Topic 4 of 6
Sources with internal resistance
A source can transfer energy internally as well as to its external load. Its e.m.f. and its terminal p.d. therefore need not be equal while it supplies current.
The e.m.f. Es is energy supplied by the source per unit charge. The terminal p.d. Vterminal is energy transferred to the external circuit per unit charge. Both are measured in volts, or J/C. The subscript s distinguishes source e.m.f. from other uses of E; see the source-energy explanation.
Model the source as an ideal e.m.f. from C to D, followed by a resistance r from D to A. Current leaves external terminal A, passes through the load and returns to C. The internal drop is Ir:
Vterminal = Es - Ir
I = Es/(R + r) for external load R
These signs describe a discharging source supplying the load. Do not apply them unchanged when an external supply drives current back into a rechargeable source. Here r is an internal resistance in ohms, not a geometrical radius.
For example, a source e.m.f. of 1.60 V and terminal p.d. of 1.50 V while supplying 0.400 A imply r = (1.60 - 1.50)/0.400 = 0.250 Ω. The terminal reading alone would not reveal the e.m.f. without the current and source model.
Read voltage and power as the load changes
A few volts across a total resistance of a few ohms suggest a current of order 1 A and powers of a few watts. A very small external resistance need not produce a large useful output: the internal heating can dominate.
Separate the source from its external terminals
D belongs to the internal model. The external terminal reading is across A/C, after the internal voltage drop.
Terminal voltage falls as current rises
The intercept gives the e.m.f.; the negative gradient gives the internal resistance. The 6 A endpoint is a model limit, not an instruction to short the source.
Load power has a maximum
Increasing current eventually increases the internal loss enough to reduce the power delivered to the load. The maximum is 9 W at 3 A.
Take Es = 6.0 V and r = 1.0 Ω, held constant. Substituting each external load gives:
| Load R / Ω | I / A | Vterminal / V |
|---|---|---|
| 5.0 | 1.0 | 5.0 |
| 1.0 | 3.0 | 3.0 |
| 0.25 | 4.8 | 1.2 |
The V-against-I graph has e.m.f. as its vertical intercept and gradient -r. Here it passes through (0 A, 6 V), (1 A, 5 V), (3 A, 3 V) and the mathematical limit (6 A, 0 V).
Use power as the rate of energy transfer to account for the two destinations:
Pinternal = I2r
Psource = EsI = Pload + Pinternal
| R / Ω | Load power / W | Internal power / W |
|---|---|---|
| 5.0 | 5.0 | 1.0 |
| 1.0 | 9.0 | 9.0 |
| 0.25 | 5.76 | 23.04 |
For this source, load power reaches 9.0 W at 3.0 A, when R = r. Reducing the load further increases current but reduces terminal p.d. enough that load power falls. Equivalently, Pload = Es2R/(R + r)2.
Maximum load power is not maximum efficiency. At R = r, equal powers go to the load and internal heating, giving 50% of source power to the load. A larger R can give a greater fraction to the load while transferring less power.
With an open circuit, I = 0, so the ideal voltmeter reads Es and both load and internal powers are zero. At the mathematical short-circuit limit R = 0, this model gives I = 6 A, terminal p.d. zero and 36 W of internal heating. This limiting calculation is not a measurement procedure.
Infer source parameters from measurements
Vary a suitable external load while measuring total current and source-terminal p.d. across A/C. Choose ranges that include the expected values, retain instrument resolution and avoid source heating or depletion that changes Es or r during the record. A high-resistance voltmeter makes its own loading small.
A generated example with I = 0, 0.1, 0.2, 0.3, 0.4 A and V = 6.0, 5.9, 5.8, 5.7, 5.6 V gives intercept 6.0 V and gradient -1.0 V/A, hence r = 1.0 Ω. For actual readings, inspect the spread and possible drift before adopting one straight-line source model; do not force the intercept to an assumed e.m.f.