**We already know how resistance, current and voltage behave for single resistors. But what about more complex resistor combinations?**

In practice, most circuits contain more than one resistor. How do resistance, current and voltage behave in more complex circuits? Let's first have a look at the basic resistor combinations, that we can make with two resistors. At the end of this tutorial, we will take a look at a more complex example with more than two resistors. Of course there also circuits with more than just resistors, but that's out of scope for this tutorial.

Our first simple resistor combination consist of two resistors in series. The question is what will actually happen? What is their combined resistance and how do current and voltage behave?

Do you remember the experiment in which we first calculated and then measured the resistance of a pencil lead? My pencil lead was 17,5 cm long. What would happen if I use a saw and saw the pencil into two parts. Let's say one part with a length of 10 cm and one with a length of 7,5 cm. If we put both pieces together, in other words connect them in series, we would expect them to have the same combined resistance as the full length pencil lead. But, is this true? And what is the resistance of our two individual parts?

Let's have a look at the formulas.
First, we have the two lengths of our pencil lead parts:

\(l_1 = 10 cm\)

\(l_2 = 7,5 cm\)

With this two lengths, the resistivity \(\rho\) and the cross-section \(A\) we can calculate the resistance for both parts.
I won't do this here, I just give the formulas:

\(R_1 = {{p \cdot l_1} \over A}\) and \(R_2 = {{p \cdot l_2} \over A}\)

How does these two resistances relate to the resistance of the whole pencil lead? Let me show you, by rewriting the formula for the combined resistance of both parts:

\(l = l_1 + l_2\)

\(R = {{p \cdot l} \over A} = {{p \cdot (l_1 + l_2)} \over A} = (l_1 + l_2) \cdot {p \over A} = l_1 \cdot {p \over A} + l_2 \cdot {p \over A} = R_1 + R_2\)

We now know that the resistance of both parts resistance adds up to the one of the full length pencil lead. For both parts in series we indeed get the same resistance as we had for the whole pencil lead. The question is is this also true for arbitrary resistors? In the case of our pencil lead both parts share a common cross-section and resistivity. What about other resistors? Well, if we want to we could replace an arbitrary resistor by a pencil lead with our cross-section and resistivity. We would just need to adjust the length to match the resistance value of the resistor we want to replace. For this pencil lead resistor our formula does obviously apply. We can therefore conclude that our formula applies to just any resistor.

We could also divide our pencil in more than two parts. In the end we would come to the same conclusion: The common resistance of multiple resistors in series is the sum of their individual resistances.
The general formula for resistors in series is:

\(R = R_1 + R_2 + R_3 + ...\)

Another important question is how current and voltage behave for resistors in series. We can calculate the combined resistance using our formula. Using this combined resistance and our battery voltage we can calculate the current. As there is only one path and electrons are neither created nor consumed, this current has to be the same for all resistors. In conclusion, we can simply use ohm's law, the common current and the individual resistance values, to calculate the voltage drop over each resistor.

Let's look at an example:

First we calculate our combined resistance:

\(R_1 = 100 Ω\)

\(R_2 = 1 kΩ\)

\(R = R_1 + R_2 = 100 Ω + 1 kΩ = 1.1 kΩ\)

Next, we calculate the common current flowing through all the resistors:

\(U = 9 V\)

\(I = {U \over R} = {{9 V} \over {1.1 kΩ}} = 8.18 mA\)

And as a final step, we can then calculate the voltage drop over each of the resistors:

\(U_1 = R_1 \cdot I = 100 Ω \cdot 8.18 mA = 0.82 V\)

\(U_2 = R_2 \cdot I = 1 kΩ \cdot 8.18 mA = 8.18 V\)

As you can see, the total voltage is divided between both resistors. Both voltage drops will add up to our total voltage of 9V. One use case of resistors in series are voltage dividers, in which we make use of this fact. We will cover them in the next tutorial.

The second simple resistor combination consist of two resistors in parallel:

You might be familiar with the saying that electrons always take the path of the least resistance. This is not true, at least not if you want to apply it to our resistor combination in such a simple fashion. Electrons have no path finding algorithm. As electrons repel each other they will always spread over the whole conductive material. Of course, there is also the interaction with the atoms in the conductive material. Not all paths are equally conductive. Following ohm's law, less current flows through paths with a higher resistance and more through paths with less resistance. The result is however not the sole use of the path of with the least resistance, but the use of all paths at the same time. The electron flow splits up between the different paths.

When we look at our pencil lead example and the resistance formula, we can see that the combined resistance of two resistors in parallel is actually less than the resistance of each individual resistor. By adding the second pencil lead in parallel, we add more conductive material and create an additional path for the current to flow through. When looking at the resistance formula, we increase the cross-section. Let's see if we can infer a formula for resistors in parallel, as we did for resistors in series.

Our overall cross-section is now the sum of the cross-section of both leads:

\(A = A_1 + A_2\)

We further get the following two equations, for the resistance values of our pencil leads:

\(R_1 = {{p \cdot l} \over A_1}\)

\(R_2 = {{p \cdot l} \over A_2}\)

And now again we take a look at the formula for the combined resistance and rewrite it for our needs:

\(R = {{p \cdot l} \over {A_1 + A_2}} = {1 \over {{A_1 + A_2} \over {p \cdot l}}} = {1 \over {{A_1 \over {p \cdot l}} + {A_2 \over {p \cdot l}}}} = {1 \over {{1 \over R_1} + {1 \over R_2}}}\)

Yes, this awful looking formula is indeed the formula for the combined resistance and with the same reasoning, we used for the resistors in series we can say that this applies for arbitrary resistors. We can extend this formula for more than two resistors, as well:

\(R = {1 \over {{1 \over R_1} + {1 \over R_2} + {1 \over R_3} + ...}}\)

What about current and voltage? We know that have the same voltage applied to both resistors. Calculate the current that flows through each resistor is really easy this time. We just use ohm's law:

\(I_1 = {U \over R_1}\)

\(I_2 = {U \over R_2}\)

We can also use ohm's law to calculate the current for the combined resistance. The calculated result is equal to the sum of individual currents flowing through the two resistors.

