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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

The voltage across an inductor and the current through the inductor are related by \(v_{\mathrm{L}}=L \Delta i / \Delta t .\) Suppose that $i(t)=I \sin \omega t .\( (a) Write an expression for \)v_{\mathrm{L}}(t) .$ [Hint: Use one of the relationships of Eq. \((20-7) .]\) (b) From your expression for \(v_{\mathrm{L}}(t),\) show that the reactance of the inductor is \(X_{\mathrm{L}}=\omega L_{\mathrm{r}}\) (c) Sketch graphs of \(i(t)\) and \(v_{\mathrm{L}}(t)\) on the same axes. What is the phase difference? Which one leads?
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