Let's break the problem step by step: Given: - Resistance, \( R = 5.5\,\Omega \) - Capacitance, \( C = 2\,\mathrm{F} \) - Inductance, \( L = 6\,\mathrm{H} \) - The circuit is fed by a sinusoidal voltage of variable frequency (\( f \)), and the circuit is **inductive** (so \( \omega L > \frac{1}{\omega C} \)). - We are to **find the frequency** at which the **active power equals the reactive power in magnitude**. --- ## 1. **Impedance of Series RLC circuit** The total impedance: \[ Z = R + j(\omega L - \frac{1}{\omega C}) \] where \(\omega = 2\pi f\). --- ## 2. **Active and Reactive Power** Let the **rms voltage** be \( V \). - **Active power** (real power): \[ P = V_{\text{rms}} I_{\text{rms}} \cos\phi \] where \( \cos\phi \) is the power factor. - **Reactive power**: \[ Q = V_{\text{rms}} I_{\text{rms}} \sin\phi \] - **Apparent power:** \[ S = V_{\text{rms}} I_{\text{rms}} \] We are told \( |P| = |Q| \). --- ## 3. **Power Factor Angle** The phase angle \( \phi \) is given by: \[ \tan\phi = \frac{\text{Imaginary part of } Z}{\text{Real part of } Z} = \frac{\omega L - \frac{1}{\omega C}}{R} \] - \( \cos\phi = \frac{R}{|Z|} \) - \( \sin\phi = \frac{\omega L - \frac{1}{\omega C}}{|Z|} \) --- ## 4. **Equating Active and Reactive Power** \[ |P| = |Q| \implies |\cos\phi| = |\sin\phi| \] So, either \( \cos\phi = \sin\phi \) or \( \cos\phi = -\sin\phi \). But since the circuit is **inductive** (\( \omega L > \frac{1}{\omega C} \)), the angle \( \phi \) is **positive** and between 0 and 90°. So we use: \[ \cos\phi = \sin\phi \] \[ \Rightarrow \tan\phi = 1 \implies \phi = 45^\circ \] --- ## 5. **Expressing \( \tan\phi \) in terms of frequency** Recall: \[ \tan\phi = \frac{\omega L - \frac{1}{\omega C}}{R} \] Set \( \tan\phi = 1 \): \[ \frac{\omega L - \frac{1}{\omega C}}{R} = 1 \] \[ \omega L - \frac{1}{\omega C} = R \] --- ## 6. **Plug in Values** Given: - \( R = 5.5\,\Omega \) - \( L = 6\,\mathrm{H} \) - \( C = 2\,\mathrm{F} \) \[ \omega L - \frac{1}{\omega C} = 5.5 \] \[ \omega \cdot 6 - \frac{1}{\omega \cdot 2} = 5.5 \] \[ 6\omega - \frac{1}{2\omega} = 5.5 \] --- ## 7. **Solve for \( \omega \)** Multiply both sides by \( 2\omega \): \[ 2\omega \cdot 6\omega - 1 = 2\omega \cdot 5.5 \] \[ 12\omega^2 - 1 = 11\omega \] \[ 12\omega^2 - 11\omega - 1 = 0 \] This is a quadratic equation in \( \omega \): \[ 12\omega^2 - 11\omega - 1 = 0 \] --- ## 8. **Quadratic Solution** \[ \omega = \frac{11 \pm \sqrt{(-11)^2 - 4 \cdot 12 \cdot (-1)}}{2 \cdot 12} \] \[ = \frac{11 \pm \sqrt{121 + 48}}{24} \] \[ = \frac{11 \pm \sqrt{169}}{24} \] \[ = \frac{11 \pm 13}{24} \] So: - \( \omega_1 = \frac{24}{24} = 1 \) rad/s - \( \omega_2 = \frac{-2}{24} = -\frac{1}{12} \) rad/s (not physical, discard negative frequency) So, the only physical solution is: \[ \omega = 1\, \text{rad/s} \] --- ## 9. **Convert to Frequency** \[ f = \frac{\omega}{2\pi} = \frac{1}{2\pi} \approx 0.159\,\text{Hz} \] --- ## 10. **Check that the circuit is inductive at this frequency** \[ \omega L = 1 \times 6 = 6 \] \[ \frac{1}{\omega C} = \frac{1}{1 \times 2} = 0.5 \] \[ \omega L > \frac{1}{\omega C} \implies 6 > 0.5 \] Yes, the circuit is inductive at this frequency. --- ## **Final Answer** \[ \boxed{f = \frac{1}{2\pi} \approx 0.159\,\text{Hz}} \] **At a frequency of approximately \( \boxed{0.159\,\text{Hz}} \), the active (real) power equals the reactive power in magnitude in the given RLC series circuit.**