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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

The windings of an electromagnet have inductance \(L=8.0 \mathrm{H}\) and resistance \(R=2.0 \Omega . \quad A \quad 100.0-V \quad d c\) power supply is connected to the windings by closing switch \(S_{2} .\) (a) What is the current in the windings? (b) The electromagnet is to be shut off. Before disconnecting the power supply by opening switch \(S_{2},\) a shunt resistor with resistance \(20.0 \Omega\) is connected in parallel across the windings. Why is the shunt resistor needed? Why must it be connected before the power supply is disconnected? (c) What is the maximum power dissipated in the shunt resistor? The shunt resistor must be chosen so that it can handle at least this much power without damage. (d) When the power supply is disconnected by opening switch \(S_{2},\) how long does it take for the current in the windings to drop to \(0.10 \mathrm{A} ?\) (e) Would a larger shunt resistor dissipate the energy stored in the electromagnet faster? Explain.
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)
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?
When the armature of an ac generator rotates at $15.0 \mathrm{rad} / \mathrm{s},\( the amplitude of the induced emf is \)27.0 \mathrm{V}$ What is the amplitude of the induced emf when the armature rotates at $10.0 \mathrm{rad} / \mathrm{s} ?$ (tutorial: generator)
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