Let's look at an example:

For our combined resistance we get:

\(R_1 = 100 Ω\)

\(R_2 = 1 kΩ\)

\(R = {1 \over {{1 \over R_1} + {1 \over R_2}}} = {1 \over {{1 \over 100 Ω} + {1 \over 1 kΩ}}} = {1 \over {{0.01 {1 \over Ω}} + {0.001 {1 \over Ω}}}} = {1 \over 0.011} Ω = 90.91 Ω\)

The overall current flowing out of our battery is:

\(U = 9 V\)

\(I = {U \over R} = {{9 V} \over {90.91 Ω}} = 99.00 mA\)

Of this current the following parts flows through each resistor:
\(I_1 = {U \over R_1} = {{9 V} \over {100 Ω}} = 90 mA\)

\(I_2 = {U \over R_2} = {{9 V} \over {1 kΩ}} = 9 mA\)

For resistors in series there was the use case of a voltage divider, but what is the sense of two resistors in parallel? Why not a single resistor?

One possible use case is to create a resistor with a non-typical value using multiple standard resistors. There is also another use case, however.
Let's have a look at the power dissipation of the two resistors:

\(P = I^2 \cdot R = (99 mA) ^ 2 \cdot 90.91 Ω = 0.891 W\)

\(P_1 = I_1^2 \cdot R_1 = (90 mA) ^ 2 \cdot 100 Ω = 0.81 W\)

\(P_2 = I_2^2 \cdot R_2 = (9 mA) ^ 2 \cdot 1 kΩ = 0.081 W\)

As we can see the power dissipation is split up between the resistors as well. The individual values add up to the total power dissipation. This means we can use multiple resistors with a low power rating together to form a resistor a higher power rating.
We could for instance use ten 1 kΩ resistors with a power rating of 0.25 W in parallel. For our total resistance we get:

\(R = {1 \over {10 \over {1 kΩ}}} = {1 \over 0.01} Ω = 100 Ω\)

This resistor combination, however, can be used up to 2.5 W of power dissipation. This is good to know if you are in need of a high power resistor once. The calculation is, of course only that easy because all resistors share a common value. If this is not the case, you have to be more careful. The load will not be equally divided, in this case. The resistor with the smallest value has to be able to cope with the highest current and power dissipation.

Till now, we only looked at simple resistor combinations. What about more complex ones? For this I have prepared another example:

How can we calculate the combined resistance? The solution is to divide the problem into multiple simpler ones. We do this by identifying multiple resistors in series and in parallel. We then step-by-step replace them in thought, with a resistor with an equivalent resistance value.
In the example we can easily spot that \(R_2\) and \(R_3\) are used in parallel. Their combined resistance is:

\(R_{23} = {1 \over {{1 \over R_2} + {1 \over R_3}}} = {1 \over {{1 \over {360 Ω}} + {1 \over {360 Ω}}}} = {1 \over {{2 \over {360 Ω}}}} = 180 Ω\)

As a next step we calculate the combined resistance for \(R_{23}\) and \(R_4\) which are used in series:

\(R_{234} = R_{23} + R_4 = 180 Ω + 39 kΩ = 39.18 kΩ\)

Finally, we can calculate the combined resistance of all resistors:

\(R = {1 \over {{1 \over R_1} + {1 \over R_{234}}}} = {1 \over {{1 \over 100 kΩ} + {1 \over 39.18 kΩ}}} = 28.15 kΩ\)

Are we right? Let's check it! The picture below shows the measured value. It matches our calculated value within the specified tolerance of 5 %.

Like we did for our simple combinations, we can also calculate the voltages and currents for this more complex one. Once again we proceed step-by-step.

Our batteries voltage is 9 V. This voltage is applied both to \(R_1\) and \(R_{234}\).
Let's first calculate the overall current and the one for both paths:

\(U = U_1 = U_{234} = 9V\)

\(I = {U \over R} = {{9 V} \over {28.15 kΩ}} = 0.32 mA\)

\(I_1 = {U_1 \over R_1} = {{9 V} \over {100 kΩ}} = 0.09 mA\)

\(I_{234} = {U_{234} \over R_{234}} = {{9 V} \over {39.18 kΩ}} = 0.23 mA\)

The current \(I_{234}\) flows through the resistor \(R_4\) as well as through \(R_{23}\). We can use this result to calculate the voltage drop over \(R_4\) and \(R_{23}\):

\(I_4 = I_{23} = I_{234} = 0.32 mA\)

\(U_{23} = R_{23} \cdot I_{23} = 180 Ω \cdot 0.23 mA = 0.04 V\)

\(U_4 = R_{4} \cdot I_{4} = 39 kΩ \cdot 0.23 mA =8.97 V\)

As \(R_2\) and \(R_3\) have the same resistance, we know that the current is split equally between them:

\(I_2 = I_3 = {0.23 mA \over 2} = 0.12 mA\)

\(U_2 = U_3 = U_{23} = 0.04 V\)

The picture below shows our final results:

You may have noticed, that not all values add up as they should. This due to rounding errors.