Activity 1.1 5b Circuit Theory Simulation Answer Key

Introduction

Activity 1.1 5B – Circuit Theory Simulation Answer Key is a cornerstone exercise in many introductory electrical‑engineering courses, especially those that use virtual labs such as LTspice, Multisim, or NI ELVIS. The activity challenges students to build, simulate, and analyze a simple resistive‑inductive‑capacitive (RLC) network, then compare the simulated results with hand‑calculated values. This article walks through the complete solution step‑by‑step, explains the underlying theory, highlights common pitfalls, and provides a ready‑to‑use answer key that instructors can adopt or adapt for their own classrooms.

1. Overview of Activity 1.1 5B

2. Theoretical Background

2.1 Series RLC Fundamentals

In a series RLC circuit powered by a DC source, the inductor initially behaves like an open circuit (current = 0) while the capacitor behaves like a short circuit (voltage = 0). As time progresses, the inductor’s current builds according to

[ i_L(t)=\frac{V_s}{R}\left(1-e^{-t/\tau}\right) ]

where the time constant (\tau = \frac{L}{R}) Surprisingly effective..

Simultaneously, the capacitor charges following

[ v_C(t)=V_s\left(1-e^{-t/(R C)}\right) ]

with a distinct time constant (\tau_C = R C). Because the two components share the same series path, the overall transient response is a second‑order system that can be under‑damped, critically damped, or over‑damped depending on the damping factor

[ \zeta = \frac{R}{2}\sqrt{\frac{C}{L}}. ]

For the values in Activity 1.1 5B:

[ \tau_L = \frac{0.Which means 2\times10^3\ \Omega}=3. 47\ \text{H}}{1.92\times10^{-4}\ \text{s}=0 Simple as that..

[ \tau_C = (1.2\times10^3\ \Omega)(10\times10^{-6}\ \text{F})=0.012\ \text{s}=12\ \text{ms} ]

[ \zeta = \frac{1.2\times10^3}{2}\sqrt{\frac{10\times10^{-6}}{0.47}} \approx 0.28. ]

Since (\zeta < 1), the circuit is under‑damped, producing a brief overshoot before settling at the steady‑state DC value Surprisingly effective..

2.2 Steady‑State DC Conditions

After a few time constants (≈5 × max({\tau_L,\tau_C}) ≈ 60 ms), the inductor behaves like a short circuit and the capacitor like an open circuit. The circuit reduces to a simple resistor, so the final current is

[ I_{\text{ss}} = \frac{V_s}{R} = \frac{12\ \text{V}}{1.2\ \text{k}\Omega}=10\ \text{mA}. ]

Correspondingly, the voltage drops become

[ V_R = I_{\text{ss}}R = 12\ \text{V},\qquad V_L = 0\ \text{V},\qquad V_C = 0\ \text{V}. ]

These values are the benchmark against which the simulation results are judged But it adds up..

3. Step‑by‑Step Simulation Procedure

3.1 Setting up the schematic

1. Open the simulator (e.g., LTspice).
2. Place a voltage source (V1) and set its DC value to 12 V.
3. Insert a resistor (R1) with 1.2 kΩ, an inductor (L1) with 470 mH, and a capacitor (C1) with 10 µF in series.
4. Connect the negative terminal of the source to ground; the circuit should form a single loop.

3.2 Configuring the analysis

• Choose Transient → Run.
• Set Stop Time to 50 ms (enough to capture the overshoot and the settling).
• Use a Maximum Timestep of 0.01 ms to obtain smooth waveforms.

3.3 Running the simulation

Press Run. The simulator will generate three waveforms:

• Current through the loop (I(R1)) – identical for all series elements.
• Voltage across the resistor (V(R1)).
• Voltage across the inductor (V(L1)) and capacitor (V(C1)) (displayed by clicking the respective components).

3.4 Extracting data

• Zoom into the region 30 ms – 50 ms; the curves should have flattened.
• Use the cursor tool to read the steady‑state current (≈10 mA).
• Record the peak overshoot of the inductor voltage; it typically reaches about +2 V before decaying.

4. Hand Calculations vs. Simulation Results

The tiny discrepancies arise from the simulator’s numerical integration method (trapezoidal rule) and the finite timestep. They are well within acceptable engineering tolerances And it works..

