Simple resistor networks

Electrical networks are often composed of simpler sub-circuits. It is often helpful to identify these sub-circuits, simplify them, and then combine them into the overall circuit. Such combinations of components are generally permissible, provided that the terminal behaviour, i.e. the behaviour between current and voltage between the connection points, does not change.

1 Series connection of resistors

In a series connection, the same current \(I\) flows through all components (see Figure 1).

Applying the mesh rule (here: counterclockwise direction of rotation) leads to the following mesh equation:

\begin {equation*} U_0 - U_1 - U_2 =0 \end {equation*}

With the help of Ohm’s law, the partial voltages \(U_1\) and \(U_2\) can be represented as the product of current and resistance:

\begin {equation*} U_0 - R_1 \cdot I - R_2 \cdot I = 0 \end {equation*} \begin {equation*} U_0 - (R_1+R_2) \cdot I = 0 \end {equation*}

In this example, resistors \(R_1\) and \(R_2\) can be combined by adding their resistance values to \(R_1 + R_2 = R_\mathrm {total}\) (Figure 2).

In general, the total voltage \(U_\mathrm {tot}\) can be summarised according to this principle as follows:

\begin {equation} U_\mathrm {tot} = \sum _{k=1}^{n} U_k = \sum _{k=1}^{n} R_k \cdot I = R_\mathrm {tot} \cdot I \label {eq:gesamtwiderstand} \end {equation}

A comparison of the coefficients of the last two terms of equation 1 yields the generally valid result for combining resistors in a series connection:

If, instead of the resistance values \(R\), we consider the conductance values \(G = 1/R\) of the resistors, the total conductance of a series connection is:

2 Parallel connection of resistors

In a parallel connection of resistors, the same voltage is applied to all components (see Figure 3):

\begin {equation*} U_\mathrm {1} = U_\mathrm {2} = U_0 \end {equation*}

By applying the node rule, it can be shown that the total current \(I_0\) before the resistors is divided into the partial currents \(I_1\) and \(I_2\):

\begin {equation} I_0 - I_1 - I_2 = 0 \label {eq:gesamtwiderstand2} \end {equation}

As with series connection in Section 4.1, it is often advantageous to combine resistors connected in parallel into a total resistance \(R_\mathrm {total}\) (see Figure 4):

With the help of Ohm’s law, the unknown partial currents \(I_1\) and \(I_2\) can be quantified from the node rule (equation 4): \begin {equation*} I_1 = \frac {U_0}{R_1}, I_2 = \frac {U_0}{R_2} \end {equation*}

Inserted into equation 4, this gives:

\begin {equation*} I_0 - \frac {U_0}{R_1} - \frac {U_0}{R_2} = 0 \end {equation*}

\begin {equation*} I_0 - \bigg ( \frac {1}{R_1}+\frac {1}{R_2} \bigg ) \cdot U_0 = 0 \end {equation*}

\begin {equation*} \rightarrow I_0 = U_0 \cdot \bigg ( \frac {1}{R_1}+\frac {1}{R_2} \bigg ) \end {equation*}

A coefficient comparison with Ohm’s law applied to Figure 4 ( \(I_0 = U_0 \cdot \frac {1}{R_\mathrm {tot}}\))

provides for \(R_\mathrm {total}\) in the case shown of two resistors connected in parallel:

\begin {equation*} \frac {1}{R_\mathrm {tot}} = \frac {1}{R_1}+\frac {1}{R_2} \end {equation*}

Multiplying out this expression gives the form frequently used in calculations:

\begin {equation} R_\mathrm {tot} = \frac {R_1\cdot R_2}{R_1 + R_2} \label {eq:r2parallel} \end {equation}

The total resistance of a parallel connection of any number of resistors can be determined using the general formula for parallel connections:

Since electrical conductance is described by the reciprocal of resistance, \(G=1/R\), the total conductance of a parallel connection of resistors is often easier to calculate:

Parallel connection of resistorsparallelschaltung

The three resistors \(R_1 = 1 \, \mathrm {k}\Omega \), \(R_2 = 10 \, \mathrm {k}\Omega \), \(R_3 = 100 \, \mathrm {k}\Omega \) are connected in parallel as shown in the following figure. What is the total resistance \(R_\mathrm {total}\) of this parallel connection?

