Let’s solve the problem step by step. Given: - Peak Voltage, \( V_0 = 54 \) V - Peak Current, \( I_0 = 3 \) A - Capacitance, \( C = 15 \mu F = 15 \times 10^{-6} \) F - Frequency, \( f = 100 \) Hz - The current leads voltage by \( \phi = 40^\circ \) ### **Step 1: Impedance Calculation** The peak current in an RLC circuit is: \[ I_0 = \frac{V_0}{Z} \] So, \[ Z = \frac{V_0}{I_0} = \frac{54}{3} = 18\; \Omega \] --- ### **Step 2: Phase Angle Relation** For a series RLC circuit, \[ \tan \phi = \frac{X_C - X_L}{R} \] Given \(\phi = 40^\circ\) (current leads voltage), so: \[ \tan 40^\circ = \frac{X_C - X_L}{R} \] \[ \tan 40^\circ \approx 0.8391 \] --- ### **Step 3: Express \(X_C\) and \(X_L\)** \[ X_C = \frac{1}{\omega C} \] \[ X_L = \omega L \] where \( \omega = 2\pi f = 2\pi \times 100 = 628.32 \) rad/s. So, \[ X_C = \frac{1}{628.32 \times 15 \times 10^{-6}} = \frac{1}{0.0094248} \approx 106.14~\Omega \] --- ### **Step 4: Impedance in Terms of R, \(X_L\), \(X_C\)** \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] But since the current **leads**, \(X_C > X_L\) (circuit is more capacitive). \[ Z = \sqrt{R^2 + (X_L - X_C)^2} = 18 \] --- ### **Step 5: Set Up Equations** Let \(X_L - X_C = -A\) (since \(X_C > X_L\)), so \(A = X_C - X_L\). From above, \[ \tan 40^\circ = \frac{A}{R} \implies 0.8391 = \frac{A}{R} \implies A = 0.8391 R \] And from the impedance equation: \[ Z^2 = R^2 + (X_L - X_C)^2 = R^2 + (-A)^2 = R^2 + A^2 = 18^2 \] \[ R^2 + (0.8391 R)^2 = 324 \] \[ R^2 + 0.7041 R^2 = 324 \] \[ 1.7041 R^2 = 324 \] \[ R^2 = \frac{324}{1.7041} = 190.21 \] \[ R = \sqrt{190.21} = 13.79~\Omega \] --- ### **Step 6: Calculate \(A = X_C - X_L\)** \[ A = 0.8391 \times 13.79 = 11.57~\Omega \] So, \[ X_C - X_L = 11.57 \] \[ X_L = X_C - 11.57 = 106.14 - 11.57 = 94.57~\Omega \] --- ### **Step 7: Calculate Inductance \(L\)** \[ X_L = \omega L \implies L = \frac{X_L}{\omega} = \frac{94.57}{628.32} = 0.1505~\text{H} \] --- ## **Final Answers:** \[ \boxed{ \begin{aligned} R &= 13.8~\Omega \\ L &= 0.151~\text{H} \end{aligned} } \] (Rounded to three significant digits.)

Let's carefully work through this problem step-by-step, providing detailed explanations and calculations. --- ### **Given Data:** - Peak voltage, \( V_0 = 54\, \text{V} \) - Peak current, \( I_0 = 3\, \text{A} \) - Capacitance, \( C = 15\, \mu F = 15 \times 10^{-6}\, \text{F} \) - Frequency, \( f = 100\, \text{Hz} \) - Phase angle between current and voltage, \( \phi = 40^\circ \) (current leads voltage) --- ### **Step 1: Find the total impedance \( Z \)** The peak voltage and current relate as: \[ Z = \frac{V_0}{I_0} = \frac{54}{3} = 18\, \Omega \] --- ### **Step 2: Determine the phase angle relation** In an RLC circuit, the phase angle \( \phi \) is related to the impedance components by: \[ \tan \phi = \frac{X_L - X_C}{R} \] Since the current **leads** the voltage by \( 40^\circ \), the circuit is more capacitive, implying: \[ X_C > X_L \] Thus, \[ \tan 40^\circ = \frac{X_C - X_L}{R} \] Calculate \( \tan 40^\circ \): \[ \tan 40^\circ \approx 0.8391 \] --- ### **Step 3: Calculate \( X_C \) and \( X_L \)** - **Capacitive reactance**: \[ X_C = \frac{1}{\omega C} \] where \[ \omega = 2 \pi f = 2 \pi \times 100 \approx 628.32\, \text{rad/sec} \] \[ X_C = \frac{1}{628.32 \times 15 \times 10^{-6}} = \frac{1}{0.0094248} \approx 106.14\, \Omega \] - **Inductive reactance**: \[ X_L = \omega L \] --- ### **Step 4: Express \( R \) in terms of \( X_C \) and \( X_L \)** From the phase angle: \[ 0.8391 = \frac{X_C - X_L}{R} \] Rearranged: \[ X_C - X_L = 0.8391 R \] --- ### **Step 5: Use impedance to find \( R \) and \( X_L \)** The total impedance magnitude: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] Given \( Z = 18\, \Omega \): \[ 18^2 = R^2 + (X_L - X_C)^2 \] But from above: \[ X_L - X_C = -A \] where \[ A = X_C - X_L = 0.8391 R \] So, \[ ( X_L - X_C ) = -A = -0.8391 R \] Plug into the impedance equation: \[ 324 = R^2 + (-0.8391 R)^2 \] \[ 324 = R^2 + 0.7041 R^2 \] \[ 324 = (1 + 0.7041) R^2 = 1.7041 R^2 \] Solve for \( R \): \[ R^2 = \frac{324}{1.7041} \approx 190.21 \] \[ R = \sqrt{190.21} \approx 13.79\, \Omega \] --- ### **Step 6: Find \( X_L \)** Recall: \[ A = X_C - X_L = 0.8391 R \] Calculate: \[ A = 0.8391 \times 13.79 \approx 11.56\, \Omega \] Since \( X_C - X_L = A \): \[ X_L = X_C - A = 106.14 - 11.56 \approx 94.58\, \Omega \] --- ### **Step 7: Calculate the inductance \( L \)** \[ X_L = \omega L \Rightarrow L = \frac{X_L}{\omega} \] \[ L = \frac{94.58}{628.32} \approx 0.1505\, \text{H} \] --- ### **### Final Results:** \[ \boxed{ \begin{aligned} \textbf{Resistance, } R &= \mathbf{13.8\, \Omega} \\ \textbf{Inductance, } L &= \mathbf{0.151\, \text{H}} \end{aligned} } \] --- **Summary:** - Resistance \( R \approx 13.8\, \Omega \) - Inductance \( L \approx 0.151\, \text{H} \) This detailed calculation shows how impedance, reactances, and phase relationships determine the unknown circuit parameters.

In a series RLC circuit, the impedance combines resistance and reactances, with the phase angle between current and voltage determined by their ratio. When the current leads the voltage, the circuit is predominantly capacitive, allowing calculation of resistance and inductance from the impedance magnitude and phase angle using trigonometric relationships and reactance formulas.