5. Common Mistakes and How to Avoid Them

1. Incorrect component values – Double‑check the unit prefixes (kΩ, mH, µF). A misplaced decimal quickly leads to a completely different time constant.
2. Forgetting to ground the negative terminal – Without a reference node, the simulator may treat the circuit as floating, producing undefined results.
3. Using an AC analysis instead of Transient – AC sweeps ignore the time‑domain charging behavior of the capacitor and inductor, so the overshoot will never appear.
4. Too large a timestep – Setting the maximum timestep to 1 ms masks the rapid rise of the inductor current, making the overshoot appear smaller or disappear.
5. Reading the wrong cursor – In series circuits the current is identical through every element; ensure you are measuring the loop current, not the voltage across a single component when you need current.

6. Frequently Asked Questions (FAQ)

Q1. Why does the inductor voltage show a brief positive spike at the start?
Because the current cannot change instantaneously through an inductor. The sudden application of 12 V forces the inductor to develop a voltage proportional to (L \frac{di}{dt}), resulting in the observed spike.

Q2. If I replace the DC source with a 60 Hz AC source, will the circuit still be under‑damped?
With sinusoidal excitation, the concept of damping changes to frequency response. The circuit will exhibit a resonant peak near its natural frequency (\omega_0 = 1/\sqrt{LC}). For the given values, (\omega_0 \approx 1/\sqrt{0.47 \times 10 \times 10^{-6}} \approx 1460\ \text{rad/s}) (≈ 233 Hz). At 60 Hz the circuit behaves mostly resistive, and the “damping” term is less relevant.

Q3. How many time constants are needed before I can consider the circuit settled?
A rule of thumb is 5 × the largest time constant. Here the larger constant is (\tau_C = 12\ \text{ms}); therefore, after about 60 ms the response is within 1 % of the final value.

Q4. Can I use the same answer key for a parallel RLC configuration?
No. A parallel network follows a different differential equation, and the steady‑state current splits among branches. You would need to recompute the equivalent impedance and adjust the answer key accordingly.

Q5. What if my simulation shows a negative current after the overshoot?
That indicates an under‑damped system with a damping factor lower than the one calculated, often caused by an incorrect component value (e.g., a smaller resistance). Verify the resistor value and rerun the analysis.

7. Extending the Activity

To deepen understanding, instructors can ask students to:

1. Vary the resistor (e.g., 500 Ω, 2 kΩ) and observe how the damping factor changes from under‑damped to critically damped.
2. Add a second capacitor in parallel with the first and recalculate the effective capacitance, then discuss the impact on the charging curve.
3. Replace the DC source with a pulse generator (10 ms width, 12 V amplitude) and compare the transient response to the step input.
4. Export the waveform data to a spreadsheet, fit an exponential curve, and extract the experimental time constants for verification.

These extensions reinforce the link between analytical math and simulation practice, a skill set highly valued in modern engineering curricula.

8. Complete Answer Key (Ready for Publication)

Below is a concise, copy‑ready answer key that can be handed to students or posted on a learning management system Most people skip this — try not to..

Scoring rubric (sample):

• Correct schematic: 5 pts
• Proper transient settings: 5 pts
• Accurate steady‑state measurements (within 2 %): 10 pts
• Correct peak voltage identification (within 5 %): 5 pts
• Demonstrated KVL verification: 5 pts
• Written explanation of damping factor: 5 pts
• Bonus (extension tasks): up to 5 pts

Total possible: 40 pts.

9. Conclusion

Activity 1.Think about it: by following the step‑by‑step guide, students can confidently construct the schematic, run a transient analysis, and reconcile the simulated waveforms with hand‑derived equations. 1 5B offers a compact yet powerful illustration of how circuit theory translates into real‑world behavior when examined through simulation. The answer key presented here not only supplies the numeric benchmarks but also reinforces conceptual understanding—particularly the role of the damping factor and the significance of time constants No workaround needed..

Through careful attention to component values, simulation settings, and measurement techniques, learners will develop a solid foundation for more complex analyses, such as frequency response, non‑linear elements, and mixed‑signal environments. Practically speaking, instructors can reuse the provided key, adapt the numbers, or expand the experiment to keep the material fresh and challenging. At the end of the day, mastering this activity equips students with the analytical mindset and practical skills essential for any future work in electronics, power systems, or embedded‑device design.

The official docs gloss over this. That's a mistake.