Question 494 of 949
In a series RLC circuit, what effect does increasing the inductance (L) have on the resonant frequency (fâ) 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.