A series RLC circuit has power factor 0.764 and impedance 182 Q at 442 Hz. (a) What's the resistance? (b) If the inductance is 25.0 mH, what's the resonant frequencv? (1 point)

Sure, let's solve the problem step by step.

Part (a): Finding the Resistance

Given:

• Power factor (pf) = 0.764
• Impedance (Z) = 182 Ω

The power factor is the cosine of the phase angle (θ) between the voltage and the current. Therefore, $\cos(θ) = \text{pf}$ $\cos(θ) = 0.764$

The impedance (Z) in a series RLC circuit is given by: $Z = \sqrt{R^2 + (X_L - X_C)^2}$

Where:

• $R$ is the resistance
• $X_L$ is the inductive reactance
• $X_C$ is the capacitive reactance

The power factor is also given by: $\text{pf} = \frac{R}{Z}$

Rearranging to solve for $R$: $R = \text{pf} \times Z$ $R = 0.764 \times 182$ $R \approx 139.048 \, \Omega$

So, the resistance $R$ is approximately $139.048 \, \Omega$.

Part (b): Finding the Resonant Frequency

Given:

• Inductance (L) = 25.0 mH = 25.0 × 10^-3 H

The resonant frequency $f_0$ in a series RLC circuit is given by: $f_0 = \frac{1}{2\pi\sqrt{LC}}$

We need to find the capacitance $C$. However, the problem does not provide $C$ directly. But we can use the given frequency (442 Hz) to find $C$ if we assume the circuit is at resonance at this frequency. At resonance, $X_L = X_C$.

The inductive reactance $X_L$ is given by: $X_L = 2\pi f L$ $X_L = 2\pi \times 442 \times 25.0 \times 10^{-3}$ $X_L \approx 69.5 \, \Omega$

At resonance, $X_L = X_C$, so: $X_C = \frac{1}{2\pi f C}$ $69.5 = \frac{1}{2\pi \times 442 \times C}$

Solving for $C$: $C = \frac{1}{2\pi \times 442 \times 69.5}$ $C \approx 5.2 \times 10^{-7} \, F$

Now, using this $C$ to find the resonant frequency $f_0$: $f_0 = \frac{1}{2\pi\sqrt{LC}}$ $f_0 = \frac{1}{2\pi\sqrt{25.0 \times 10^{-3} \times 5.2 \times 10^{-7}}}$ $f_0 \approx 442 \, \text{Hz}$

So, the resonant frequency $f_0$ is approximately 442 Hz, which matches the given frequency, confirming our calculations.