8867 / 2027
Circuits overview

Lesson 4 of 6

Sources with internal resistance

  • I can explain how source internal resistance affects terminal p.d. and output power.Syllabus

    Syllabus 8867, 9(g). Explain how source internal resistance affects terminal p.d. and output power.

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A cell with internal resistance drives a current through a variable resistor. The resistor is turned down to a smaller resistance. What happens to the terminal p.d. of the cell?

  1. It rises, because a larger current needs a larger terminal p.d. across the load.
  2. It stays equal to the e.m.f.
  3. It falls, because the larger current loses more volts inside the cell.
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It falls, because the larger current loses more volts inside the cell.

A smaller external resistance gives a larger current. The lost volts Ir inside the cell grow with the current, so less of the e.m.f. appears across the terminals.

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:

Es = Ir + Vterminal
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

Separate the source from its external terminalsThe dashed source boundary contains a six-volt ideal e.m.f. between C and the private node D, and one-ohm internal resistance between D and A. A five-ohm external load connects A to C. The voltmeter taps are A and C, not D and C. Negligible meter current gives one ampere through the load, five volts at the external terminals and one volt across the internal resistance.+-VDACr = 1.0 ΩE = 6.0 VSource5.0 ΩLoad1.0 AV between A and C = 5.0 V

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

Terminal voltage falls as current risesFor the supplied e.m.f. of six volts and one-ohm internal resistance, terminal p.d. is six minus current in volts when current is in amperes. The line contains zero amperes and six volts, one and five, three and three, and six and zero. The last point is a mathematical short-circuit limit, not a suggested measurement.01234560246Terminal p.d. / VCurrent I / A

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

Load power has a maximumLoad power equals six I minus I squared in watts for current from zero to six amperes. This downward parabola starts and ends at zero and has its maximum of nine watts at three amperes. The one-ampere load receives five watts and the 4.8-ampere load receives 5.76 watts. Maximum load power is not maximum efficiency.01234560369Load power / WCurrent I / A(3 A, 9 W)

Increasing current eventually increases the internal loss enough to reduce the power delivered to the load. The maximum is 9 W at 3 A.

The terminal measurement spans A/C and excludes the ideal source's private node D. The two graphs use the same 6.0 V e.m.f. and 1.0 ohm internal-resistance model, with current horizontal and voltage or load power on separate vertical axes.

Take Es = 6.0 V and r = 1.0 Ω, held constant. Substituting each external load gives:

Supplied source model: load, current and terminal p.d.
Load R / ΩI / AVterminal / V
5.01.05.0
1.03.03.0
0.254.81.2

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:

Pload = IVterminal = I(Es - Ir)
Pinternal = I2r
Psource = EsI = Pload + Pinternal
Power destinations for the same three loads
R / ΩLoad power / WInternal power / W
5.05.01.0
1.09.09.0
0.255.7623.04

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.

Try it step by step

Work it out in steps: Current I = E / (R + r) in A, then Terminal p.d. V = IR in V, then Lost volts Ir in V.

  1. Problem 1. A battery of e.m.f. 12 V and internal resistance 0.50 ohm is connected to a 5.5 ohm resistor. Find the current, the terminal p.d. and the lost volts.
  2. Problem 2. A cell of e.m.f. 9.0 V and internal resistance 1.0 ohm is connected to an 8.0 ohm resistor. Find the current, the terminal p.d. and the lost volts.
  3. Problem 3. A cell of e.m.f. 6.0 V and internal resistance 0.80 ohm is connected to a 2.2 ohm resistor. Find the current, the terminal p.d. and the lost volts.
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Problem 1.

Current I = E / (R + r) in A: I = 12 / (5.5 + 0.50) = 2.0 A

Terminal p.d. V = IR in V: V = 2.0 x 5.5 = 11 V

Lost volts Ir in V: Ir = 2.0 x 0.50 = 1.0 V (check: 11 + 1.0 = 12 V)

Problem 2.

Current I = E / (R + r) in A: I = 9.0 / (8.0 + 1.0) = 1.0 A

Terminal p.d. V = IR in V: V = 1.0 x 8.0 = 8.0 V

Lost volts Ir in V: Ir = 1.0 x 1.0 = 1.0 V

Problem 3.

Current I = E / (R + r) in A: I = 6.0 / (2.2 + 0.80) = 2.0 A

Terminal p.d. V = IR in V: V = 2.0 x 2.2 = 4.4 V

Lost volts Ir in V: Ir = 2.0 x 0.80 = 1.6 V

Check your understanding

A discharging source has e.m.f. 6.0 V and internal resistance 1.0 ohm. It supplies a 1.0 ohm external load. What is the power account?

  1. The load receives 9 W and the internal resistance receives 9 W.
  2. 18 W reaches the load because source power is e.m.f. times current.
  3. The load receives 9 W with no internal transfer, so this is maximum efficiency.
Show the answer

The load receives 9 W and the internal resistance receives 9 W.

I = 6.0/(1.0 + 1.0) = 3.0 A. Each 1.0 ohm resistance receives I2 R = 9 W, summing to the 18 W source conversion.

Practise exam questions

Write your answer first, then tick the mark points it makes.

Exam-style question

Explain why the terminal potential difference of a cell is less than its e.m.f. when it supplies a current. [2]

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Exam-style question

A student varies the current from a cell and plots terminal p.d. V against current I. The line is straight, meets the V-axis at 1.50 V and passes through (0.50 A, 1.10 V). Use the graph to find the e.m.f. and the internal resistance of the cell. [3]

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Exam-style question

Data-based question: A portable speaker battery pack [15]

A portable speaker runs from a rechargeable battery pack labelled "5000 mA h". A student connects the pack to a variable resistor and measures the terminal potential difference V for different currents I.

Table 1. Terminal p.d. of the battery pack
I / A0.501.002.003.004.00
V / V7.327.247.086.926.76

At normal volume the speaker draws 1.20 A. At full volume it draws 4.0 A.

  1. (a) Explain why the terminal p.d. falls as the current increases. [2]
  2. (b) Use the data to determine the e.m.f. E and the internal resistance r of the pack. [3]
  3. (c) Express the charge stored, 5000 mA h, in coulombs. [1]
  4. (d) Calculate how long the speaker can play at normal volume on a full charge. [1]
  5. (e) At normal volume, calculate the power delivered to the speaker, the power wasted in the pack, and the efficiency of the pack. [3]
  6. (f) The maker says the pack "is just as efficient at full volume". Use the data to evaluate this claim. [3]
  7. (g) The pack is recharged with a constant current of 2.0 A. Estimate the charging time and state one assumption you made. [2]

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