Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 24 - Alternating-Current Circuits - Problems and Conceptual Exercises - Page 870: 70

Answer

(a) The resonance frequency will decrease. (b) $77.5Hz$

Work Step by Step

(a) As we know, $f=\frac{1}{2\pi \sqrt{LC}}$. This equation shows us that resonance frequency is inversely proportional to L and C. That is why the resonance frequency will decrease, if we double L and C. (b) We know that $f=\frac{1}{2\pi \sqrt{(2L)(2C)}}$ $\implies f=\frac{1}{2\pi \sqrt{4LC}}$ $\implies f=\frac{1}{2}(\frac{1}{2\pi \sqrt{LC}})$ $f=\frac{1}{2}f_{\circ}$ We plug in the known values to obtain: $f=\frac{1}{2}(155)=77.5Hzz$
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