# Lesson : Series and Parallel combination circuit 2

In previous lessons, we studied one combination of series and parallel circuits. We are going to see a second example of a series and parallel combination circuit. Ohm’s law, Kirchhoff’s laws and equivalent circuits are going to be used to solve this problem. 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 2:

In this example, let’s assume $R_1$ = 1kΩ (1000), $R_2$ = 2kΩ (2000), $R_3$ and $R_4$ = 3kΩ (3000). We want to find $V_1$, $V_2$ and $V_3$. We also want to find $I_1$, $I_2$, $I_3$ and $I_4$. $I_1$ is the current flowing into $R_1$, $I_2$ is the current flowing into $R_2$ and so on.

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

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 : $Req=\cfrac{1}{\cfrac{1}{R_2}+\cfrac{1}{R_3}}$ $Req=\cfrac{1}{\cfrac{1}{2000\Omega}+\cfrac{1}{3000\Omega}}$ $Req=1200\Omega$

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

We have a series circuit with three resistors which can be simplify again to a single resistor circuit using equivalent circuit which would be equal to the three resistors added together. $R_{eqq} = R_1+R_{eq}+R_3 = 1000\Omega+1200\Omega+3000\Omega = 5200\Omega$

We can now calculate the current flowing into the circuit with Ohm’s law : $I_{tot} = I_1 = I_3 = \cfrac{12V}{Reqq} = \cfrac{12V}{5200\Omega} \approx 2.31 mA$ $\approx 2.31 mA = 0.00231A$

(Note : This approximation will slightly change the value below if we were to keep all number but to simplify the lessons, the answers will be rounded.)

We can now calculate $V_1$ and $V_3$ with Ohm’s law : $V_1 = R_1 \cdot I_1 = 1000\Omega \cdot 2.31mA = 2.31V$ $V_3 = R_3 \cdot I_3 = 3000\Omega \cdot 2.31mA = 6.93V$

With Kirchhoff’s voltage law (often called KVL), we can find $V_2$. Kirchhoff’s voltage law or loop (mesh) rule states that the sum of voltages in a closed loop will be equal to 0. Voltage gain in a loop (in our case the voltage source) will have a positive value and voltage drop will have a negative value. $V_{DC} - V_1 - V_2 - V_3 = 0$ $12V - 2.31V - V_2 - 6.93V = 0$

If we isolate $V_2$ : $12V- 2.31V - 6.93V = V_2$ $2.76V = V_2$

We can used Ohm’s law to calculate $I_2$ and $I_3$ : $I_2 = \cfrac{V_2}{R_2} = \cfrac{2.76V}{2000\Omega} = 1.38mA = 0.00138A$ $I_3 = \cfrac{V_2}{R_3} = \cfrac{2.76V}{3000\Omega} = 920uA = 0.920mA = 0.000920A$

We have completed the second example of Series and Parallel combination circuit. The next lessons will be exercises and a cheat sheet to help you solve the exercises.