Question Detail
Question
The resonant frequency of a circuit is 1000 KHz. If each of capacitance and inductance in the circuit is reduced by 5, the resonant frequency will be:
Correct Answer
Option D is the correct answer.
Detailed Explanation
Here are the steps to solve this type of problem step-by-step:
Step 1: Recall the Resonant Frequency Formula
The resonant frequency $f$ of an $LC$ circuit depends on inductance $L$ and capacitance $C$:
$$f = \frac{1}{2\pi \sqrt{L C}}$$
Step 2: Express the New Values in Terms of the Old Ones
- Both $L$ and $C$ are reduced by a factor of 5: $$L' = \frac{L}{5}$$ $$C' = \frac{C}{5}$$
- Multiply inside the square root: $\frac{L}{5} \times \frac{C}{5} = \frac{LC}{25}$
- Take the square root of the denominator: $\sqrt{\frac{LC}{25}} = \frac{\sqrt{LC}}{5}$
- Multiply your initial resonant frequency ($1000\text{ kHz}$) by $5$ to get the new frequency!
Hint
Recall the formula for resonant frequency. Consider how changes in L and C affect the frequency. The wording 'reduced by 5' likely implies a change by a factor, which leads to a division of the original frequency to match the options.