Sure! Let's break down the **step-by-step solution** for both questions based on the provided circuit and instructions. --- ## **Question 1: Sinusoidal Steady State Analysis** ### **Given Circuit** - \( R = 10 \Omega \) - \( L = 3.05 \, \text{H} \) - \( C = 0.02 \, \text{F} \) - \( V_{in}(t) = 100 \sin (2\pi \cdot 50t) \), \( V_{pp} = 100 \, V \), \( f = 50 \, \text{Hz} \) --- ### **Part 1a: Plot \( V_{in}(t) \) and \( V_{o}(t) \), Measure Phase Shift** #### **Steps:** 1. **Draw Circuit in PSPICE:** - Series connection: AC Voltage Source (\( V_{in} \)) → \( R \) → \( L \) → \( C \) → Ground. 2. **Set Source:** - Sinusoidal source: Amplitude = 100V, Frequency = 50Hz. 3. **Run Transient Simulation:** - Total time: At least 2-3 periods (\( T = 1/f = 20\text{ms} \)), so set simulation time to 60ms. 4. **Plot \( V_{in}(t) \) and \( V_{o}(t) \):** - \( V_{o}(t) \) is voltage across the capacitor. 5. **Measure Time Difference (\( \Delta t \)) Between Peaks:** - Use cursor to measure time between corresponding peaks of \( V_{in}(t) \) and \( V_{o}(t) \). 6. **Calculate Phase Shift:** \[ \Delta \theta = 360^\circ \times f \times \Delta t \] #### **Discuss Results:** - Comment on whether \( V_{o}(t) \) leads or lags \( V_{in}(t) \), and by how many degrees. --- ### **Part 1b: Plot \( V_{in}(t) \) and \( i_0(t) \), Measure Phase Shift** #### **Steps:** 1. **Plot \( V_{in}(t) \) and \( i_0(t) \) (current through series circuit):** - Use marker to display current (\( i_0(t) \)). 2. **Measure Time Difference and Calculate Phase Shift:** - Repeat previous procedure; discuss whether current leads or lags the input voltage. --- ### **Part 2: Change Input Frequency** - Repeat above steps for three new frequencies: 1. \( f_1 = \frac{\text{ID}}{500} \) Hz 2. \( f_2 = \frac{\text{ID}}{200} \) Hz 3. \( f_3 = \frac{\text{ID}}{100} \) Hz #### **For each frequency:** - Draw plots for \( V_{in}(t) \) and \( V_{o}(t) \) - Compare and discuss phase shift and amplitude changes. --- ## **Question 2: Second-Order RLC Circuit Analysis** ### **Given Example:** - **ID number:** 1217598 - \( L_4 = 7\, \text{H} \) - \( C_3 = 5 \) (so \( 5 \times 10 = 50\, \text{mF} \)) - \( R_2 = 9\,\Omega \), \( R_1 = 8\,\Omega \) --- ### **Steps:** #### **a) Construct PSPICE Schematic:** - **Series RLC:** - 20V DC source, switch in series with: - \( R_1 = 8\Omega \) - \( R_2 = 9\Omega \) - \( L = 7\,\text{H} \) - \( C = 50\,\text{mF} \) #### **b) Start Simulation at t = 1s** - **Switch**: Closed up to \( t = 1s \), open at \( t = 1s \). #### **c) Simulate and Plot:** - Plot \( i(t) \): Current through the inductor. - Plot \( v_C(t) \): Voltage across the capacitor. - Use markers to show the transient rise and fall. --- ### **Example Calculation for Question 2:** #### **Before t = 1s:** - Circuit in steady state; inductor behaves as a short, capacitor as open. - Find initial values for \( i(1^-) \) and \( v_C(1^-) \). #### **After t = 1s:** - Switch opens; circuit becomes a series RLC with initial conditions set by \( t=1s \) values. - Use PSPICE to simulate the transient response. --- ### **Final Answers (General Steps):** 1. **Construct the circuit in PSPICE as described.** 2. **Set up the simulation parameters (transient, time = 0 to 2s, switch opens at 1s).** 3. **Run the simulation, plot the required waveforms.** 4. **Use the simulation markers to show phase shift (Q1) and transient response (Q2).** 5. **Write your observations and discuss the effect of frequency (Q1) and RLC values (Q2).** --- **If you want an actual PSPICE schematic or output plot, let me know your specific software and I can guide you further! If you want to see the hand calculations for steady-state or transient analysis, specify which part.** --- ### **Summary Table for Your Example (Q2)** | Parameter | Value | |-----------|-------| | \(L\) | 7 H | | \(C\) | 50 mF | | \(R_1\) | 8 Ω | | \(R_2\) | 9 Ω | | \(V_{source}\) | 20V | --- **Let me know if you want the analytical solution for any circuit step!**

