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Ideal and Practical Voltage Source

Important Formula

Formula: Practical (Real) Voltage Source

$$U = \frac{\class{blue}{U_0}}{1 + \frac{R_{\text i}}{R}}$$ $$U = \frac{\class{blue}{U_0}}{1 + \frac{R_{\text i}}{R}}$$ $$\class{blue}{U_0} ~=~ \left( 1 + \frac{ R_{\text i} }{ R } \right) \, U$$ $$R_{\text i} = \left( \frac{\class{blue}{U_0}}{U} - 1 \right) \, R$$ $$R = \left( \frac{\class{blue}{U_0}}{U} - 1 \right)^{-1} \, R_{\text i}$$

Rearrange formula

What do the formula symbols mean?

Terminal Voltage

$$ U $$ Unit $$ \mathrm{V} $$

Electrical voltage of an ideal voltage source MINUS the voltage across the internal resistance $R_{\text i}$.

It can be seen from the formula that the terminal voltage depends on both the internal resistance of the voltage source and the load resistance.

Source voltage

$$ \class{blue}{U_0} $$ Unit $$ \mathrm{V} $$

Voltage of an ideal voltage source - by definition, it has infinite resistance.

Internal resistance

$$ R_{\text i} $$ Unit $$ \mathrm{\Omega} $$

Resistance of a practical (real) voltage source.

Load resistance

$$ R $$ Unit $$ \mathrm{\Omega} $$

Electrical resistance connected to the terminals (to which the terminal voltage is applied). The terminal voltage and source voltage are equal when the internal resistance is zero.

An ideal voltage source has no internal resistance, meaning it provides a voltage $U_0$ (referred to as open-circuit voltage or source voltage) that remains independent of the resistance $R$ connected to the terminals of the voltage source. This implies that the voltage across the resistance $R$ is always the voltage $U_0$.

Circuit with an Ideal Voltage Source

To keep the voltage $U_0$ constant, the current $I$ can become arbitrary depending on the choice of resistance. This can result in very high currents that may damage the circuit.

A practical voltage source, on the other hand, has an internal resistance $R_{\text i}$ that limits the current $I$ in the circuit. Applying Kirchhoff's voltage law results in a voltage $U$ across the resistor $R$, which may not necessarily be equal to the source voltage $U_0$:

$$ \begin{align} U ~=~ U_0 - R_{\text i} \, I \end{align} $$

Practical voltage source with a load resistor.

So that a practical voltage source corresponds as closely as possible to the ideal voltage source, the internal resistance $R_{\text i}$ must be chosen as small as possible so that the second term in 2 is virtually eliminated and then applies: $ U \approx U_0 $.

Let's continue considering a practical voltage source. Apply a voltage $U$ across a measurement resistor $R$. The relationship between voltage and resistance is described by Ohm's Law:

$$ \begin{align} U ~=~ R \, I \end{align} $$

Circuit with a voltmeter

To measure this voltage at the measuring resistor, a voltmeter is connected in parallel to the measuring resistor. This voltmeter has an internal resistance $ R_{\text v} $. As a result, the voltage value at the measuring resistor changes, because now a fraction $I_{\text R}$ of the total current $I = I_{\text R} + I_{\text v} $ flows through the measuring resistor and a fraction $I_{\text v}$ through the voltmeter:

$$ \begin{align} U ~&=~ R_{\text{ges}} \, I ~~\leftrightarrow \\\\
\frac{U}{R_{\text{ges}}} ~&=~ I_{\text R} + I_{\text V} \end{align} $$

The voltage at the measuring resistor and voltmeter are of course the same $U = R \, I_{\text R}$ and $U = R_{\text V} \, I_{\text V}$, so that the total resistance can be specified with Eq. 2:

$$ \begin{align} R_{\text{ges}} = \frac{R \, R_{\text V}}{R + R_{\text V}} \end{align} $$

This means that the measured voltage is different from the actual voltage 1 without a voltmeter:

$$ \begin{align} U ~&=~ I \, \frac{R \, R_{\text V}}{R + R_{\text V}} \\\\
&=~ I \, \frac{R}{\frac{R}{R_{\text V}} + 1} \end{align} $$

The measured voltage is lower than the actual voltage due to $ R \geq R \, R_{\text V} / (R + R_{\text V}) $. In order to be able to reliably measure a voltage at the measuring resistor $R$, the voltage source must have the highest possible internal resistance $R_{\text V} $.

Difference between real and ideal voltage source

In contrast to a real voltage source, an ideal voltage source has an infinite internal resistance $R_{\text V} \rightarrow \infty $. An ideal voltage source therefore does not distort the actual voltage across the measuring resistor.

Is it allowed to touch the electrode of a welding machine during arc welding?

Arc welding uses a voltage source that specifies an open-circuit voltage $U_0$ that is applied between the workpiece and the welding electrode. A real voltage source also has an internal resistance $ R_{\text i} $. The welding resistance $ R $ is the electrical resistance between the electrode and the workpiece. This resistance increases when the electrode is moved further away from the workpiece and decreases when the electrode is moved towards the workpiece. If the electrode is brought so close to the workpiece that the welding resistance is very low, an arc with a high current $I$ is created between the electrode and the workpiece. Sounds dangerous at first.

Voltage source with an internal resistance and the resistance $R$ between the electrode and the workpiece.

As shown in the illustration, the following relationship applies according to the mesh current method and Ohm's law:

$$ \begin{align} U ~=~ \frac{U_0}{\frac{R_{\text i}}{R} + 1 } \end{align} $$

Here $U$ is the voltage between the electrode and the workpiece.

Case 1: If the electrode is removed from the workpiece, the welding resistance $ R $ becomes high. This means that in 1 the quotient $ R_{\text i} / R$ is small and in the limiting case: U = U_0.

However, since the current $ I = \frac{U_0}{R} $ becomes very small with a large $ R $, the electrode can be touched.

Example: When not welding

Let $U_0 = 50 \, \text{V} $, $ R_{\text i} = 1 \, \Omega $ and $ R = 100 \, \text{k}\Omega $. Then the welding voltage is $ U = 50 \, \text{V} $. And the current $ I = 0.5 \, \text{mA} $

Case 2: When the electrode is brought closer to the workpiece, the welding resistance $ R $ becomes small (so small that an arc is created). With a small $ R $, the ratio $ R_{\text i} / R $ in 1 becomes large. However, this makes $U$ small. Consequently, it would not be dangerous to touch the electrode during welding because the voltage is small.

Example: During welding

Let $U_0 = 50 \, \text{V} $, $ R_{\text i} = 1 \, \Omega $ and $ R = 0.3 \, \Omega $. Then the welding voltage is just $ U = 11.5 \, \text{V} $. But the current $ I = 38 \, \text{A} $ between the electrode and the workpiece.