An LC resonant circuit: If L is increased while keeping C fixed, what happens to the resonant frequency f0 = 1/(2π√(LC))?

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Multiple Choice

An LC resonant circuit: If L is increased while keeping C fixed, what happens to the resonant frequency f0 = 1/(2π√(LC))?

Explanation:
The resonant frequency of an LC circuit is determined by how quickly the circuit can exchange energy between the inductor and capacitor. It’s given by f0 = 1/(2π√(LC)). With the capacitor fixed, increasing the inductance makes the product LC larger, so the square root √(LC) grows. Since f0 is the reciprocal of that, the frequency drops. In other words, f0 is inversely proportional to the square root of L when C is fixed. If L is doubled, f0 decreases by a factor of √2; if L increases fourfold, f0 halves. This doesn’t eliminate resonance—the circuit still resonates, but at a lower frequency.

The resonant frequency of an LC circuit is determined by how quickly the circuit can exchange energy between the inductor and capacitor. It’s given by f0 = 1/(2π√(LC)). With the capacitor fixed, increasing the inductance makes the product LC larger, so the square root √(LC) grows. Since f0 is the reciprocal of that, the frequency drops. In other words, f0 is inversely proportional to the square root of L when C is fixed. If L is doubled, f0 decreases by a factor of √2; if L increases fourfold, f0 halves. This doesn’t eliminate resonance—the circuit still resonates, but at a lower frequency.

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