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Question 494 of 949

In a series RLC circuit, what effect does increasing the inductance (L) have on the resonant frequency (f₀) of the circuit?

  • Increases the resonant frequency
  • Decreases the resonant frequency
  • Has no effect on the resonant frequency
  • Doubles the resonant frequency

Correct Answer: B

Explanation
### Correct Option: B. Decreases the resonant frequency ### Detailed Explanation: In a series RLC (Resistor, Inductor, Capacitor) circuit, the resonant frequency (\(f_0\)) is the frequency at which the circuit naturally oscillates when not driven by an external source. The formula for the resonant frequency in a series RLC circuit is given by: \[ f_0 = \frac{1}{2\pi\sqrt{LC}} \] Where: - \(f_0\) is the resonant frequency in hertz (Hz), - \(L\) is the inductance in henries (H), - \(C\) is the capacitance in farads (F). #### Step-by-Step Analysis: 1. **Understanding the Formula**: The resonant frequency is inversely proportional to the square root of the product of inductance (\(L\)) and capacitance (\(C\)). This means that as \(L\) increases, the term \(\sqrt{LC}\) also increases. 2. **Effect of Increasing Inductance**: If we increase the inductance \(L\), the value of \(\sqrt{LC}\) becomes larger. Since \(f_0\) is inversely related to \(\sqrt{LC}\), an increase in \(L\) leads to a decrease in \(f_0\). 3. **Mathematical Interpretation**: - If \(L\) is doubled, for example, the new resonant frequency \(f_0'\) can be calculated as: \[ f_0' = \frac{1}{2\pi\sqrt{(2L)C}} = \frac{1}{2\pi\sqrt{2}\sqrt{LC}} = \frac{f_0}{\sqrt{2}} \] This shows that the resonant frequency decreases when \(L\) increases. 4. **Physical Interpretation**: In a physical sense, increasing the inductance means that the circuit can store more magnetic energy. This increased energy storage leads to a slower oscillation frequency, hence a lower resonant frequency. ### Why Other Options Are Incorrect: - **Option A: Increases the resonant frequency**: This is incorrect because increasing \(L\) increases the denominator in the formula for \(f_0\), which results in a smaller value for \(f_0\). - **Option C: Has no effect on the resonant frequency**: This is incorrect because \(L\) directly influences the resonant frequency. An increase in \(L\) will always affect \(f_0\). - **Option D: Doubles the resonant frequency**: This is incorrect as well. Doubling \(L\) does not double \(f_0\); instead, it reduces \(f_0\) by a factor of \(\sqrt{2}\), as shown in the mathematical interpretation. ### Common Pitfalls: - **Confusing Inductance and Capacitance**: Remember that inductance and capacitance have opposite effects on resonant frequency. Increasing capacitance would increase \(f_0\), while increasing inductance decreases it. - **Misunderstanding the Formula**: Ensure you understand how the formula is derived and how each component affects the resonant frequency. ### Revision Summary: - The resonant frequency \(f_0\) in a series RLC circuit is given by \(f_0 = \frac{1}{2\pi\sqrt{LC}}\). - Increasing the inductance \(L\) decreases the resonant frequency \(f_0\). - The relationship is inversely proportional; as \(L\) increases, \(\sqrt{LC}\) increases, leading to a lower \(f_0\). - Always differentiate the effects of inductance and capacitance on resonant frequency to avoid confusion.
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