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Chapter 20: Problem 37

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?

Short Answer

Expert verified
Answer: The voltage amplitude across the secondary coil is approximately 28.33V, and the current amplitude in the primary coil is approximately 0.53A.

Step by step solution

01

Determine the Voltage-Turns Ratio

To calculate the voltage across the secondary coil, we first need to find the turns ratio between the primary and secondary coils. The turns ratio is the ratio of the number of turns on the primary coil to the number of turns on the secondary coil: $$ Turns \ Ratio = \frac{N_{primary}}{N_{secondary}} $$ With \(N_{primary} = 1800\) turns and \(N_{secondary} = 300\) turns: $$ Turns \ Ratio = \frac{1800}{300} = 6 $$
02

Calculate the Voltage Amplitude Across the Secondary Coil

Now, using the turns ratio, we can find the voltage amplitude across the secondary coil using the voltage-turns ratio formula: $$ V_{secondary} = \frac{V_{primary}}{Turns \ Ratio} $$ With \(V_{primary} = 170 \mathrm{V}\) and \(Turns \ Ratio = 6\): $$ V_{secondary} = \frac{170 \mathrm{V}}{6} \cong 28.33 \mathrm{V} $$ So, the voltage amplitude across the secondary coil is approximately \(28.33 \mathrm{V}\).
03

Calculate the Current-Turns Ratio

Now, let's determine the current-turns ratio, which is the inverse of the turns ratio: $$ Current \ Ratio = \frac{1}{Turns \ Ratio} = \frac{1}{6} $$
04

Calculate the Current Amplitude in the Primary Coil

We can now use the current-turns ratio to find the current amplitude in the primary coil. The current in the secondary coil is given as \(I_{secondary} = 3.2 \mathrm{A}\). Using the current-turns ratio formula: $$ I_{primary} = Current \ Ratio \times I_{secondary} $$ With \(Current \ Ratio = \frac{1}{6}\) and \(I_{secondary} = 3.2 \mathrm{A}\): $$ I_{primary} = \frac{1}{6} \times 3.2\mathrm{A} \cong 0.53 \mathrm{A} $$ So, the current amplitude in the primary coil is approximately \(0.53 \mathrm{A}\). In conclusion, the voltage amplitude across the secondary coil is approximately \(28.33 \mathrm{V}\), and the current amplitude in the primary coil is approximately \(0.53 \mathrm{A}\).

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Most popular questions from this chapter

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.

An ideal solenoid has length \(\ell\). If the windings are compressed so that the length of the solenoid is reduced to \(0.50 \ell,\) what happens to the inductance of the solenoid?
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?
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 \(\left(N_{2} / N_{1}\right) ?\)
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