Hello, I am confused on this question, the answer is attached as well but I am having difficulty with converting the complex numbers and the reasoning behind these steps. Thanks. **Problem 1**(a) Determine the equivalent impedance seen by the source \( V_s \) in Figure when: \[ V_s(t) = 10 \cos(4000t + 60^\circ) \, V, \quad R_1 = 800 \, \Omega, \quad R_2 = 500 \, \Omega, \quad L = 200 \, mH, \quad C = 70 \, nF \](b) Determine the impedance when \( v_s(t) = 10 \cos(2000t + 60^\circ) \, (V) \)---**Explanation of Figure P3.47**The diagram shows an AC circuit with a voltage source, \( V_s \), connected to two resistors, \( R_1 \) and \( R_2 \), an inductor \( L \), and a capacitor \( C \). The components \( R_2 \), \( L \), and \( C \) are in parallel with each other. The voltage source is connected to one terminal of \( R_1 \) and the parallel combination of \( R_2 \), \( L \), and \( C \). **Analysis:**1. \( X_L = \omega L = 800 \, \Omega \Rightarrow Z_L = +j \cdot X_L = +j \cdot 800 \, \Omega \)2. \( X_C = \frac{1}{\omega C} = 3571 \, \Omega \Rightarrow Z_C = -j \cdot X_C = -j \cdot 3571 \, \Omega \)3. \( Z_{eq1} = Z_{R2} + Z_L = R_2 + jX_L = 500 + j \cdot 800 \, \Omega = 943.4 \angle 57.99^\circ \)4. \( Z_{eq2} = \frac{Z_{eq1} \cdot R_1}{Z_{eq1} + R_1} = 442.918 + j \cdot 219.71 = 494.43 \angle 26.39^\circ \)5. \( Z_{eq} = \frac{Z_{eq2} \cdot Z_C}{Z_{eq2} + Z_C} = 494.27 + j \cdot 168.88 = 522.3105 \angle 18.85^\circ \)In this analysis, the impedances are combined in sequential steps to derive the equivalent impedance \( Z_{eq} \). Each calculation combines resistances and reactances through impedance transformations and polar mathematical operations.

b) To find the equivalent impedance as seen from the source Vs

Answered: (Problem 1) (a) Determine the…

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Hello, I am confused on this question, the answer is attached as well but I am having difficulty with converting the complex numbers and the reasoning behind these steps. Thanks.

**Problem 1**

(a) Determine the equivalent impedance seen by the source \( V_s \) in Figure when:

\[ V_s(t) = 10 \cos(4000t + 60^\circ) \, V, \quad R_1 = 800 \, \Omega, \quad R_2 = 500 \, \Omega, \quad L = 200 \, mH, \quad C = 70 \, nF \]

(b) Determine the impedance when \( v_s(t) = 10 \cos(2000t + 60^\circ) \, (V) \)

---

**Explanation of Figure P3.47**

The diagram shows an AC circuit with a voltage source, \( V_s \), connected to two resistors, \( R_1 \) and \( R_2 \), an inductor \( L \), and a capacitor \( C \). The components \( R_2 \), \( L \), and \( C \) are in parallel with each other. The voltage source is connected to one terminal of \( R_1 \) and the parallel combination of \( R_2 \), \( L \), and \( C \).

**Analysis:**

1. \( X_L = \omega L = 800 \, \Omega \Rightarrow Z_L = +j \cdot X_L = +j \cdot 800 \, \Omega \)

2. \( X_C = \frac{1}{\omega C} = 3571 \, \Omega \Rightarrow Z_C = -j \cdot X_C = -j \cdot 3571 \, \Omega \)

3. \( Z_{eq1} = Z_{R2} + Z_L = R_2 + jX_L = 500 + j \cdot 800 \, \Omega = 943.4 \angle 57.99^\circ \)

4. \( Z_{eq2} = \frac{Z_{eq1} \cdot R_1}{Z_{eq1} + R_1} = 442.918 + j \cdot 219.71 = 494.43 \angle 26.39^\circ \)

5. \( Z_{eq} = \frac{Z_{eq2} \cdot Z_C}{Z_{eq2} + Z_C} = 494.27 + j \cdot 168.88 = 522.3105 \angle 18.85^\circ \)

In this analysis, the impedances are combined in sequential steps to derive the equivalent impedance \( Z_{eq} \). Each calculation combines resistances and reactances through impedance transformations and polar mathematical operations.

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This doesn't match the answer on the answer key I submitted with it - I was hoping to know the rationale behind the work and help with converting the complex numbers. Thanks.

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