# Lesson : Combination Circuit Example

Before we dive into this example, you can find a cheat sheet for Ohm’s law, Power law, series and parallel circuit at this page. You can find the cheat sheet for Voltage Divider, Kirchhoff’s laws and combination circuit at this page. This will be useful to solve the exercises and to understand this example. The example will be solve by following the steps of the cheat sheet.

Example 1:

In this example, we want to find every missing values. Let’s assume $R_1$ = 4kΩ (4000), $R_2$ = 6kΩ (6000), $R_3$ = 12kΩ (12000) and $R_4$ = 1kΩ (1000). We want to find $V_1$ and $V_2.$ 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)

1. Verify if you can use Ohm’s law anywhere to find some information. If you can, use it to find unknown voltages, currents or resistances. Otherwise, move to step 2. In the example above, we can use Ohm’s law to find $I_4$ on $R_4$ since it is directly connected to the voltage source and we have the resistance of $R_4.$

$I_4 = \cfrac{V_4}{R_4} = \cfrac{12V}{1000\Omega} = 12mA = 0.012A$

2. Simplify the circuit until you get a simple circuits like a series, parallel or a voltage divider circuit. You can simplify it multiples times to reach a simple circuits. In this example, we will go to the most simplified circuit which is a one resistor circuit. You could stop when you have a parallel circuit to avoid more steps but both approach are acceptable. In our example, $R_2$ and $R_3$ are in parallel and we can simplify the circuit by replacing $R_2$ and $R_3$ by $R_{eq1}.$

It is a parallel circuit so the generic formula for the equivalent resistor of a parallel circuit is :

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

We have two resistors in parallel ($R_2$ and $R_3$). The formula will be :

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

$R_{eq1}=\cfrac{1}{\cfrac{1}{6000\Omega}+\cfrac{1}{12000\Omega}}$

$R_{eq1}=4000\Omega$

We can simplify the circuit again. In our circuit above, $R_1$ and $R_{eq1}$ are in series and we can simplify the circuit by replacing $R_1$ and $R_{eq1}$ by $R_{eq2}.$ We have a series circuit with two resistors which can be simplify again to a single resistor using equivalent circuit which would be equal to the two resistors added together.

$R_{eq2}=R_{eq1}+R_1 = 4000\Omega + 4000\Omega = 8000\Omega$

You could stop at this point since we have a basic parallel circuit but we are going to simplify once again to get the most simple circuit we can have which is a one resistor circuit. In our circuit above, $R_{eq2}$ and $R_4$ are in parallel and we can simplify the circuit by replacing $R_{eq2}$ and $R_4$ by $R_{eq3}.$

We have two resistors in parallel ($R_{eq2}$ and $R_4$). The formula will be :

$R_{eq3}=\cfrac{1}{\cfrac{1}{R_{eq2}}+\cfrac{1}{R_4}}$

$R_{eq3}=\cfrac{1}{\cfrac{1}{8000\Omega}+\cfrac{1}{1000\Omega}}$

$R_{eq3}=888.889\Omega$

3. Use Ohm’s law or voltage divider formula to find unknown voltages, currents or resistances.

$I_{tot}=\cfrac{V_{DC}}{R_{eq3}} = \cfrac{12V}{888.889\Omega} = 13.5mA = 0.0135A$

4. De-simplify the circuit (one times if you did multiples simplification) and add the information you found like currents, voltages and resistances.

5. Use Ohm’s law to find the missing information in the de-simplify circuit.

We already have $I_{4} = 0.012A = 12mA$. We know $R_{eq2}=8000\Omega$ and the voltage across it is 12V. We can use Ohm’s law to find $I_{eq2}.$

$I_{eq2}=\cfrac{V_{DC}}{R_{eq2}} =\cfrac{12V}{8000\Omega}= 1.5mA = 0.0015A$

6. If the circuit is still not completely de-simplified, restart from step 4 until you get the original circuit. Step 4 : De-simplify the circuit (one times if you did multiples simplification) and add the information you found like currents, voltages and resistances.

7. We know the current in $R_1$ and $R_{eq1}$ is $I_{eq2}.$ We can use Ohm’s law to find the missing information in the de-simplify circuit. We can find $V_1$ and $V_2.$

$I_{eq2} = I_1$

$V_1 = R_1 \cdot I_1 = 4000\Omega \cdot 0.0015A = 6V$

$V_2 = R_{eq1} \cdot I_1 = 4000\Omega \cdot 0.0015A = 6V$

8. If the circuit is still not completely de-simplified, restart from step 4 until you get the original circuit. Step 4 : De-simplify the circuit (one times if you did multiples simplification) and add the information you found like currents, voltages and resistances. This is the last time we will do this since we now have the original circuit.

9. Use Ohm’s law and Power law if needed to find the missing information in the original circuit. We are only missing $I_{2}$ and $I_{3}$. We know $V_{2}$ so we can calculate $I_{2}$ and $I_{3}$ with Ohm’s law.

$I_{2}=\cfrac{V_2}{R_2} =\cfrac{6V}{6000\Omega}= 1mA = 0.001A$

$I_{3}=\cfrac{V_2}{R_3} =\cfrac{6V}{12000\Omega}= 500uA =0.5mA = 0.0005A$

Do note that you can validate your currents values by comparing $I_{tot}$ with the current in the first branch and second branch.

First branch :

$I_{1} = I_2 + I_3 = 1.5mA$

Second branch :

$I_4 = 12mA$

Total :

$I_{tot} = I_1 + I_4 = 13.5mA$

As we can see, all equations are true which is a good indicator that we didn’t make errors in the process. We have now completed the example and the next lesson will be exercises similar to this example.