Real generator
U = E − r · I
| Quantity | Symbol | Unit |
|---|---|---|
| EMF | E | V |
| Internal resistance | r | Ω |
| Terminal voltage | U | V |
Elec Class
learn electricity hands-on
Lesson · chapter 7 of 15
A real voltage source (battery, cell) has an EMF E, the voltage it produces with no load, and an internal resistance r: when it delivers a current I, its terminal voltage drops to U = E − r × I. A load such as a motor does the opposite: U = E' + r' × I.
The result is worked out before your eyes, line by line, just like on the board.
1I'm looking for
2I enter the known values
Formula used U = E − r · I
3The result
Terminal voltage U11 V
A source with an EMF of 12 V and an internal resistance of 500 mΩ delivering 2 A has 11 V across its terminals.
The definition, the key points to remember and everyday examples.
A real generator has an EMF E and an internal resistance r: when it delivers I, its terminal voltage drops to U = E − r·I. At no load U = E; in short circuit I_sc = E / r.
An active load (motor, charging battery) has a back-EMF E' and an internal resistance r': U = E' + r'·I. The power E'·I becomes mechanical or chemical, r'·I² is lost as heat. In a loop with generator and load: I = (E − E') / (r + r' + R).
A voltage source is like a pump pushing water around a circuit. The pressure it creates is E; but the water first has to go through the narrow pump itself and loses pressure there, more so as the flow increases: that's r × I. With no flow, all the pressure is available (U = E). Where it falls short: internal resistance isn't a component you can remove; it increases as the battery wears out or when it's cold.
| Quantity | Symbol | Unit |
|---|---|---|
| EMF | E | V |
| Internal resistance | r | Ω |
| Terminal voltage | U | V |
| Quantity | Symbol | Unit |
|---|---|---|
| Back-EMF | E' | V |
| Internal resistance | r' | Ω |
| Quantity | Symbol | Unit |
|---|---|---|
| Current | I | A |
| Sum of loop resistances | ΣR | Ω |
Four steps to solve a problem, then an example solved line by line.
Replace each source with E in series with r, and each load with E' in series with r'.
Current leaves a source through its + terminal and enters a load through its + terminal: that sets the signs.
I = (ΣE − ΣE') / ΣR, adding up all the resistances in the loop, internal ones included.
U = E − r × I for the source, U = E' + r' × I for the load; then check that E × I = U × I + r × I².
A cell of EMF E = 9 V and internal resistance r = 1.5 Ω supplies a resistor R = 28.5 Ω. What current does it deliver, what is its terminal voltage? What would the short-circuit current be?
Generalised Ohm's law: I = E / (r + R) = 9 / (1.5 + 28.5) = 0.3 A.
U = E − r · I = 9 − 1.5 × 0.3 = 8.55 V: the internal resistance "eats" 0.45 V.
In short circuit (R = 0): I_sc = E / r = 9 / 1.5 = 6 A, and all the power E·I heats the cell.
It's the voltage it produces when it delivers no current, written E and measured in volts. Despite its name, it isn't a force. A new AA battery has an EMF of about 1.5 V.
Subtract the drop across the internal resistance from the EMF: U = E − r × I. A 12 V, 0.5 Ω battery delivering 2 A gives U = 12 − 0.5 × 2 = 11 V.
Because the voltmeter draws almost no current: it measures U ≈ E. The wear shows up in the internal resistance, which only acts under load; to test a battery, measure its voltage while it powers the device.
It's the voltage E' the motor produces itself as it turns, which opposes the supply voltage. It is proportional to speed: zero at start-up, hence the high inrush current I = U / r'.
In series, the EMFs and the internal resistances add up: 4 cells of 1.5 V and 0.2 Ω give 6 V and 0.8 Ω. In parallel, identical cells keep the same EMF and their internal resistance is divided by their number.
When the load resistance equals its internal resistance (R = r): it then delivers E² / (4r), but half the energy heats the source. With E = 12 V and r = 2 Ω, a 2 Ω load gets 3 A and 18 W.
Because a current flows through its internal resistance r and causes a drop r·I. The larger the current (or the older the cell, with a larger r), the more the voltage collapses.
Take two points of the U = f(I) characteristic (for instance at no load and under load): r is the slope −ΔU/ΔI. Never with an ohmmeter, which must not be connected to a source.
From the simplest to the most complete, each problem comes with a step-by-step solution.
Record the U = f(I) characteristic of a 4.5 V battery feeding a rheostat (6 to 8 points, without going below a few ohms), plot the line and deduce E (the intercept) and r (the slope); compare a new battery with a worn-out one.
Using Elec Class in class