Certainly! Below is a detailed step-by-step solution for **Question 1** and **Question 2**, including calculations, explanations, and final results. --- # **Question 1: Sinusoidal Steady State Analysis** ## **Given Data:** - Input Voltage: \( V_{in}(t) = 100 \sin(2\pi \times 50 t) \) - Frequency: \( f = 50\, \text{Hz} \) - Resistance: \( R = 10\, \Omega \) - Inductance: \( L = 3.05\, \text{H} \) - Capacitance: \( C = 0.02\, \text{F} \) --- ## **Part 1a: Calculate Theoretical Phase Shift and Transient Response** ### **Step 1: Find the Impedance of the series RLC circuit** \[ Z = R + j \left( \omega L - \frac{1}{\omega C} \right) \] where \[ \omega = 2\pi f = 2\pi \times 50 = 100\pi \approx 314.16\, \text{rad/sec} \] Calculate: \[ X_L = \omega L = 314.16 \times 3.05 \approx 958.68\, \Omega \] \[ X_C = \frac{1}{\omega C} = \frac{1}{314.16 \times 0.02} = \frac{1}{6.2832} \approx 0.159\, \Omega \] ### **Step 2: Calculate the Total Impedance** \[ Z = 10 + j (958.68 - 0.159) = 10 + j 958.52\, \Omega \] Magnitude: \[ |Z| = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{10^2 + 958.52^2} \approx \sqrt{100 + 918273} \approx 958.52\, \Omega \] Phase angle: \[ \phi = \arctan \left( \frac{X_L - X_C}{R} \right) = \arctan \left( \frac{958.52}{10} \right) \approx \arctan(95.852) \approx 89.4^\circ \] ### **Step 3: Calculate the amplitude of the output voltage across the capacitor** The impedance of the capacitor: \[ Z_C = \frac{1}{j \omega C} = -j 0.159\, \Omega \] The voltage across the capacitor: \[ V_{o}(t) \text{ amplitude} = V_{in} \times \frac{1/ j \omega C}{Z} = V_{in} \times \frac{1}{j \omega C} \div Z \] But more straightforwardly, the steady-state voltage across the capacitor: \[ V_{C} = V_{in} \times \frac{1 / j \omega C}{Z} = V_{in} \times \frac{1/ j \omega C}{|Z| e^{j \phi}} \] Since the phase of \( V_{C} \) relative to \( V_{in} \): \[ \angle V_{C} = - \arctan \left( \frac{X_C}{R} \right) = - \arctan \left( \frac{0.159}{10} \right) \approx -0.91^\circ \] **Note:** Because \( X_L \gg R \), the circuit behaves almost like an inductor at this frequency, and the voltage across the capacitor lags the input voltage by approximately \( \arctan \left( \frac{X_C}{X_L} \right) \). --- ## **Part 1b: Measure and Calculate the Phase Shift** ### **Time difference between peaks (\( \Delta t \))** - Suppose from simulation, the measured time difference between the peaks of \( V_{in}(t) \) and \( V_{o}(t) \) is \( \Delta t \). - Let's assume, for example, \( \Delta t = 0.0005\, s \) (this is an example; actual measurement depends on simulation). ### **Calculate phase shift:** \[ \text{Phase shift } (\Delta \theta) = 360^\circ \times f \times \Delta t \] \[ \Rightarrow \Delta \theta = 360^\circ \times 50\, \text{Hz} \times 0.0005\, s = 360^\circ \times 0.025 = 9^\circ \] **Result:** The output voltage across the capacitor lags the input voltage by approximately **9°**. --- ## **Discussion:** - The theoretical phase shift (~89.4°) indicates that the capacitor voltage lags the input significantly at 50Hz. - The measured phase shift (assumed 9° here) is much smaller, indicating the actual circuit may not be purely ideal or the measurement is taken at a different point. - Typically, at high \( X_L \), the circuit behaves like an inductor, and the capacitor voltage lags the input. --- ## **Part 2: Response at Different Frequencies** ### **Calculate new frequencies:** - Suppose your ID is **12345** for demonstration. \[ f_1 = \frac{12345}{500} \approx 24.69\, \text{Hz} \] \[ f_2 = \frac{12345}{200} \approx 61.72\, \text{Hz} \] \[ f_3 = 12345\, \text{Hz} \] ### **For each frequency:** - Repeat the impedance calculations. - Observe how the phase shift and amplitude change. **Expected trend:** - At lower frequencies (~25Hz), \( X_L \) and \( X_C \) are smaller, leading to less phase shift. - At higher frequencies (~12345Hz), \( X_L \) becomes enormous, and the circuit behaves more like an inductor, increasing phase lag. --- # **Question 2: Second-Order RLC Circuit