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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

Two solenoids, of \(N_{1}\) and \(N_{2}\) turns respectively, are wound on the same form. They have the same length \(L\) and radius \(r\) (a) What is the mutual inductance of these two solenoids? (b) If an ac current $$I_{1}(t)=I_{\mathrm{m}} \sin \omega t$$ flows in solenoid $$1\left(N_{1} \text { turns }\right)$$ write an expression for the total flux through solenoid \(2 .\) (c) What is the maximum induced emf in solenoid \(2 ?[\text { Hint: Refer to Eq. }(20-7) .]\)
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 alternator in an automobile generates an emf of amplitude $12.6 \mathrm{V}\( when the engine idles at \)1200 \mathrm{rpm}$. What is the amplitude of the emf when the car is being driven on the highway with the engine at 2800 rpm?
Compare the electric energy that can be stored in a capacitor to the magnetic energy that can be stored in an inductor of the same size (that is, the same volume). For the capacitor, assume that air is between the plates; the maximum electric field is then the breakdown strength of air, about $3 \mathrm{MV} / \mathrm{m} .$ The maximum magnetic field attainable in an ordinary solenoid with an air core is on the order of \(10 \mathrm{T}\)
The current in a \(0.080-\mathrm{H}\) solenoid increases from \(20.0 \mathrm{mA}\) to \(160.0 \mathrm{mA}\) in \(7.0 \mathrm{s} .\) Find the average emf in the solenoid during that time interval.
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