Question

A transistor-oscillator using a resonant circuit with an inductor L (of negligible resistance) and a capacitor C in series produce oscillations of frequency f. If L is doubled and C is changed to 4C, the frequency will be :

• A. \(\frac{\text{f}}{4}\)
• B. 8f
• C. \(\frac{\text{f}}{2\sqrt 2}\)
• D. 2f

Answer 1

The Correct Option is C

The question asks us to find the new frequency of oscillation when the inductance L is doubled and the capacitance C is changed to 4C in a resonant circuit. The original frequency of oscillation is given by:

f = \frac{1}{2\pi \sqrt{LC}}

With the changes, the new inductance becomes 2L and the new capacitance becomes 4C. The new frequency f_{\text{new}} can be calculated using the formula:

f_{\text{new}} = \frac{1}{2\pi \sqrt{(2L)(4C)}}

Let's simplify this expression step-by-step:

1. Substitute the new values into the formula: f_{\text{new}} = \frac{1}{2\pi \sqrt{8LC}}
2. Simplify \sqrt{8LC}: \sqrt{8LC} = \sqrt{8} \cdot \sqrt{LC}
3. We know \sqrt{8} = 2\sqrt{2}. Thus: f_{\text{new}} = \frac{1}{2\pi \cdot (2\sqrt{2}) \cdot \sqrt{LC}}
4. Notice that the original frequency f = \frac{1}{2\pi \sqrt{LC}}, so: f_{\text{new}} = \frac{1}{2\sqrt{2}} \cdot f or, f_{\text{new}} = \frac{f}{2\sqrt{2}}

Therefore, the new frequency of oscillation when L is doubled and C is increased to 4C is \frac{f}{2\sqrt{2}}, which matches the given correct answer.