Let's break down the solution step by step as requested: --- ### **Step 1: Calculate the Impedance (Z)** The impedance of an RLC series circuit is given by: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] Where: - \( R = 200\, \Omega \) - \( L = 230\, \text{mH} = .230\, \text{H} \) - \( C = 70.\, \mu\text{F} = 70. \times 10^{-6}\, \text{F} \) Assume the source frequency is not given, so let's use the standard method for writing the answer in terms of angular frequency \( \omega \). - Inductive reactance: \( X_L = \omega L \) - Capacitive reactance: \( X_C = \frac{1}{\omega C} \) So, \[ Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2} \] **Explanation:** The impedance formula combines the resistance, inductive reactance, and capacitive reactance to find the total opposition to current in the circuit. Since frequency (\( \omega \)) isn't given, the answer remains in terms of \( \omega \). --- ### **Step 2: Calculate the Current in the Circuit (I)** The maximum current is given by: \[ I = \frac{V_{max}}{Z} \] Given \( V_{max} = 36\,V \), the current is: \[ I = \frac{36}{Z} \] Substitute the value of \( Z \) from above: \[ I = \frac{36}{\sqrt{200^2 + (\omega \times .230 - \frac{1}{\omega \times 70 \times 10^{-6}})^2}} \] **Explanation:** The maximum current is the maximum emf divided by the total impedance. Since frequency (\( \omega \)) is unknown, the answer remains in terms of \( \omega \). --- ## **Final Answers** (i) **Impedance of the circuit:** \[ Z = \sqrt{200^2 + (\omega \times .230 - \frac{1}{\omega \times 70 \times 10^{-6}})^2}\ \Omega \] (ii) **Current in the circuit:** \[ I = \frac{36}{\sqrt{200^2 + (\omega \times .230 - \frac{1}{\omega \times 70 \times 10^{-6}})^2}}\ \text{A} \] *(where \(\omega\) is the angular frequency of the AC source)* If you know the frequency, you can substitute \( \omega = 2\pi f \) and calculate the numerical values.