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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 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?
A coil of wire is connected to an ideal \(6.00-\mathrm{V}\) battery at \(t=0 .\) At \(t=10.0 \mathrm{ms},\) the current in the coil is \(204 \mathrm{mA}\) One minute later, the current is 273 mA. Find the resistance and inductance of the coil. [Hint: Sketch \(I(t) .]\)
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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.
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.
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