Analysis** ## **Given Data:** Suppose your student ID is **1217598**: - \( L_4 = 7\, \text{H} \) - Last 4 digits: 598 - \( C_3 = 5 \Rightarrow 50\, \text{mF} \) (since \( C_3 \) is in mF) - \( R_1 = 8\, \Omega \) - \( R_2 = 9\, \Omega \) --- ## **Part a): Constructing the Circuit** - The circuit is a series RLC with switch. - At \( t=1\,s \), switch opens. --- ## **Part b): Transient Response** ### **Initial Conditions:** - Before switch opens: - Steady-state DC implies: - Inductor acts as short circuit. - Capacitor acts as open circuit. - Current at \( t=1^- \): \[ i(1^-) = \frac{V_{s}}{R_{total}} \quad \text{(assuming simple series with supply)} \] - Voltage across capacitor: \[ v_{C}(1^-) = V_{s} \quad \text{(since capacitor is open circuit at steady state)} \] Suppose source voltage \( V_s = 20\, V \). ### **Calculations:** - **Initial current:** \[ i(1^-) = \frac{20V}{R_1 + R_2} = \frac{20}{8 + 9} = \frac{20}{17} \approx 1.176\, A \] - **Initial capacitor voltage:** \[ v_{C}(1^-) = V_s = 20\, V \] --- ### **Transient Response after switch opens** The circuit becomes an RLC with initial conditions: - \( i(1^+) = i(1^-) = 1.176\, A \) - \( v_{C}(1^+) = v_{C}(1^-) = 20\, V \) The governing differential equation: \[ L \frac{d^2 i}{dt^2} + R \frac{di}{dt} + \frac{1}{C} i = 0 \] where: \[ R = R_1 + R_2 = 17\, \Omega \] \[ L = 7\, H \] \[ C = 50\, \text{mF} = 0.05\, F \] Calculate damping factor: \[ \alpha = \frac{R}{2L} = \frac{17}{2 \times 7} \approx 1.214\, \text{rad/sec} \] Calculate natural frequency: \[ \omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{7 \times 0.05}} = \frac{1}{\sqrt{0.35}} \approx 1.692\, \text{rad/sec} \] Since \( \alpha > \omega_0 \), the circuit is **overdamped**. --- ### **Solution of the differential equation:** \[ i(t) = A e^{s_1 t} + B e^{s_2 t} \] where: \[ s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} \] Calculate: \[ \sqrt{\alpha^2 - \omega_0^2} = \sqrt{(1.214)^2 - (1.692)^2} = \sqrt{1.473 - 2.86} = \sqrt{-1.387} \] Since the square root is imaginary, the system is **underdamped**; correction needed. Actually, because \( \alpha = 1.214 \) and \( \omega_0=1.692 \), with \( \alpha < \omega_0 \), the damping is **underdamped**: \[ \text{Corrected: } \alpha = 1.214\, \text{rad/sec}, \quad \omega_0 = 1.692\, \text{rad/sec} \] Calculate: \[ \omega_d = \sqrt{\omega_0^2 - \alpha^2} = \sqrt{(1.692)^2 - (1.214)^2} \approx \sqrt{2.86 - 1.473} = \sqrt{1.387} \approx 1.178\, \text{rad/sec} \] The solution: \[ i(t) = e^{-\alpha t} \left( C_1 \cos \omega_d t + C_2 \sin \omega_d t \right) \] Use initial conditions at \( t=1\,s \): \[ i(1) = 1.176\,A \] \[ v_C(1) = 20\,V \] Express \( v_C(t) \): \[ v_C(t) = \frac{1}{C} \int i(t) dt + V_{C}(0) \] Alternatively, for initial analysis, focus on \( i(t) \), and use the relation: \[ v_C(t) = \frac{1}{C} \int i(t) dt \] --- ## **Final notes:** - The transient analysis shows oscillatory decay. - Exact constants \( C_1, C_2 \) require solving with initial conditions. --- # **Summary of Final Results:** | Part | Key Results | |--------|--------------| | **Q1 (Theoretical Phase Shift at 50Hz)** | ~89.4° lag of \( V_{o} \) relative to \( V_{in} \) | | **Q1 (Measured Phase Shift)** | Approx. 9° (from assumed simulation) | | **Q1 (At different frequencies)** | Phase shift decreases at lower frequencies, increases at higher frequencies | | **Q2 Initial Conditions** | \( i(1) \approx 1.176\,A \), \( v_{C}(1) = 20\,V \) | | **Q2 Transient Response** | Overdamped or underdamped depending on parameters; oscillatory decay expected | --- # **Conclusion:** - The sinusoidal steady-state analysis confirms the high reactance at 50Hz causes significant phase lag. - Changing frequency alters phase and amplitude, as expected. - The transient analysis shows the circuit's response after switching states, requiring solving second-order differential equations with initial conditions. --- If you need actual simulation data or further detailed numerical solutions, I can assist with those as well!

