Let's solve the problem step by step. ### **Given:** - \( V_{\text{rms}} = 200 \) V - \( f = \frac{160}{\pi} \) Hz - \( L = 0.15 \) H - \( C = 25~\mu\text{F} = 25 \times 10^{-6} \) F - \( V_{R(\text{rms})} = 45 \) V --- ## **Step 1: Find the angular frequency (\(\omega\))** \[ \omega = 2\pi f = 2\pi \left(\frac{160}{\pi}\right) = 320 \text{ rad/s} \] --- ## **Step 2: Find current \( I \) (rms)** \[ V_R = I R \implies I = \frac{V_R}{R} \] But we don't know \(R\) yet. So, let's move on and find \(R\) first. --- ## **Step 3: Find the impedance \( Z \) of the circuit** \[ I = \frac{V}{Z} \] \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] where: - \(X_L = \omega L\) - \(X_C = \frac{1}{\omega C}\) Let's calculate both: \[ X_L = 320 \times 0.15 = 48~\Omega \] \[ X_C = \frac{1}{320 \times 25 \times 10^{-6}} = \frac{1}{0.008} = 125~\Omega \] So, \[ X_L - X_C = 48 - 125 = -77~\Omega \] --- ## **Step 4: Express Z in terms of R** \[ Z = \sqrt{R^2 + (-77)^2} = \sqrt{R^2 + 5929} \] --- ## **Step 5: Express I in terms of R** \[ I = \frac{V}{Z} = \frac{200}{\sqrt{R^2 + 5929}} \] But we also have: \[ I = \frac{V_R}{R} = \frac{45}{R} \] Set equal: \[ \frac{200}{\sqrt{R^2 + 5929}} = \frac{45}{R} \] --- ## **Step 6: Solve for R** Cross multiply: \[ 200R = 45\sqrt{R^2 + 5929} \] \[ \frac{200R}{45} = \sqrt{R^2 + 5929} \] \[ \left(\frac{200R}{45}\right)^2 = R^2 + 5929 \] \[ \frac{40000R^2}{2025} = R^2 + 5929 \] \[ 40000R^2 = 2025R^2 + 2025 \times 5929 \] \[ 40000R^2 - 2025R^2 = 2025 \times 5929 \] \[ 37975R^2 = 2025 \times 5929 \] \[ R^2 = \frac{2025 \times 5929}{37975} \] \[ R = \sqrt{ \frac{2025 \times 5929}{37975} } \] Calculate: - \(2025 \times 5929 = 12,010,725\) - \(37975\) So, \[ R = \sqrt{ \frac{12,010,725}{37,975} } \approx \sqrt{316.4} \approx 17.8~\Omega \] --- ## **Step 7: Find I (rms)** \[ I = \frac{45}{17.8} \approx 2.53~\text{A} \] --- ## **Step 8: Find \(V_L\) and \(V_C\) (rms)** \[ V_L = I X_L = 2.53 \times 48 \approx 121.4~\text{V} \] \[ V_C = I X_C = 2.53 \times 125 \approx 316.3~\text{V} \] --- ## **Final Answers:** **a) \( I \) (rms):** \[ \boxed{2.53~\text{A}} \] **b) \( R \):** \[ \boxed{17.8~\Omega} \] **c) \( V_L \) (rms):** \[ \boxed{121.4~\text{V}} \] **d) \( V_C \) (rms):** \[ \boxed{316.3~\text{V}} \] --- Let me know if you'd like a full calculation breakdown for any step!

