1 pf the data also show that the imaginary part of

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ; ω = 2π 19 × 106 s−1 1 + ω 2 C 2 R2 which is easily solved for C , yielding the result C ￿ 25.1 pF. The data also show that the imaginary part of the reactance is approximately Xm = 810Ω at f = 18 MHz. Hence Z = j ωL + 810 ￿ ω L − ω CR2 ; ω = 2π 18 × 106 s−1 1 + ω 2 C 2 R2 which is easily solved for L, yielding L ￿ 9.3 µH. 3 3-6 Solution a. The impedance of the inductor is ZL = or Z=R 1 ||(R + j ω L) j ωC 1 (1 − ω 2 LC )2 + ω 2 R2 C 2 + j ωL (1 − ω 2 LC ) − R2 C/L . (1 − ω 2 LC )2 + ω 2 R2 C 2 The resonant frequency is found by setting the imaginary part of the impedance to zero, and solving for ω . The result is ￿ 1 R2 ωp = − 2. LC L ￿ 2 1 1 L Clearly, ωp ￿ √LC only if R2 ￿ LC , or when R is small...
View Full Document

Ask a homework question - tutors are online