Explain voltage sag inside a source

The source has an inside and an outside

A source labeled 6 V need not maintain 6 V while delivering current. Here an ideal emf of 6 V sits in series with an internal resistance r and an external load R. Emf is energy supplied per charge, measured in volts, not a mechanical force. Terminal voltage is the potential difference across the source’s external terminals, also across the load in this circuit.

The dashed boundary contains the ideal source and internal resistor. The dotted span between terminals is a voltage-measurement annotation, not a wire that shorts them. Real internal resistance can summarize several physical effects; the drawing does not claim there is a removable resistor inside every battery.

Find one point that satisfies two conditions

The source relation is V terminal = emf − I r. The load relation is V terminal = I R. Here I is discharge current in amperes, R and r are resistances in ohms, and all voltage terms are in volts. Combining them gives I = emf/(R + r). These steady-state relations are developed in OpenStax: electromotive force.

The solid source line slopes downward as current increases; the dashed load line slopes upward. Their intersection is the state satisfying both equations. The graph keeps the same 0–60 mA and 0–6 V scales. It is a relation between possible states, not a trace of a source changing over time. Line segments include theoretical states outside the selectable load presets.

At the default R = 200 Ω and r = 40 Ω, current is 6/240 = 0.025 A = 25 mA. The internal drop is 0.025 × 40 = 1 V, leaving 5 V across the load. Check that the point lies on both lines, then explain why using 6/200 would overestimate this current.

Change the load, keeping the source model fixed

Before moving the load slider, predict both current and terminal voltage at the two extremes with r = 40 Ω.

Load	Current	Terminal voltage
500 Ω	11.111 mA	5.556 V
100 Ω	42.857 mA	4.286 V

Capture both. Lower load resistance draws more current, causing a larger internal drop. Terminal voltage falls even though the emf stays at 6 V. Returning to the original setting returns the same calculation: this model has no depletion or history.

Now set r = 0. The source line becomes horizontal, terminal voltage is always 6 V for the allowed loads, and internal dissipation vanishes. This recovers the earlier ideal-source assumption. The zero-current intercept is an unloaded limit; no control opens the circuit or performs an actual open-circuit measurement.

Reconcile the power before discussing efficiency

At the default setting, emf-source power is (6 V)(0.025 A) = 0.150 W. The load receives (5 V)(0.025 A) = 0.125 W. Internal dissipation is I²r = (0.025)² × 40 = 0.025 W. The two destinations sum to source output. Energy is transferred and dissipated, not created or destroyed by the bookkeeping.

The load fraction is P load/P source = R/(R + r), or 83.33% at the default. This is the model’s electrical delivery efficiency, not battery capacity, lifetime or a complete chemical efficiency. At r = 0 it is 100% only because every modeled loss is absent.

With r = 40 Ω, increasing R from 100 to 500 Ω raises the fraction from 71.43% to 92.59%, while load power falls from approximately 0.1837 to 0.0617 W. A larger percentage does not necessarily deliver more useful power. The model cannot calculate a temperature rise from internal dissipation without a thermal model.

Transfer to another source

Outside the fixed-emf browser, suppose emf = 9 V, r = 50 Ω and R = 250 Ω. Calculate current, terminal voltage, load power, internal power and the load fraction.

Check the nonideal-source transfer

I = 9/300 = 0.030 A. Terminal voltage is 0.030 × 250 = 7.5 V, also 9 − 0.030 × 50. Load power is 0.225 W; internal power is 0.045 W; their sum is 0.270 W = 9 × 0.030. The load fraction is 0.225/0.270 = 83.33%. None of these values gives a discharge duration without information about stored energy and changing source behavior.