Engineering Frequency of operation and Q factor of a coil

[cnh1995]

Homework Statement

Select the correct option.

Relevant Equations

Q= X/R
X=wL
P=I*I*R

My reasoning:
Initial Q factor Q₁= X₁/R
After doubling the frequency, reactance also doubles.
So, Q2= 2(X1)/R.

Hence, Q factor doubles (option a is out).

Now, since it is mentioned that the coil has high Q factor, I assumed X >> R and hence, total impedance is approximately equal to the reactance X.
So, as frequency doubles, impedance also doubles, and current is halved.
Therefore, P decreases by 4 times.
So, d) should be the correct option.

But the book gives a) as the correct one.
Is there a mistake in my solution or is the book wrong?

 

[phyzguy]

I think your analysis is correct and you're right that (d) is correct. Any way to double check the book or check with your professor?

 

[cnh1995]

Thanks for the confirmation.

Any way to double check the book or check with your professor?

No, I am preparing for an exam by myself. Maybe I'll write to the publisher next week.