Set up the equation for \(R_\mathrm {total}\) according to equation 6:

\begin {equation*} R_\mathrm {tot} = \frac {1}{\sum _{n} \frac {1}{R_n}} = \frac {1}{\frac {1}{R_1} + \frac {1}{R_2}+ \frac {1}{R_3}} \end {equation*}

Inserting the numerical values and solving using a calculator:

\begin {equation*} R_\mathrm {tot} = \frac {1}{\frac {1}{1\cdot 10^3 \, \Omega } + \frac {1}{10\cdot 10^3 \, \Omega } + \frac {1}{100 \cdot 10^3 \, \Omega }} \end {equation*}

\begin {equation*} \rightarrow R_\mathrm {tot} = 900,9 \, \Omega \end {equation*}

3 Voltage divider on resistors

One application of connecting two resistors in series is the voltage divider. It is used to divide an input voltage \(U_0\) into two smaller partial voltages, thus providing a precisely defined output voltage \(U_2\).

3.1 The unbalanced voltage divider

First, the voltage divider shown in Figure 5 is unloaded, i.e. no load resistor is connected to the open terminals. All components are traversed by the same current \(I\):

\begin {equation*} I = \frac {U_1}{R_1} = \frac {U_2}{R_2} \end {equation*}

By combining the series connection of resistors, the following result is obtained:

\begin {equation*} I = \frac {U_0}{R_1+R_2} \end {equation*}

Equating the two voltage terms yields the ratio equation:

\begin {equation*} \frac {U_2}{R_2} = \frac {U_0}{R_1+R_2} \end {equation*}

Isolating the output voltage on the left side of the equal sign ultimately leads to the general equation for the unloaded voltage divider:

Unloaded voltage dividerSpannungsteiler

An input voltage of \(U_0 = 24\) V is to be reduced to \(U_2 = 6\) V using a voltage divider. What should the ratio of the resistances \(R_2\) to \(R_1\) be? How large should the resistances \(R_1\) and \(R_2\) be if the total current is to be \(I = 10\) mA?

\begin {equation*} \frac {U_2}{U_0} = \frac {6 \, \mathrm {V}}{24 \, \mathrm {V}} = 0,25 = \frac {R_2}{R_1+R_2} \end {equation*}

\begin {equation*} \rightarrow R_2 = 0.25 (R_1+R_2) \end {equation*}

\begin {equation*} 0.75 \, R_2 = 0.25 \, R_1 \end {equation*}

\begin {equation*} R_1 = 3 \, R_2 \end {equation*}

The resistance \(R_1\) must therefore be selected to be three times greater than the resistance \(R_2\).

\begin {equation*} I = \frac {U_2}{R_2} \rightarrow R_2 = \frac {U_2}{I} = \frac {5 \, \mathrm {V}}{0.01 \mathrm {A}} = 500 \, \Omega \end {equation*}

\begin {equation*} R_1 = 3 \cdot R_2 = 3 \cdot 500 \, \Omega = 1500 \, \Omega \end {equation*}

3.2 The loaded voltage divider

If the output side of the voltage divider is loaded as shown in Figure 6, i.e. a load resistor \(R_\mathrm {L}\) is added, then - from the perspective of the voltage source - this results in a parallel connection of the resistors \(R_2\) and \(R_\mathrm {L}\).

If the equivalent resistance \(R_\mathrm {new}\) of this parallel connection is determined via \(\frac {1}{R_\mathrm {new}} = \frac {1}{R_2} + \frac {1}{R_\mathrm {L}}\), the ratio of the resistances of the voltage divider changes, and consequently also the output voltage. This change must be taken into account when designing a loaded voltage divider.

Loaded voltage dividerSpannungsteiler

A voltage divider is designed to reduce an input voltage of \(U_0 = 20\) V to an output voltage of \(U_2 = 5\) V. The voltage divider is loaded with \(R_\mathrm {L} = 200 \, \Omega \). The resistance \(R_1 = 300 \, \Omega \) is specified.

How large must \(R_2\) be in order to achieve the desired output voltage?

