FE Electrical & Computer Practice Question Solution included after investigation
CASE E-019 · FE

The Voltage Divider That Collapsed Under Load

The divider measured exactly 6 V with no load. Connect one load, and the output falls to 4 V.

▶ INTERACTIVE CASE FILE • 6–10 MIN • CLICK TO INVESTIGATE

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

Case E-019 engineering failure visual
HOW THIS WORKS

Think like
a real engineer.

Follow the investigation process used in the field — 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 12 V source feeds a divider made from a 10 kΩ top resistor and a 10 kΩ lower resistor. The midpoint measures 6 V with no load. A 10 kΩ load is then connected from the midpoint to ground, and the output falls to 4 V. Diagnose why the unloaded calculation no longer applies and choose a correction that restores 6 V for this known load.

Circuit Analysis Electrical & Computer Design Verification Root Cause Analysis Safe Correction
E-019
INVESTIGATION WORKSPACE WORK THROUGH THE CASE — YOUR CHOICES MATTER.
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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 · ENGINEERING VERDICT

What actually failed — and what should change.

What changed

The 10 kΩ load paralleled the 10 kΩ lower resistor, reducing the effective lower leg to 5 kΩ.

What was missed

The unloaded divider formula was applied after the load changed the network.

Corrected model

Reduce the actual loaded network first, then calculate Vout from the resulting divider ratio.

QUICK ANSWER

A finite load changes a voltage divider because it becomes part of the lower leg. Here, a 10 kΩ load in parallel with the 10 kΩ lower resistor gives 5 kΩ, so Vout = 12 × 5/(10+5) = 4 V instead of 6 V.

ENGINEERING TAKEAWAY

Before applying the voltage-divider equation, include the connected load in the circuit model. Reduce the loaded lower leg to its equivalent resistance, then calculate the divider output.

CASE QUESTIONS

Questions this case should settle.

Why does a voltage divider output drop under load?

A load connected from the output to ground is in parallel with the lower divider resistor. That reduces the effective lower resistance and changes the divider ratio.

What is 10 kΩ in parallel with 10 kΩ?

The equivalent resistance is 5 kΩ.

How do you calculate a loaded voltage divider?

First combine the load with the resistor it parallels, then use that equivalent resistance in the voltage-divider equation.

NEXT FE PRACTICE QUESTION

The Parallel Load That Tripped the Supply

Circuit Analysis · 6–10 min

Continue Practice →
WHAT YOU’LL LEARN

Real failures. Lasting skills.

Each case is designed to build the judgment, analysis, and confidence you need for engineering exams — 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?

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Are the cases aligned with engineering exams?

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What happens after I complete a case?

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