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Learn through experiments

More current, less useful power?

Can drawing more current make a power source less effective?

Entering midway? Revisit Hit the target without hitting the ceiling for the preceding worked example, or start here and use the explanation below.

1 · Try

Start with a concrete question

Original synthetic circuit: an ideal voltage source has a series internal resistance and supplies one resistive load.

Try the initial inputs, then write a prediction before changing one setting. All calculations run locally on your CPU.

Go directly to the experiment · Review basic mathematics

2 · Predict

What do you expect?

With 2 ohms of internal resistance, predict whether reducing the load from 6 to 1 ohm always increases useful power.

3 · Experiment

Change one thing. Inspect what happens.

  1. Keep 12 V and 2 ohms internal resistance; compare load values 6, 2 and 1.
  2. Find a load with the largest useful power and compare internal loss at that setting.
  3. Set internal resistance to zero and explain why the behavior changes.
  4. Choose a setting for low internal loss, then explain the useful-power trade-off.
Experiment inputs
Inputs
Your settingsMechanism
Calculation belowEvidence
Compare outcomes
Current1.5 A
Load voltage9 V
Useful power13.5 W
Internal loss4.5 W

Compare the current results

Every bar starts at zero and uses the same numeric scale in this chart. The upper bound automatically adjusts to the largest current value (at least 1), so compare the numbers when changing inputs. Units and quantities come from each label; different kinds of quantity are not interchangeable. Displayed values are rounded to at most six significant digits.
013.5
  • Useful power (W)13.5
    Useful power (W): 13.5Horizontal bar from zero to 13.5, on the shared zero to 13.5 numeric axis.
  • Internal loss (W)4.5
    Internal loss (W): 4.5Horizontal bar from zero to 4.5, on the shared zero to 13.5 numeric axis.

Follow the calculation

  1. Total resistance = 6 + 2 = 8 ohms.
  2. Current = voltage / resistance = 12 / 8 = 1.50 A.
  3. Load voltage = 1.50 × 6 = 9.00 V.
  4. Useful power = current² × load resistance = 13.50 W; internal loss = 4.50 W.
  5. Supply power 18.00 W = useful 13.50 + loss 4.50.

Reducing load resistance draws more current, but internal resistance wastes power and reduces the voltage delivered to the load.

Captures keep your inputs, prediction and observation. Results are recalculated from those inputs. Resetting inputs keeps your captures and challenge history.

4 · Understand

Explain the mechanism

What the experiment explains

Original synthetic circuit: an ideal voltage source has a series internal resistance and supplies one resistive load.

Resistance limits current. The same current passes through the load and the internal resistance, so both convert electrical energy into heat.

At the defaults current is 12 / 8 = 1.5 A. Useful power is 1.5² × 6 = 13.5 W and internal loss is 1.5² × 2 = 4.5 W.

For positive internal resistance the maximum load power occurs when load and internal resistance match. That point is only 50% efficient in this model; maximum power and maximum efficiency are different objectives.

Read the relationship

I = V / (R + r); useful power = I²R; internal loss = I²r. I is amperes, V volts, R load resistance and r internal resistance in ohms; power is watts.

Use the trace as evidence

The trace recomputes from your current controls. Change one control at a time, record what changed, and compare the result with your prediction. The bars use the same quantities as the numerical trace; a larger bar is not automatically a better outcome.

Boundaries of the model

  • Ideal constant voltage and resistance; no temperature changes, current limits, or battery chemistry.
  • This is a virtual circuit only; it is not an instruction to connect physical power sources.

Continue with Why double speed needs more than double space to investigate a different mechanism.

5 · Apply

Try a fresh case

A 9 V source has 1 ohm internal resistance and a 2 ohm load. How much power reaches the load in watts?

This is a public practice check. Retries support learning; matching a number does not certify independent proficiency.

Take the investigation further

Choose any extension. These are project prompts for you to carry out; they are not automatically assessed.

School investigation

Check power conservation for three settings using supply power = voltage × current.

College investigation

Differentiate the load-power expression and verify the maximum-power condition.

Workplace investigation

Compare a maximum-power design with a thermal-loss constraint and document missing physical assumptions.

6 · Continue

Keep the evidence, choose your next step

Completion records your own decision that you have worked through this investigation. Practice checks and independent project evidence are separate.