FE Electrical & Computer Practice Question Solution included after investigation
CASE E-021 Β· FE

The RLC Circuit That Peaked Instead of Filtering

Both reactive components looked like opposition. At resonance, they canceled and the current doubled the protection limit.

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Inspect, commit, prove, fix, and sign off.

Case E-021 engineering failure visual
HOW THIS WORKS

Think like
a real engineer.

Follow the investigation process used in the field β€” in five guided steps.

1

Inspect

Review the problem, diagram, and evidence.

2

Commit

Choose your hypothesis.

3

Prove

Run calculations and test your idea.

4

Fix

Select and validate a safe correction.

5

Sign Off

See the full debrief and key takeaways.

Case Brief

SAME PRINCIPLES. HIGHER STANDARDS.
PROBLEM STATEMENT

A 10 V rms source drives a series RLC circuit with R = 20 Ξ©, L = 100 mH, and C = 100 Β΅F. The branch is protected for 0.25 A. The technician assumes the inductor and capacitor both add opposition, but near 50.3 Hz the current rises to 0.50 A and protection operates. Find the resonance condition, explain why the reactive terms do not add, and choose a safe correction while keeping the source, L, and C unchanged.

Linear Systems Electrical & Computer Design Verification Root Cause Analysis Safe Correction
E-021
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DEBRIEF (LOCKED)

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Find the root cause, confirm the fix, and see how this connects to the exam.

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CASE CLOSED Β· ENGINEERING VERDICT

What actually failed β€” and what should change.

What tripped protection

Series resonance reduced the impedance to about 20 Ξ©, producing 0.50 A.

What was missed

Inductive and capacitive reactances have opposite signs and cancel at resonance.

Corrected model

Use Z = R + j(XL - XC); at resonance, |Z| = R and I = V/R.

QUICK ANSWER

In a series RLC circuit, impedance is Z = R + j(XL - XC). At resonance XL = XC, so the reactive terms cancel and the impedance magnitude becomes R. With 10 V rms and 20 Ξ©, the resonant current is 0.50 A.

ENGINEERING TAKEAWAY

Use Z = R + j(XL - XC). At series resonance XL = XC, the reactive part is zero, |Z| = R, and current reaches its maximum value V/R.

CASE QUESTIONS

Questions this case should settle.

What happens to current at series resonance?

The net reactance becomes zero, the impedance magnitude is minimized to the series resistance, and current reaches its maximum value for a fixed source voltage.

What is the resonant frequency for 100 mH and 100 Β΅F?

f0 = 1/(2Ο€βˆšLC) β‰ˆ 50.33 Hz.

Do inductive and capacitive reactance add at resonance?

Their magnitudes are equal, but their signs are opposite, so they cancel in the series impedance.

NEXT FE PRACTICE QUESTION

The Transfer Function That Settled at 5, Not 20

Linear Systems Β· 5–10 min

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WHAT YOU’LL LEARN

Real failures. Lasting skills.

Each case is designed to build the judgment, analysis, and confidence you need for engineering exams β€” and beyond.

  • Apply core engineering concepts to real-world problems
  • Practice structured troubleshooting and analysis
  • Strengthen exam-ready thinking through realistic scenarios
COMMON QUESTIONS
Do I need prior knowledge to use these cases?

Basic subject familiarity helps, but every case is designed to teach through the investigation itself.

How long does a case take?

Most cases are designed for a focused 5–10 minute investigation.

Are the cases aligned with engineering exams?

Each case is mapped to a verified exam, subject, topic, and misconception before publication.

What happens after I complete a case?

The sealed debrief unlocks with the root cause, corrected reasoning, fix, and takeaway.