Equivalent resistance

Sometimes before we start calculating a task related to an electrical circuit, it can be simplified by replacing several resistors (or other elements, e.g., in the case of analyzing AC circuits using symbolic methods) with one that describes the equivalent resistance (in the case of analyzing AC circuits, the equivalent impedance).

There are a number of tasks where calculating the equivalent resistance is necessary for other reasons - compare, for example, Thevenin's Method or Norton's Method.

Series Connection

If the elements are connected in series:

1. The current flowing through them has the same value
2. The equivalent resistance is the sum of the resistances of the individual elements
\begin{aligned} &R_z=R_1+R_2+...\\ &R_z=\sum{R_i} \end{aligned}

It is very important to note that the key to determining whether the elements are connected in series is to establish whether the same current flows through them; this will be more evident in the examples

Parallel Connection

If the elements are connected in parallel:

1. The voltage across them has the same value
2. The equivalent conductance is the sum of the conductances of the individual elements
\begin{aligned} &G_z=G_1+G_2+...\\ &G_z=\sum{G_i} \end{aligned}
3. The reciprocal of the equivalent resistance is the sum of the reciprocals of the resistances of the individual elements
\begin{aligned} &\frac{1}{R_z}=\frac{1}{R_1}+\frac{1}{R_2}+...\\ &\frac{1}{R_z}=\sum{\frac{1}{R_i}}\\ \end{aligned}

It is very important to note that the key to determining whether the elements are connected in parallel is to establish whether the same voltage is present across them

For clarity - it is very rare that it is truly beneficial to use the formula with conductance, but it helps to see the relationship that occurs for such a connection of elements

In practice, it is also worth remembering the formula for two elements connected in parallel:

\begin{aligned} &\frac{1}{R_z}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{R_2}{R_1\cdot R_1 }+\frac{R_1}{R_2\cdot R_1 }\\ &\frac{1}{R_z}=\frac{R_1+R_2}{R_1\cdot R_2}\\ &R_z=\frac{R_1\cdot R_2}{R_1+R_2}\\ \end{aligned}

If you ever have doubts about what is in the denominator and what is in the numerator, remember that in the end, you must get the correct unit - \(\Omega\), so multiplication MUST be in the numerator because:

\begin{aligned} &\frac{\Omega\cdot \Omega}{\Omega + \Omega}=\frac{\Omega^2}{\Omega}=\Omega\\ \end{aligned}

and not:

\begin{aligned} &\frac{\Omega+\Omega}{\Omega \cdot \Omega}=\frac{\Omega}{\Omega^2}=\frac{1}{\Omega}\\ \end{aligned}

Mixed Connection (Series-Parallel)

Personally, I really dislike this term :-) In practice, a mixed connection still needs to be reduced to the superposition of series and parallel connections, so I see no particular reason to distinguish this as a separate case.

However, at this point, I would like to make a small digression - every equivalent resistance is calculated with respect to two points (terminals); it is not some absolute value but merely a simplification that we use to make our lives easier

Equivalent Resistance from Kirchhoff's Laws

A VERY useful approach to thinking about equivalent resistance, especially in more complicated and non-obvious cases that sometimes arise, e.g., in tasks involving Thevenin's Method, is to approach it from the perspective of Kirchhoff's laws.

Let's imagine for a moment that there is a voltage \(U_{AB}\) between terminals AB with respect to which we need to calculate the equivalent resistance, which causes a current \(I\)

It is obvious that \(R_z=\frac{U_{AB}}{I}\)

Data:

\begin{aligned} R_1 & =30 \\ R_2 & =20 \\ R_3 & =10 \end{aligned}

Using the formulas for series and parallel connections:

\begin{aligned} &R_{12}=\frac{R_1 \cdot R_2}{R_1+R_2}=12\\ &R_{z A B}=R_{12}+R_3=22 \end{aligned}

Using Kirchhoff's laws:

\begin{aligned} & I=I_1+I_2 \\ & I_1 \cdot R_1=I_2 \cdot R_2 \\ & U_{A B}=I_2 \cdot R_2+I \cdot R_3 \\ & R_{z A B}=\frac{U_{A B}}{I} \end{aligned}

We calculate:

\begin{aligned} & I_1=\frac{I_2 \cdot R_2}{R_1} \\ & I=\frac{I_2 \cdot R_2}{R_1}+I_2 \\ & I_2=\frac{R_1 \cdot I}{R_2+R_1} \\ & U_{A B}=\frac{R_1 \cdot I}{R_2+R_1} \cdot R_2+I \cdot R_3 \\ & R_{z A B}=\frac{\frac{R_1 \cdot I}{R_2+R_1} \cdot R_2+I \cdot R_3}{I}=\frac{R_1}{R_2+R_1} \cdot R_2+R_3=22 \end{aligned}

And it is clear that this is not actually calculating the equivalent resistance, but in more complex cases, this way of thinking can be very helpful, although the calculations themselves will usually be more complicated