Let's carefully work through this problem step-by-step, providing detailed calculations at each stage. --- ### **Given Data:** - Total rms voltage supplied to the circuit: \( V = 200\,\text{V} \) - Frequency: \( f = \frac{160}{\pi}\,\text{Hz} \) - Inductance: \( L = 0.15\,\text{H} \) - Capacitance: \( C = 25\,\mu\text{F} = 25 \times 10^{-6}\,\text{F} \) - rms voltage across the resistor: \( V_{R} = 45\,\text{V} \) --- ### **Step 1: Calculate the angular frequency \(\omega\)** \[ \omega = 2\pi f = 2\pi \times \frac{160}{\pi} = 2 \times 160 = 320\,\text{rad/sec} \] --- ### **Step 2: Find the reactances \(X_L\) and \(X_C\)** \[ X_L = \omega L = 320 \times 0.15 = 48\,\Omega \] \[ X_C = \frac{1}{\omega C} = \frac{1}{320 \times 25 \times 10^{-6}} = \frac{1}{0.008} = 125\,\Omega \] --- ### **Step 3: Express the total impedance \(Z\)** In an RLC series circuit, the total impedance is: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] The circuit's total rms current \(I\) is: \[ I = \frac{V}{Z} \] The rms voltage across the resistor is: \[ V_{R} = I R \] Given \(V_{R} = 45\,\text{V}\), we have: \[ I = \frac{V_{R}}{R} = \frac{45}{R} \] But also, \[ I = \frac{V}{Z} = \frac{200}{Z} \] Set these equal: \[ \frac{200}{Z} = \frac{45}{R} \] which implies: \[ Z = \frac{200 R}{45} = \frac{40 R}{9} \] --- ### **Step 4: Relate \(Z\) to \(R\) and reactances** From earlier, \[ Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + (-77)^2} = \sqrt{R^2 + 5929} \] Set equal to the expression from the voltage ratio: \[ \sqrt{R^2 + 5929} = \frac{40 R}{9} \] Square both sides: \[ R^2 + 5929 = \left(\frac{40 R}{9}\right)^2 = \frac{1600 R^2}{81} \] Multiply through by 81 to clear denominator: \[ 81 R^2 + 81 \times 5929 = 1600 R^2 \] Calculate \(81 \times 5929\): \[ 81 \times 5929 = 481,089 \] So: \[ 81 R^2 + 481,089 = 1600 R^2 \] Bring all to one side: \[ 1600 R^2 - 81 R^2 = 481,089 \] \[ (1600 - 81) R^2 = 481,089 \] \[ 1519 R^2 = 481,089 \] Solve for \(R^2\): \[ R^2 = \frac{481,089}{1519} \approx 316.4 \] Thus, \[ R = \sqrt{316.4} \approx 17.8\,\Omega \] --- ### **Step 5: Find the circuit current \(I\)** \[ I = \frac{V}{Z} = \frac{200}{Z} \] From earlier, \[ Z = \frac{40 R}{9} = \frac{40 \times 17.8}{9} \approx \frac{712}{9} \approx 79.1\,\Omega \] Now, \[ I = \frac{200}{79.1} \approx 2.53\,\text{A} \] --- ### **Step 6: Calculate the voltages across \(L\) and \(C\)** \[ V_L = I X_L = 2.53 \times 48 \approx 121.4\,\text{V} \] \[ V_C = I X_C = 2.53 \times 125 \approx 316.3\,\text{V} \] --- ### **### Final Answers:** | **Quantity** | **Value** | |----------------|------------| | **a) rms current \(I\)** | **2.53 A** | | **b) Resistance \(R\)** | **17.8 Ω** | | **c) Voltage across inductor \(V_L\)** | **121.4 V** | | **d) Voltage across capacitor \(V_C\)** | **316.3 V** | --- ### **Summary:** - The circuit's resistance is approximately **17.8 Ω**. - The rms current in the circuit is approximately **2.53 A**. - The voltage across the inductor is approximately **121.4 V**. - The voltage across the capacitor is approximately **316.3 V**. Let me know if you'd like further details or clarifications!

In an RLC series circuit, the impedance combines resistance and reactances, determining the circuit's current and voltage distribution. The voltages across inductor and capacitor can greatly exceed the source voltage due to reactive effects, especially near resonance.