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The resonant frequency of a circuit is f. If the capacitance is made 4 times the initial value, then the resonant frequency will becomes
- a)f/2
- b)2f
- c)f
- d)f/4
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The resonant frequency of a circuit is f. If the capacitance is made 4...
f = 2π/√LC
∴ f' = 2π/√4LC = 1/2 x 2π/√LC = f/2
∴ f' = 2π/√4LC = 1/2 x 2π/√LC = f/2
Most Upvoted Answer
The resonant frequency of a circuit is f. If the capacitance is made 4...
Understanding Resonant Frequency
The resonant frequency of a circuit, particularly in an LC circuit, is defined by the formula:
\[ f = \frac{1}{2\pi\sqrt{LC}} \]
where:
- \( f \) is the resonant frequency,
- \( L \) is the inductance,
- \( C \) is the capacitance.
Effect of Increasing Capacitance
When the capacitance \( C \) is increased to four times its initial value, we can denote the new capacitance as \( C' = 4C \). Substituting this into the formula for resonant frequency gives:
\[ f' = \frac{1}{2\pi\sqrt{L \cdot 4C}} \]
This can be simplified as:
\[ f' = \frac{1}{2\pi\sqrt{4LC}} \]
Calculating the New Frequency
Since \( \sqrt{4} = 2 \), we can write:
\[ f' = \frac{1}{2\pi \cdot 2\sqrt{LC}} = \frac{1}{2} \cdot \frac{1}{2\pi\sqrt{LC}} \]
Thus, we have:
\[ f' = \frac{f}{2} \]
Conclusion
Therefore, increasing the capacitance by a factor of four results in the resonant frequency being halved. The correct answer is:
Option A: \( \frac{f}{2} \)
The resonant frequency of a circuit, particularly in an LC circuit, is defined by the formula:
\[ f = \frac{1}{2\pi\sqrt{LC}} \]
where:
- \( f \) is the resonant frequency,
- \( L \) is the inductance,
- \( C \) is the capacitance.
Effect of Increasing Capacitance
When the capacitance \( C \) is increased to four times its initial value, we can denote the new capacitance as \( C' = 4C \). Substituting this into the formula for resonant frequency gives:
\[ f' = \frac{1}{2\pi\sqrt{L \cdot 4C}} \]
This can be simplified as:
\[ f' = \frac{1}{2\pi\sqrt{4LC}} \]
Calculating the New Frequency
Since \( \sqrt{4} = 2 \), we can write:
\[ f' = \frac{1}{2\pi \cdot 2\sqrt{LC}} = \frac{1}{2} \cdot \frac{1}{2\pi\sqrt{LC}} \]
Thus, we have:
\[ f' = \frac{f}{2} \]
Conclusion
Therefore, increasing the capacitance by a factor of four results in the resonant frequency being halved. The correct answer is:
Option A: \( \frac{f}{2} \)
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