First, the required total value \(R_\mathrm {new}\) of the parallel connection of \(R_\mathrm {L}\) and \(R_2\) must be calculated:

\begin {equation*} U_2 = U_0 \cdot \frac {R_\mathrm {neu}}{R_1+R_\mathrm {neu}} \end {equation*}

\begin {equation*} R_\mathrm {neu} = \frac {5 \, \mathrm {V}}{20 \, \mathrm {V}} \cdot (R_1 + R_\mathrm {neu}) \end {equation*}

\begin {equation*} 0.75 \cdot R_\mathrm {neu} = 0.25 \cdot R_1 = 75 \, \Omega \end {equation*}

\begin {equation*} \rightarrow R_\mathrm {neu} = 100 \, \Omega \end {equation*}

The resistance \(R_1\) can be calculated using the formula for parallel circuits:

\begin {equation*} 100 \, \Omega = \frac {R_\mathrm {L} \cdot R_2}{R_\mathrm {L} + R_2} = \frac {200 \, \Omega \cdot R_2}{200 \, \Omega + R_2} \end {equation*}

\begin {equation} R_2 \cdot \frac {200 \, \Omega }{100 \, \Omega } = 2 \cdot R_2 = 200 \, \Omega + R_2 \end {equation}

\begin {equation*} \rightarrow R_2 = 200 \, \Omega \end {equation*}

4 Current dividers in resistor networks

A parallel connection of two resistors, as shown in Figure 4, divides a current \(I_0\) flowing in a circuit into two partial currents \(I_1\) and \(I_2\). Such a circuit is called a current divider. The current \(I_0\) flowing in the circuit is divided by the resistors \(R_1\) and \(R_2\) according to the formula:

All components of this circuit have the identical output voltage \(U_0\). Using Ohm’s law, this voltage across resistors \(R_1\) and \(R_2\) can also be described as the product of the resistance value and the current:

\begin {equation*} U_0 = I_1 \cdot R_1 = I_2 \cdot R_2 \end {equation*}

By combining resistors \(R_1\) and \(R_2\) to form \(R_\mathrm {total}\) as described in Figure 4, the total current \(I_0\) can be determined as a function of the output voltage \(U_0\):

\begin {equation*} U_0 = I_0 \cdot R_\mathrm {tot} \end {equation*}

Equating the two current terms yields the following relationship between \(I_0\) and \(I_2\)

\begin {equation*} I_2 \cdot R_2 = I_0 \cdot R_\mathrm {tot} \end {equation*}

The parallel connection \(R_\mathrm {total}\) can be calculated directly from the two resistance values \(R_1\) and \(R_2\) as shown in equation 5:

\begin {equation*} I_2 \cdot R_2 = I_0 \cdot \frac {R_1 \cdot R_2}{R_1 + R_2} \end {equation*}

The reduction of \(R_2\) results in the division ratio of the electric current at the current divider:

5 Special operating conditions for active two-pole systems

Active two-pole circuits can be grouped together as equivalent voltage sources or equivalent current sources. This can be particularly useful for simplifying electrical networks.

5.1 Short-circuit and no-load test on the replacement voltage source

The real source of tension

As already known from Module 3 Electrical Components, a real voltage source consists of an ideal voltage source \(U_0\) and an internal resistance \(R_\mathrm {i}\).

    

If the real voltage source is unloaded, i.e. no component is connected to the output terminals, no current \(I\) flows (see Figure 8). \begin {equation*} I = 0 \end {equation*}

\begin {equation*} \rightarrow U_\mathrm {Ri} = R_\mathrm {i} \cdot I = 0 \end {equation*}

Since the voltage across the internal resistance \(R_\mathrm {i}\) is 0 volts, the following applies to the terminal voltage \(U\):

\begin {equation*} U = U_0 \end {equation*}

If the real voltage source is loaded with a resistor \(R_\mathrm {a}\) as shown in Figure 9, a current \(I > 0\) flows through the line.

Consequently, a voltage \(U_\mathrm {Ri} = R_\mathrm {i} \cdot I\) drops across the internal resistance \(R_\mathrm {i}\), and the terminal voltage is

\begin {equation*} U = U_0 - U_\mathrm {Ri} \end {equation*}

To reduce the voltage drop within the voltage source, the internal resistance \(R_\mathrm {i}\) should be kept as low as possible:

\begin {equation*} U_0 << I \cdot R_\mathrm {i} \end {equation*}

The replacement voltage source

Each active two-terminal network with at least one source and any number of resistors can be combined into a equivalent voltage source (Figure 10). As with a real voltage source, it is sufficient to determine an ideal voltage source \(U_0\) and a combined internal resistance \(R_\mathrm {i}\) to fully characterise the equivalent voltage source. Externally, it has completely equivalent terminal behaviour to the original circuit at its two connection terminals.

