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Chapter 21: Problem 69

An x-ray machine uses \(240 \mathrm{kV}\) rms at \(60.0 \mathrm{mA}\) ms when it is operating. If the power source is a \(420-\mathrm{V}\) rms line, (a) what must be the turns ratio of the transformer? (b) What is the rms current in the primary? (c) What is the average power used by the \(x\) -ray tube?

Short Answer

Expert verified
Answer: The turns ratio of the transformer is approximately 1:571.43, and the average power used by the x-ray tube is 7.2 kW.

Step by step solution

01

Understand the relationship between voltage and turns in a transformer

In a transformer, the turns ratio is directly proportional to the ratio of voltages across the primary and secondary windings. The equation is given by: $$ \frac{V_{primary}}{V_{secondary}} = \frac{N_{primary}}{N_{secondary}} $$ where \(V_{primary}\) and \(V_{secondary}\) are the primary and secondary voltages, and \(N_{primary}\) and \(N_{secondary}\) are the primary and secondary turns, respectively.
02

Calculate the turns ratio

Using the given values in the problem, we can calculate the turns ratio: $$ \frac{420V}{240kV} = \frac{N_{primary}}{N_{secondary}} $$ Here 240kV = 240,000V, so we can rewrite the equation as: $$ \frac{420}{240,000} = \frac{N_{primary}}{N_{secondary}} $$ Now, we can find the turns ratio. Divide both sides by \(420\): $$ \frac{N_{primary}}{N_{secondary}} = \frac{1}{571.43} $$ So the turns ratio is approximately \(1:571.43\).
03

Calculate the rms current in the primary

The power transferred to the secondary side is equal to the power transferred from the primary side for an ideal transformer. Thus, by using the formula for power as \(P = IV\) (current times voltage), we have: $$ I_{primary} V_{primary} = I_{secondary} V_{secondary} $$ Given values of the problem are \(V_{primary} = 420V\), \(V_{secondary} = 240,000V\), and \(I_{secondary} = 60.0 mA\). Replace the values and solve for \(I_{primary}\): $$ I_{primary} \times 420 = 60.0 \times 10^{-3} \times 240,000 $$ Now, find \(I_{primary}\) by dividing both sides by 420: $$ I_{primary} = \frac{60.0 \times 10^{-3} \times 240,000}{420} = 34.29A $$ The rms current in the primary is approximately \(34.29A\).
04

Calculate the average power used by the x-ray tube

To find the average power used by the x-ray tube, first, we need to understand that rms values are given for both voltage and current. The formula for the average power in terms of rms values is: $$ P_{avg} = \frac{1}{2} × V_{rms} \times I_{rms} $$ Here, \(V_{rms} = 240kV\) and \(I_{rms} = 60mA\). Replace the values and solve for \(P_{avg}\): $$ P_{avg} = \frac{1}{2} \times 240,000 \times 60 \times 10^{-3} = 7200 \mathrm{W} $$ The average power used by the x-ray tube is \(7200W\) or \(7.2kW\).

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Most popular questions from this chapter

For a particular \(R L C\) series circuit, the capacitive reactance is $12.0 \Omega,\( the inductive reactance is \)23.0 \Omega,$ and the maximum voltage across the \(25.0-\Omega\) resistor is \(8.00 \mathrm{V}\) (a) What is the impedance of the circuit? (b) What is the maximum voltage across this circuit?
In an RLC circuit, these three clements are connected in series: a resistor of \(20.0 \Omega,\) a \(35.0-\mathrm{mH}\) inductor, and a 50.0 - \(\mu\) F capacitor. The ac source of the circuit has an rms voltage of \(100.0 \mathrm{V}\) and an angular frequency of $1.0 \times 10^{3} \mathrm{rad} / \mathrm{s} .$ Find (a) the reactances of the capacitor and inductor, (b) the impedance, (c) the rms current, (d) the current amplitude, (e) the phase angle, and (f) the rims voltages across each of the circuit elements. (g) Does the current lead or lag the voltage? (h) Draw a phasor diagram.
A 40.0 -mH inductor, with internal resistance of \(30.0 \Omega\) is connected to an ac source $$\varepsilon(t)=(286 \mathrm{V}) \sin [(390 \mathrm{rad} / \mathrm{s}) t]$$ (a) What is the impedance of the inductor in the circuit? (b) What are the peak and rms voltages across the inductor (including the internal resistance)? (c) What is the peak current in the circuit? (d) What is the average power dissipated in the circuit? (e) Write an expression for the current through the inductor as a function of time.
The FM radio band is broadcast between \(88 \mathrm{MHz}\) and $108 \mathrm{MHz}$. What range of capacitors must be used to tune in these signals if an inductor of \(3.00 \mu \mathrm{H}\) is used?
(a) What is the reactance of a \(10.0-\mathrm{mH}\) inductor at the frequency \(f=250.0 \mathrm{Hz} ?\) (b) What is the impedance of a series combination of the \(10.0-\mathrm{mH}\) inductor and a \(10.0-\Omega\) resistor at $250.0 \mathrm{Hz} ?$ (c) What is the maximum current through the same circuit when the ac voltage source has a peak value of \(1.00 \mathrm{V} ?\) (d) By what angle does the current lag the voltage in the circuit?
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