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SCIENCE · INVESTIGATE AND EXPLAIN

Close a source, internal-loss and load-power ledger

An ideal emf E supplies energy per charge inside the source.

Learning outcome: Compare loads below, equal to and above the internal resistance. Retain terminal voltage, source power, each resistor power and load fraction. Differentiate the load-power expression if choosing the optional maximum challenge, then contrast the result with efficiency.

Starting recap

Integrate current or resistance, preserve node topology and derive transient state with a signed energy account.

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Try

With E = 12 V, r = 2 ohm and R = 4 ohm, I = 2 A and terminal voltage is 8 V. Source power is 24 W, split into 8 W internal and 16 W load. Using 12 V rather than 8 V in load power would give the wrong account.

Open the experiment

Predict

Which voltage belongs in V²/R for the external load?

Experiment

Compare loads below, equal to and above the internal resistance. Retain terminal voltage, source power, each resistor power and load fraction. Differentiate the load-power expression if choosing the optional maximum challenge, then contrast the result with efficiency.

Close a nonideal source power account

Worked teaching model · synthetic values · Ideal emf 12 V in series with selected internal resistance and one load. All parameters fixed; load not a real bulb temperature model. Virtual positive resistances only, no physical short-circuit experiment.

Reset keeps your prediction, captures and practice history.

Load power for the fixed internal resistance

power (W)load resistance (ohm)11217.99190

Axes rescale independently. Curves join sampled coordinates; a joining segment is not additional measured evidence. Read physical units rather than screen angles or apparent slopes.

Load power for the fixed internal resistance: numerical coordinates
load resistance (ohm)power (W)
116
1.1833316.8153
1.3666717.363
1.5517.7108
1.7333317.9082
1.9166717.9919
2.117.9893
2.2833317.9212
2.4666717.8035
2.6517.6483
2.8333317.4649
3.0166717.2607
3.217.0414
3.3833316.8114
3.5666716.5743
3.7516.3327
3.9333316.0889
4.1166715.8445
4.315.6009
4.4833315.3591
4.6666715.12
4.8514.8841
5.0333314.652
5.2166714.4239
5.414.2001
5.5833313.9809
5.7666713.7663
5.9513.5564
6.1333313.3512
6.3166713.1508
6.512.955
6.6833312.7639
6.8666712.5773
7.0512.3952
7.2333312.2175
7.4166712.0442
7.611.875
7.7833311.7099
7.9666711.5489
8.1511.3917
8.3333311.2383
8.5166711.0886
8.710.9424
8.8833310.7998
9.0666710.6605
9.2510.5244
9.4333310.3916
9.6166710.2618
9.810.135
9.9833310.0111
10.16679.89004
10.359.77167
10.53339.65595
10.71679.5428
10.99.43213
11.08339.32387
11.26679.21795
11.459.1143
11.63339.01286
11.81678.91355
128.81633
Calculated values under the declared model
QuantityValueUnit
Circuit current2A
Loaded terminal voltage8V
Source power supplied24W
Internal resistor power8W
Load power16W
Power residual0W
Load-power fraction0.666667ratio

Terminal voltage falls during discharge through internal resistance; source power splits between internal and external components.

Loop balance gives emf = I(r + R). Therefore terminal voltage is emf − Ir = IR, and source power equals I²r + I²R. Optional differentiation shows maximum load power at R = r, which is not maximum efficiency. Comparing brightness by power requires comparable bulbs and is qualitative, not a universal luminous-output calibration.

Derive before checking. Retain contrasting captures and explain one assumption that could fail; calculated evidence is not actual laboratory work.

Terminal voltage falls during discharge through internal resistance; source power splits between internal and external components.

Loop balance gives emf = I(r + R). Therefore terminal voltage is emf − Ir = IR, and source power equals I²r + I²R. Optional differentiation shows maximum load power at R = r, which is not maximum efficiency. Comparing brightness by power requires comparable bulbs and is qualitative, not a universal luminous-output calibration.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Distinguish emf from loaded voltage

An ideal emf E supplies energy per charge inside the source. A simple nonideal battery model places internal resistance r in series. With load R, loop balance gives I = E/(r + R), while terminal voltage is E − Ir = IR during discharge. At zero current the terminal value approaches emf. The displayed model does not describe chemical aging or a safe physical power supply.

Track each energy destination

Source power is EI. Internal and load resistors receive I²r and I²R, whose sum equals EI. The equivalent forms VI, I²R and V²/R use the voltage across the particular resistor, not the source emf indiscriminately. Thermal conversion can change real component temperature; our stated fixed parameters exclude that feedback.

Compare power and efficiency

Optional differentiation of load power E²R/(r + R)² gives a maximum at R = r. At that point half the source power is lost internally, so maximum delivered power is not maximum load fraction. For comparable idealized bulbs, larger dissipated power suggests greater brightness qualitatively; no universal conversion from watts to perceived brightness is supplied.

Respect omitted and mechanical elements

A wire’s resistance may be negligible compared with other large resistances, but not when it is the only element limiting current. An ideal zero-resistance load directly across an ideal voltage source makes the model singular rather than predicting a realizable finite current. For a separate ideal motor packet, 3 V at 2 A gives 6 W electrical input; if mechanical conversion is stipulated 75%, mechanical output is 4.5 W and the remaining 1.5 W needs an explicit loss destination.

Worked original case

With E = 12 V, r = 2 ohm and R = 4 ohm, I = 2 A and terminal voltage is 8 V. Source power is 24 W, split into 8 W internal and 16 W load. Using 12 V rather than 8 V in load power would give the wrong account.

Investigation

Compare loads below, equal to and above the internal resistance. Retain terminal voltage, source power, each resistor power and load fraction. Differentiate the load-power expression if choosing the optional maximum challenge, then contrast the result with efficiency.

Changed-case practice

Try a new practice case

For E = 12 V, r = 3 ohm and R = 3 ohm, what is load power in watts?

This public, unsaved check is not a secure assessment. Explain the source integral, symmetry or charge boundary in the retained course evidence workspace.

Sources and evidence boundary

Original teaching checked 31 October 2026 against the current official framework, Unit 11 and OpenStax electrical energy and power. All diagrams and numerical records are virtual, supplied or generated. No physical wiring, battery connection, short circuit, meter operation or electrical handling is requested. Public feedback and retained self-reports do not certify independent reasoning, actual laboratory participation or institutional credit.

Apply

Which voltage belongs in V²/R for the external load?

Public practice checks a specific claim. It does not establish independent mastery.

Which source, topology, initial condition or instrument assumption would invalidate this inference?

Continue

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