To solve the problem, we need to find the frequency at which the current in a series L-C-R circuit is maximum, given that the current is the same at two different frequencies (4 kHz and 9 kHz).
### Step-by-Step Solution:
1. **Understanding the Circuit**:
- In a series L-C-R circuit, the current depends on the impedance, which is affected by the resistance (R), inductive reactance (X_L), and capacitive reactance (X_C).
- The current is maximum at resonance frequency, where the inductive reactance equals the capacitive reactance (X_L = X_C).
2. **Given Frequencies**:
- We have two frequencies: \( f_1 = 4 \, \text{kHz} \) and \( f_2 = 9 \, \text{kHz} \).
- The current at these frequencies is the same.
3. **Using Reactance Relations**:
- The reactance of the capacitor is given by \( X_C = \frac{1}{\omega C} \) and the reactance of the inductor is \( X_L = \omega L \), where \( \omega = 2\pi f \).
- Since the current is the same at both frequencies, we can set up the equation:
\[
\frac{1}{\omega_1 C} - \omega_1 L = \frac{1}{\omega_2 C} - \omega_2 L
\]
- Here, \( \omega_1 = 2\pi f_1 \) and \( \omega_2 = 2\pi f_2 \).
4. **Simplifying the Equation**:
- Rearranging gives us:
\[
\frac{1}{\omega_1} - \frac{1}{\omega_2} = L(\omega_1 - \omega_2)
\]
- We can factor out \( \frac{1}{C} \) and \( L \) from both sides.
5. **Finding Resonance Frequency**:
- The resonance frequency \( \omega_0 \) is given by:
\[
\omega_0 = \sqrt{\omega_1 \omega_2}
\]
- Substitute \( \omega_1 = 2\pi \times 4 \, \text{kHz} \) and \( \omega_2 = 2\pi \times 9 \, \text{kHz} \).
6. **Calculating**:
- First, calculate \( \omega_1 \) and \( \omega_2 \):
\[
\omega_1 = 2\pi \times 4000 \approx 25132 \, \text{rad/s}
\]
\[
\omega_2 = 2\pi \times 9000 \approx 56549 \, \text{rad/s}
\]
- Now, find \( \omega_0 \):
\[
\omega_0 = \sqrt{(2\pi \times 4000)(2\pi \times 9000)} = \sqrt{(25132)(56549)} \approx 2\pi \times 6000 \, \text{rad/s}
\]
7. **Final Frequency**:
- Convert back to frequency:
\[
f_0 = \frac{\omega_0}{2\pi} \approx 6 \, \text{kHz}
\]
### Conclusion:
The current in the circuit is maximum at a frequency of **6 kHz**.
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