Lesson : Series and Parallel combination circuit 1

In Chapter 1, we studied series and parallel circuits. These circuits can actually be combine to create more complex circuits. These circuits are a bit more difficult to analyse but we can analyse them with Kirchhoff’s laws and Ohm’s law. We are going to start with a series circuit followed by a parallel circuit. Do note that the approach used here won’t use complex algebra and this will be seen in a different chapter. Understanding the basics of how combine circuits works is more important.

Example 1:

In this example, let’s assume $R_1$ = 1kΩ (1000), $R_2$ and $R_3$ = 4kΩ (4000). We want to find $V_1$ and $V_2$ . We also want to find $I_1$ , $I_2$ and $I_3$ .

(Note : k means you multiply the number by a thousand)

Example 1 : Series and Parallel Combination

There is multiples ways you could solve this example with Ohm’s law, Kirchhoff’s laws and equivalent circuit. We are going to see one way only. The first thing we are going to do is to find the equivalent circuit of $R_2$ and $R_3$ . It is a parallel circuit so the generic formula for the equivalent resistor of a parallel circuit is :

$Req=\cfrac{1}{\cfrac{1}{R_1}+\cfrac{1}{R_2}+\cfrac{1}{R_n}+\dots}$

In our case, we have two resistors in parallel ( $R_2$ and $R_3$ ). The formula will be :

$Req=\cfrac{1}{\cfrac{1}{R_2}+\cfrac{1}{R_3}}$

$Req=\cfrac{1}{\cfrac{1}{4000}+\cfrac{1}{4000}}$

$Req=2000$

This would give us the circuit shown below if we replace $R_2$ and $R_3$ with the equivalent resistor we found:

Example 1 : Series and Parallel Combination with Req

We have a series circuit with two resistors which is basically a voltage divider, we can use the voltage divider formula to find $V_1$ :

$V_1 = 12V \cdot \cfrac{R_{1}}{(R_{1}+R_{eq})}$

$V_1 = 12V \cdot \cfrac{1000\Omega}{(1000\Omega+2000\Omega)}$

$V_1 = 4V$

Kirchhoff’s voltage law states that the sum of voltages in a closed loop will be equal to 0. We can find the following formula and calculate $V_2$ :

$12V-V_1-V_2=0$

$V_1$ and $V_2$ are voltage drop so they have a negative value.

$12V-4V-V_2=0$

$12V-4V=V_2$

$V_2=8V$

We can now work with the original circuit with $R_2$ and $R_3$ since we have found $V_1$ and $V_2$ . For $I_2$ and $I_3$ , we can use Ohm’s law to find the current flowing into $R_2$ and $R_3$ .

$I_2=\cfrac{V_2}{R_2}=\cfrac{8V}{4000\Omega}=0.002A=2mA$

Since $R_3$ is the same as $R_2$ and the voltage across is the same, the current will be the same.

$I_3=0.002A=2mA$

Kirchhoff’s current law or nodal rule states that the sums of currents flowing into any node or junction in a circuit will be equal to 0. In others words, the sums of current flowing into the node and out of the node will be equal. We can find $I_1$ by analyzing the following node:

Example 1 : Node analysis on Series and Parallel Combination

$I_1$ is flowing into the node while $I_2$ and $I_3$ are flowing out of the node. Knowing this, we can find the following formula :

$I_1=I_2+I_3$

$I_1=2mA+2mA=4mA$

We have completed the first example of Series and Parallel combination circuit. There is multiples way to analyze this circuit but we have decide to show this method since we used both Kirchhoff’s laws from previous lessons. The next lessons will display a different combination of Series and Parallel circuits with an example and a detailed solution like in this lesson.