Let's solve each problem step by step. --- # **Part 1A: Find the value of the capacitance, C** ### **Conceptual Introduction** For the current \( i(t) \) to be **in phase** with the source voltage \( v(t) \), the **total impedance** seen by the source must be **purely real** (i.e., the imaginary part must be zero). This means the total reactance (inductive + capacitive) must cancel out. --- ### **Given Values** - Source voltage: \( v(t) = 60 \cos(300t + 30^\circ) \) V - Components: - \( R_1 = 10\,\Omega \) - \( R_2 = 20\,\Omega \) - \( L = 60\,\text{mH} = 0.06\,\text{H} \) - \( C = ? \) (to be found) - Angular frequency: \( \omega = 300\,\text{rad/s} \) --- ### **Step-by-Step Solution** #### **1. Write the total impedance:** The total impedance is: \[ Z_{\text{total}} = R_1 + \left[ \left( R_2 + j\omega L \right) \parallel \left( \frac{1}{j\omega C} \right) \right] \] But from the diagram, the \( 20\,\Omega \), \( C \), and \( 60\,\text{mH} \) are all **in series**; then this series combination is in series with \( 10\,\Omega \): \[ Z_{\text{total}} = 10 + 20 + j\omega L + \frac{1}{j\omega C} \] \[ Z_{\text{total}} = 30 + j\omega L + \frac{1}{j\omega C} \] #### **2. Substitute values:** \[ Z_{\text{total}} = 30 + j(300 \times 0.06) + \frac{1}{j300C} \] \[ Z_{\text{total}} = 30 + j(18) + \frac{1}{j300C} \] Recall that \( \frac{1}{j300C} = -j \frac{1}{300C} \): \[ Z_{\text{total}} = 30 + j18 - j\frac{1}{300C} \] \[ Z_{\text{total}} = 30 + j\left(18 - \frac{1}{300C}\right) \] #### **3. Set imaginary part to zero:** For the impedance to be purely real: \[ 18 - \frac{1}{300C} = 0 \] #### **4. Solve for \( C \):** \[ 18 = \frac{1}{300C} \] \[ C = \frac{1}{18 \times 300} \] \[ C = \frac{1}{5400} \] \[ C \approx 1.85 \times 10^{-4} \, \text{F} = 185\,\mu\text{F} \] --- ### **Final Answer for 1A** \[ \boxed{C = 185\,\mu\text{F}} \] --- # **Part 1B: Find \( v_0(t) \) in the Network** ### **Conceptual Introduction** This is a parallel AC circuit. We are to find the voltage \( v_0(t) \) across the 5 Ω resistor. Since all components are in parallel, \( v_0(t) \) is the same across each branch. --- ### **Given Values** - Source: \( i_s(t) = 5 \cos(10t) \) A - \( R_1 = 10\,\Omega \) (parallel) - \( L_1 = 400\,\text{mH} = 0.4\,\text{H} \) (parallel) - \( C_1 = 100\,\text{mF} = 0.1\,\text{F} \) (parallel) - \( C_2 = 40\,\text{mF} = 0.04\,\text{F} \) (parallel with \( R_2 \)) - \( R_2 = 5\,\Omega \) (parallel with \( C_2 \)) - \( \omega = 10\,\text{rad/s} \) --- ### **Step-by-Step Solution** #### **1. Find Impedances** - \( Z_{R1} = 10\,\Omega \) - \( Z_{L1} = j\omega L_1 = j(10)(0.4) = j4\,\Omega \) - \( Z_{C1} = \frac{1}{j\omega C_1} = \frac{1}{j(10)(0.1)} = \frac{1}{j1} = -j1\,\Omega \) - \( Z_{R2 \parallel C2} \): - \( Z_{R2} = 5\,\Omega \) - \( Z_{C2} = \frac{1}{j\omega C_2} = \frac{1}{j(10)(0.04)} = \frac{1}{j0.4} = -j2.5\,\Omega \) - Parallel combination: \[ Z_{R2 \parallel C2} = \frac{Z_{R2} \cdot Z_{C2}}{Z_{R2} + Z_{C2}} = \frac{5 \cdot (-j2.5)}{5 - j2.5} \] \[ = \frac{-j12.5}{5 - j2.5} \] To simplify, multiply numerator and denominator by the conjugate of the denominator: \[ = \frac{-j12.5(5 + j2.5)}{(5 - j2.5)(5 + j2.5)} \] \[ = \frac{-j12.5 \cdot 5 + (-j12.5)(j2.5)}{25 + 12.5} \] \[ = \frac{-j62.5 + 31.25}{37.5} \] \[ = \frac{31.25 - j62.5}{37.5} \] \[ = 0.833 - j1.67\,\Omega \] #### **2. Find Total Parallel Admittance** Admittance \( Y = \frac{1}{Z} \): - \( Y_{R1} = \frac{1}{10} = 0.1\,S \) - \( Y_{L1} = \frac{1}{j4} = -j0.25\,S \) - \( Y_{C1} = \frac{1}{-j1} = j1\,S \) - \( Y_{R2 \parallel C2} = \frac{1}{0.833 - j1.67} \) Let's calculate \( Y_{R2 \parallel C2} \): Let \( Z = 0.833 - j1.67 \): \[ |Z|^2 = 0.833^2 + 1.67^2 = 0.694 + 2.789 = 3.483 \] \[ \frac{1}{Z} = \frac{0.833 + j1.67}{3.483} = 0.239 + j0.48\,S \] #### **3. Total Admittance** \[ Y_{\text{total}} = Y_{R1} + Y_{L1} + Y_{C1} + Y_{R2 \parallel C2} \] \[ Y_{\text{total}} = 0.1 - j0.25 + j1 + 0.239 + j0.48 \] \[ = (0.1 + 0.239) + [-j0.25 + j1 + j0.48] \] \[ = 0.339 + j(1 - 0.25 + 0.48) \] \[ = 0.339 + j1.23 \] #### **4. Find Total Impedance** \[ Z_{\text{total}} = \frac{1}{Y_{\text{total}}} \] \[ = \frac{1}{0.339 + j1.23} \] Let's find the magnitude and phase: - Magnitude: \( \sqrt{0.339^2 + 1.23^2} = \sqrt{0.115 + 1.5129} = \sqrt{1.6279} \approx 1.276 \) - Angle: \( \tan^{-1}\left(\frac{1.23}{0.339}\right) = \tan^{-1}(3.63) \approx 74.7^\circ \) So, \[ Y_{\text{total}} = 1.276\angle 74.7^\circ\,S \] \[ Z_{\text{total}} = \frac{1}{1.276\angle 74.7^\circ} = 0.784\angle -74.7^\circ\,\Omega \] #### **5. Find Output Voltage \( v_0(t) \)** The current source: \( i_s(t) = 5\cos(10t) \) In phasor form: \( \tilde{I}_s = 5\angle 0^\circ \) A Voltage across all branches: \[ \tilde{V}_0 = \tilde{I}_s \cdot Z_{\text{total}} \] \[ = 5\angle 0^\circ \cdot 0.784\angle -74.7^\circ = 3.92 \angle -74.7^\circ\,\text{V} \] Convert back to time domain: \[ v_0(t) = 3.92\cos(10t - 74.7^\circ)\,\text{V} \] --- ### **Final Answer for 1B** \[ \boxed{v_0(t) = 3.92\cos(10t - 74.7^\circ)\,\text{V}} \] --- **Summary of Answers:** - **1A:** \( \boxed{C = 185\,\mu\text{F}} \) - **1B:** \( \boxed{v_0(t) = 3.92\cos(10t - 74.7^\circ)\,\text{V}} \) Let me know if you want step-by-step phasor diagrams or further breakdowns!