As with a loaded real voltage source, the resulting output voltage \(U\) depends on both the internal resistance \(R_\mathrm {i}\) and the load current \(I\) (Figure 11):

\begin {equation*} U = U_0 - I \cdot R_\mathrm {i} \end {equation*}

The parameters \(U_0\) and \(R_\mathrm {i}\) required for characterisation can be determined using two tests: the short-circuit test and the no-load test.

In the no-load test, the two terminals of the equivalent voltage source are left open (Figure 12).

Da durch die offenen Klemmen der Strom \(I\) durch den Widerstand \(R_\mathrm {i}\) null Ampere beträgt, fällt auch keine Spannung über ihn ab. Die Leerlaufspannung ist also mit der Spannung \(U_0\) identisch:

In the short-circuit test, the two terminals of the active two-pole are directly short-circuited, i.e. an ideally conductive connection is established between them (Figure 13).

No output voltage \(U\) can be applied across the short circuit thus created:

\begin {equation*} U = 0 \end {equation*}

Consequently, the entire voltage generated by \(U_0\) drops across the internal resistance \(R_\mathrm {i}\). According to Ohm’s law, this determines the short-circuit current \(I_\mathrm {K}\).

\begin {equation*} U_\mathrm {Ri} = U_\mathrm {L} =R_\mathrm {Ri} \cdot I_\mathrm {K} \end {equation*}

The short-circuit current \(I_\mathrm {K}\) is therefore:

\begin {equation*} I_\mathrm {K} = \frac {U_\mathrm {L}}{R_\mathrm {i}} \end {equation*}

The internal resistance \(R_\mathrm {i}\) can therefore be calculated as follows:

Attention: \(U_\mathrm {L}\) und \(I_\mathrm {K}\) are certain quantities in two completely independent experiments that do not occur simultaneously. The formula therefore only looks like Ohm’s law at resistance \(R_\mathrm {i}\), but it is not.

5.2 Short-circuit and no-load test on the emergency power source

The real power source

A real current source also consists of an ideal current source \(U_0\) and an internal resistance \(R_\mathrm {i}\) (see Figure 14). However, in contrast to the real voltage source, this is connected in parallel rather than in series with the source.

Depending on the output voltage \(U\) that occurs, the current \(I_\mathrm {Ri} = U / R_\mathrm {Ri}\) flows through the finite internal resistance \(R_\mathrm {i}\), so the output current \(I\) is given by

In order to reduce the output current \(I\) through the internal resistance, this should be selected as large as possible:

\begin {equation*} I_0 >> \frac {U}{R_\mathrm {i}} \end {equation*}

The backup power source

As an alternative to the replacement voltage source, an active two-pole can also be represented as a replacement current source (Figure 15). It can also be fully characterised by an ideal current source \(I_0\) and a combined internal resistance \(R_\mathrm {i}\).

Similar to the equivalent voltage source, the parameters \(I_0\) and \(R_\mathrm {i}\) can also be determined here using the open-circuit and short-circuit tests.

In the open-circuit test, the entire current \(I_0\) flows through the internal resistance. The following open-circuit voltage \(U_\mathrm {L}\) is established at the terminals connected in parallel:

\begin {equation*} U_\mathrm {L} = R_\mathrm {i} \cdot I_0 \end {equation*}

In the short-circuit experiment, the two terminals of the equivalent current source are short-circuited. Since the voltage across the short circuit and thus also across the internal resistance connected in parallel is zero volts, no current flows through \(R_\mathrm {i}\).

The entire current \(I_0\) therefore flows through the short circuit:

With \(U_\mathrm {L} = R_\mathrm {i} \cdot I_0\), the internal resistance can now be determined:

5.3 Conversion of voltage and current sources

In network analysis, it can be useful to convert current sources into voltage sources or voltage sources into current sources. This works identically for real current and voltage sources, as well as for equivalent current and voltage sources. The value of the internal resistance \(R_\mathrm {i}\) remains the same, but its position in the equivalent circuit diagram changes (see Figure 5.3)

The open-circuit voltage and short-circuit current remain unchanged during the conversion of the source and are converted to the other source.

Attention: The Sign Conventions of current and voltage sources converted into each other are directed in opposite directions!

When converting a current source into a voltage source, the voltage \(U_0\) is:

The current \(I_0\) when converting a voltage source into a current source is: