Question A8.10: The frequency of resonance, f, of a tuned circuit (see Fig. ...
The frequency of resonance, f, of a tuned circuit (see Fig. A8.4) is given by the relationship:
f=\frac{1}{2 \pi \sqrt{L C}}If a tuned circuit is to be resonant at 6.25 MHz and C = 100 pF, determine the value of inductance, L .
Here we know that:
C = 100 pF = 100 × 10^{-12} F
f = 6.25 MHz = 6.25 × 10^{6} Hz
π = 3.142 (use the ‘π’ button on your calculator!)
L = ?
First we will re-arrange the formula in order to make L the subject:
Squaring both sides gives:
f^{2}=\left(\frac{1}{2 \pi \sqrt{L C}}\right)^{2}=\frac{1^{2}}{2^{2} \pi^{2}(\sqrt{L C})^{2}}=\frac{1}{4 \pi^{2} L C}Re-arranging gives:
L=\frac{1}{4 \pi^{2} f^{2} C}We can now replace f, C and π by the values that we know:
L=\frac{1}{4 \times 3.142^{2} \times\left(6.25 \times 10^{6}\right)^{2} \times 100 \times 10^{-12}} H\begin{aligned}L &=\frac{1}{39.49 \times 39.06 \times 10^{12} \times 100 \times 10^{-12}} H \\&=\frac{1}{154,248}=\frac{1}{1.54248 \times 10^{5}}=\frac{1}{1.54248} \times 10^{-5} \\ &=0.648 \times 10^{-5}=6.48 \times 10^{-6} H = 6 . 8 \mu H \end{aligned}
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