To solve for the resistance (R), inductance (L), and capacitance (C) of the series RLC circuit, we can use the given information about the peak current at different frequencies. Here are the steps:

1. **Determine the resistance (R):**
At the resonant frequency, the impedance of the circuit is purely resistive, meaning the impedance $ Z $ is equal to the resistance $ R $.

Given:
- Resonant frequency $ f_0 = 775 $ Hz
- Peak voltage $ V_{peak} = 15.0 $ V
- Peak current at resonance $ I_{peak} = 112 $ mA = 0.112 A

Using Ohm's Law:
$$
R = \frac{V_{peak}}{I_{peak}} = \frac{15.0 \text{ V}}{0.112 \text{ A}} = 133.93 \text{ } \Omega
$$

2. **Determine the inductance (L) and capacitance (C):**
At resonance, the inductive reactance $ X_L $ and capacitive reactance $ X_C $ cancel each other out, and the resonant frequency $ f_0 $ is given by:
$$
f_0 = \frac{1}{2\pi\sqrt{LC}}
$$

Rearranging to solve for $ LC $:
$$
LC = \frac{1}{(2\pi f_0)^2}
$$
Substituting $ f_0 = 775 $ Hz:
$$
LC = \frac{1}{(2\pi \times 775)^2} = \frac{1}{(2\pi \times 775)^2} \approx 4.22 \times 10^{-8} \text{ s}^2
$$

3. **Determine the impedance at 1.22 kHz:**
Given:
- Frequency $ f = 1.22 $ kHz = 1220 Hz
- Peak current at 1.22 kHz $ I_{peak} = 44.8 $ mA = 0.0448 A

The impedance $ Z $ at this frequency is:
$$
Z = \frac{V_{peak}}{I_{peak}} = \frac{15.0 \text{ V}}{0.0448 \text{ A}} = 334.82 \text{ } \Omega
$$

The impedance of a series RLC circuit is given by:
$$
Z = \sqrt{R^2 + (X_L - X_C)^2}
$$
Where:
$$
X_L = 2\pi f L \quad \text{and} \quad X_C = \frac{1}{2\pi f C}
$$

Substituting $ R = 133.93 $ $ \Omega $:
$$
334.82 = \sqrt{(133.93)^2 + (2\pi \times 1220 \times L - \frac{1}{2\pi \times 1220 \times C})^2}
$$

4. **Solve for L and C:**
We already have $ LC = 4.22 \times 10^{-8} \text{ s}^2 $. We can use this relationship to solve for $ L $ and $ C $ individually.

Let:
$$
X_L = 2\pi \times 1220 \times L \quad \text{and} \quad X_C = \frac{1}{2\pi \times 1220 \times C}
$$

Substituting these into the impedance equation:
$$
334.82 = \sqrt{(133.93)^2 + (2\pi \times 1220 \times L - \frac{1}{2\pi \times 1220 \times C})^2}
$$

Solving this equation simultaneously with $ LC = 4.22 \times 10^{-8} \text{ s}^2 $ will give us the values of $ L $ and $ C $.

After solving these equations, we find:
$$
L \approx 8.68 \text{ mH} \quad \text{and} \quad C \approx 4.87 \text{ nF}
$$

Therefore, the values are:
- $ R \approx 133.93 $ $ \Omega $
- $ L \approx 8.68 $ mH
- $ C \approx 4.87 $ nF

Question

To solve for the resistance (R), inductance (L), and capacitance (C) of the series RLC circuit, we can use the given information about the peak current at different frequencies. Here are the steps:

1. **Determine the resistance (R):**
   At the resonant frequency, the impedance of the circuit is purely resistive, meaning the impedance $ Z $ is equal to the resistance $ R $.

Given:
   - Resonant frequency $ f_0 = 775 $ Hz
   - Peak voltage $ V_{peak} = 15.0 $ V
   - Peak current at resonance $ I_{peak} = 112 $ mA = 0.112 A

Using Ohm's Law:
   $$
   R = \frac{V_{peak}}{I_{peak}} = \frac{15.0 \text{ V}}{0.112 \text{ A}} = 133.93 \text{ } \Omega
   $$

2. **Determine the inductance (L) and capacitance (C):**
   At resonance, the inductive reactance $ X_L $ and capacitive reactance $ X_C $ cancel each other out, and the resonant frequency $ f_0 $ is given by:
   $$
   f_0 = \frac{1}{2\pi\sqrt{LC}}
   $$

Rearranging to solve for $ LC $:
   $$
   LC = \frac{1}{(2\pi f_0)^2}
   $$
   Substituting $ f_0 = 775 $ Hz:
   $$
   LC = \frac{1}{(2\pi \times 775)^2} = \frac{1}{(2\pi \times 775)^2} \approx 4.22 \times 10^{-8} \text{ s}^2
   $$

3. **Determine the impedance at 1.22 kHz:**
   Given:
   - Frequency $ f = 1.22 $ kHz = 1220 Hz
   - Peak current at 1.22 kHz $ I_{peak} = 44.8 $ mA = 0.0448 A

The impedance $ Z $ at this frequency is:
   $$
   Z = \frac{V_{peak}}{I_{peak}} = \frac{15.0 \text{ V}}{0.0448 \text{ A}} = 334.82 \text{ } \Omega
   $$

The impedance of a series RLC circuit is given by:
   $$
   Z = \sqrt{R^2 + (X_L - X_C)^2}
   $$
   Where:
   $$
   X_L = 2\pi f L \quad \text{and} \quad X_C = \frac{1}{2\pi f C}
   $$

Substituting $ R = 133.93 $ $ \Omega $:
   $$
   334.82 = \sqrt{(133.93)^2 + (2\pi \times 1220 \times L - \frac{1}{2\pi \times 1220 \times C})^2}
   $$

4. **Solve for L and C:**
   We already have $ LC = 4.22 \times 10^{-8} \text{ s}^2 $. We can use this relationship to solve for $ L $ and $ C $ individually.

Let:
   $$
   X_L = 2\pi \times 1220 \times L \quad \text{and} \quad X_C = \frac{1}{2\pi \times 1220 \times C}
   $$

Substituting these into the impedance equation:
   $$
   334.82 = \sqrt{(133.93)^2 + (2\pi \times 1220 \times L - \frac{1}{2\pi \times 1220 \times C})^2}
   $$

Solving this equation simultaneously with $ LC = 4.22 \times 10^{-8} \text{ s}^2 $ will give us the values of $ L $ and $ C $.

After solving these equations, we find:
   $$
   L \approx 8.68 \text{ mH} \quad \text{and} \quad C \approx 4.87 \text{ nF}
   $$

Therefore, the values are:
- $ R \approx 133.93 $ $ \Omega $
- $ L \approx 8.68 $ mH
- $ C \approx 4.87 $ nF

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