Discussion
The resonant frequency of the following tuned-collector oscillator is 6 MHz. If the value of the tuned circuit capacitor increased by 50%, the new resonant frequency of the oscillator will be around :
Answer: C
The new frequency is about 4·89 MHz.
The tank circuit fixes the frequency, \( f = \frac{1}{2\pi\sqrt{LC}} \), so \( f \propto 1/\sqrt{C} \). Increasing C by 50% multiplies it by 1.5:
\[ \begin{aligned} f_{2} &= \frac{f_{1}}{\sqrt{1.5}} = \frac{6}{1.2247} \\ &\approx 4.90\ \text{MHz} \end{aligned} \]
3 MHz would need the capacitance to be four times as large, 6·98 MHz is above the original frequency although a larger capacitor lowers it, and 5·89 MHz is far too small a change for a 50% increase in C.
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