Certainly! While I cannot generate live plots directly here, I can guide you on how to produce the required graphs using simulation tools like PSPICE, MATLAB, or any circuit simulator. Additionally, I can provide schematic sketches and sample plots based on typical results. Below is a detailed description of each graph you need, along with example sketches and instructions on how to generate them. --- # **Graphs for Question 1** ## **1a. Plot of \( V_{in}(t) \) and \( V_{o}(t) \) (Voltage across capacitor)** ### **Expected Features:** - \( V_{in}(t) \): A sinusoid at 50Hz with amplitude 100V. - \( V_{o}(t) \): A sinusoid at the same frequency, but lagging behind \( V_{in}(t) \) due to the circuit's reactance. - The phase difference \(\Delta \theta\) can be measured from the plot. ### **Sample Sketch:** ``` Amplitude (V) | /\/\/\ /\/\/\ 100| / \ / \ | / \ / \ |---/----------\------/----------\----- Time t0 t_peak t_peak + Δt ``` - The peaks of \( V_{in} \) and \( V_o \) are separated by a small time \( \Delta t \). - Use cursors in PSPICE or MATLAB to measure \( \Delta t \). --- ## **1b. Plot of \( V_{in}(t) \) and \( i_0(t) \) (current through the circuit)** ### **Expected Features:** - \( i_0(t) \) will also be sinusoidal at 50Hz. - Its phase relative to \( V_{in}(t) \) indicates whether current leads or lags. ### **Sample Sketch:** ``` Amplitude (A or V) | /\/\/\ /\/\/\ I(t)| / \ / \ |---/--------\--------/--------\--- Time t0 t_peak t_peak + Δt ``` - The phase difference can again be measured using cursors. --- ## **How to generate these plots:** - **In PSPICE:** - Run a transient analysis. - Plot \( V_{in} \), \( V_{o} \), and \( i_0 \) on the same graph. - Use cursors to measure phase shifts. - **In MATLAB:** - Use `plot()` functions with simulated data. - Use `findpeaks()` to locate peaks and measure \( \Delta t \). --- # **Graphs for Question 2** ## **Transient Response: \( v_C(t) \) and \( i(t) \)** ### **Expected features:** - When the switch opens at \( t=1\,s \), the circuit's current and capacitor voltage will show a transient decay or oscillation. - For an underdamped RLC circuit, \( v_C(t) \) and \( i(t) \) exhibit exponential decay with sinusoidal oscillations. ### **Sample Sketch of \( v_C(t) \):** ``` Voltage across capacitor (V) | /\/\/\ /\/\/\ /\/\/\ 20V| / \ / \ / \ | / \ / \ / \ |-----/----------/----------/----------\-- Time t=1s t>1s ``` - The initial voltage is 20V at \( t=1s \), then decays oscillatory if underdamped. ### **Sample Sketch of \( i(t) \):** ``` Current (A) | /\/\/\ /\/\/\ /\/\/\ 1.2A| / \ / \ / \ | / \ / \ / \ |--/----------/----------/----------\-- Time t=1s t>1s ``` --- # **How to generate these transient plots:** - **In PSPICE:** - Set the initial conditions at \( t=1\,s \). - Run a transient analysis with sufficient time span. - Plot \( v_C(t) \) and \( i(t) \). - Zoom into the transient period just after \( t=1\,s \). - **In MATLAB:** - Use the analytical solution for \( i(t) \) and \( v_C(t) \). - Plot the functions over the time interval of interest. --- # **Summary of the plotting process:** | Step | Action | |---------|---------| | 1 | Set initial conditions (for transient) in simulation. | | 2 | Run transient analysis from \( t=0 \) to, say, 2 seconds. | | 3 | Focus on the interval just after \( t=1\,s \). | | 4 | Use cursors or data points to measure phase shifts. | | 5 | Overlay plots for comparison. | --- # **Final notes:** - Ensure the time scale is detailed enough to observe the transient behavior. - Use labels and legends for clarity. - For phase shift measurements, note the time difference between peaks, then convert to degrees. --- If you want, I can help you create sample MATLAB code or detailed instructions for PSPICE plotting. Let me know!