A `10` ohm resistance, `5 mH` coil and `10mu F` capacitor are joined in series. When a suitable frequency alternating current source is joined to this combination, the circuit resonates. If the resistance is halved, the resonance frequency

Options

A. is halved
B. is doubled
C. remains unchanged
D. in quadrupled

Explanation

To solve the problem step by step, we will first identify the relevant formulas and then apply them to find the resonance frequency of the circuit.

### Step 1: Identify the components and their values
We have:
- Resistance (R) = 10 ohms
- Inductance (L) = 5 mH = 5 × 10^(-3) H
- Capacitance (C) = 10 µF = 10 × 10^(-6) F

### Step 2: Write the formula for resonance frequency
The resonance frequency (f₀) of an RLC circuit is given by the formula:
\[ f₀ = \frac{1}{2\pi\sqrt{LC}} \]

### Step 3: Substitute the values into the formula
Now we will substitute the values of L and C into the formula:
\[ f₀ = \frac{1}{2\pi\sqrt{(5 \times 10^{-3}) \times (10 \times 10^{-6})}} \]

### Step 4: Calculate the product of L and C
Calculating the product:
\[ LC = (5 \times 10^{-3}) \times (10 \times 10^{-6}) = 5 \times 10^{-8} \]

### Step 5: Substitute LC back into the formula
Now substituting LC back into the formula for f₀:
\[ f₀ = \frac{1}{2\pi\sqrt{5 \times 10^{-8}}} \]

### Step 6: Calculate the square root
Calculating the square root:
\[ \sqrt{5 \times 10^{-8}} = \sqrt{5} \times 10^{-4} \]
(approximately \(\sqrt{5} \approx 2.236\))

### Step 7: Substitute the square root back into the formula
Now substituting this back into the formula:
\[ f₀ = \frac{1}{2\pi(2.236 \times 10^{-4})} \]

### Step 8: Calculate the final frequency
Calculating the frequency:
\[ f₀ \approx \frac{1}{2 \times 3.14 \times (2.236 \times 10^{-4})} \]
\[ f₀ \approx \frac{1}{1.403 \times 10^{-3}} \]
\[ f₀ \approx 712.4 \text{ Hz} \]

### Step 9: Analyze the effect of halving the resistance
The problem states that if the resistance is halved (from 10 ohms to 5 ohms), we need to determine the effect on the resonance frequency.

### Conclusion
Since the resonance frequency is independent of the resistance in an RLC circuit, halving the resistance does not change the resonance frequency. Therefore, the resonance frequency remains the same.

### Final Answer
The resonance frequency remains unchanged when the resistance is halved.

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