In a series resonant L–C–R circuit, if L is increased by 25% and C is decreased by 20%, then the resonant frequency will –

Options

A. Increase by `10%`
B. Decrease by `10%`
C. Remain unchanged
D. Increase by `2.5%`

Answer

To solve the problem regarding the change in resonant frequency of a series resonant L-C-R circuit when the inductance (L) is increased by 25% and the capacitance (C) is decreased by 20%, we can follow these steps:

### Step 1: Understand the formula for resonant frequency
The resonant frequency (f) of a series L-C-R circuit is given by the formula:
\[ f = \frac{1}{2\pi\sqrt{LC}} \]
Where:
- \( f \) is the resonant frequency,
- \( L \) is the inductance,
- \( C \) is the capacitance.

### Step 2: Calculate the new values of L and C
1. **Increase L by 25%**:
\[ L' = L + 0.25L = 1.25L \]

2. **Decrease C by 20%**:
\[ C' = C - 0.20C = 0.80C \]

### Step 3: Substitute the new values into the resonant frequency formula
Now we can substitute the new values of L and C into the resonant frequency formula:
\[ f' = \frac{1}{2\pi\sqrt{L'C'}} = \frac{1}{2\pi\sqrt{(1.25L)(0.80C)}} \]

### Step 4: Simplify the expression
We can simplify the expression for the new resonant frequency:
\[ f' = \frac{1}{2\pi\sqrt{1.25 \cdot 0.80 \cdot LC}} = \frac{1}{2\pi\sqrt{1.00 \cdot LC}} = \frac{1}{2\pi\sqrt{LC}} \]
This shows that:
\[ f' = f \]

### Step 5: Conclusion
Since the new resonant frequency \( f' \) is equal to the original resonant frequency \( f \), we conclude that the resonant frequency remains unchanged.

### Final Answer
The resonant frequency will remain unchanged.

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