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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)

Example 2 : 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 :


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




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

Example 2 : Series and Parallel Combination with Req

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.

Example 2 : Series and Parallel Combination with Reqq

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.

Example 2 : Series and Parallel Combination Kirchhoff’s Voltage Law (KVL)

 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.