrm{A},\) and \(\mathrm{s}\) (b) Show that the only combination that has units of seconds is \(L / R\)
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 TV tube requires a 20.0 -kV-amplitude power supply. (a) What is the turns ratio of the transformer that raises the 170 -V-amplitude household voltage to \(20.0 \mathrm{kV} ?\) (b) If the tube draws 82 W of power, find the currents in the primary and secondary windings. Assume an ideal transformer.
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