Lesson : Series and Parallel Circuits Examples

Before we dive into multiples examples, you can find a cheat sheet for Ohm’s law, Power law, series and parallel circuit at this page. This will be useful to solve the exercises and to understand all the examples.

Example 1: 

In this example, we want to find every missing values.

Example 1 : Series circuit

There is multiple way we could do this but we will start by calculating the voltage V in the left circuit. We know that the resistor are in series so the voltage is divide between both resistors. If we add V1 and V2, we will get the voltage V of the source. We can use Ohm’s law to get V1 and V2:

V_1 = R_1 \cdot I = 1000 \Omega \cdot 0.01 A = 10 V

V_2 = R_2 \cdot I = 1500 \Omega \cdot 0.01 A = 15 V

V = V_1 + V_2 = 10 V + 15 V = 25V

For the equivalent circuit, we know that the voltage source is the same in both circuits (V=25V). The current in both circuits is also the same (I=10mA). We only need to calculate the equivalent resistor (Req). We have a formula for this:

R_{eq} = R_1 + R_2 + R_n + \dots

We only have two resistances so the formula would be:

R_{eq} = R_1 + R_2 = 1000 \Omega + 1500 \Omega = 2500 \Omega

Example 2: 

In this example, we want to find every missing values.

Example 2 : Parallel circuit

In this example, we have multiples ways of resolving this problem. We are going to solve it using the equivalent circuit. We have the total current and the value of every resistors so we could simplify the circuit to find the voltage of the D.C voltage source. We need to calculate Req and use Ohm’s law once we have Req.

R_{eq} = \cfrac{1}{\cfrac{1}{1k \Omega} + \cfrac{1}{2.5k \Omega} + \cfrac{1}{5k \Omega}} = 625 \Omega

V = R_{eq} \cdot I_{tot} = 625 \Omega \cdot 2 A = 1250 V

Once we have the voltage, we can calculate the current flowing in the three resistors using Ohm’s law.

I_{1} = \cfrac{V}{R_{1}} = \cfrac{1250 V}{1k \Omega} = 1.25 A

I_{2} = \cfrac{V}{R_{2}} = \cfrac{1250 V}{2.5k \Omega} = 0.5 A

I_{3} = \cfrac{V}{R_{3}} = \cfrac{1250 V}{5k \Omega} = 0.25 A

We can verify our answers because

I_{tot} = I_{1} + I_{2} + I_{3}

2 A = 1.25 A + 0.5 A + 0.25 A

We can see that our answers are correct. Verifying your answers is always a good idea when working with series and parallel circuits.

Example 3: 

In this example, we want to find every missing values. Do note that we want to do this without calculating the current. We only need to find V_{1}, V_{2} and V_{3} . We also want to calculate the power dissipate in every resistors.

Example 3 : Series circuit

In series circuit, the voltage is divide between all the resistors. To calculate the voltage across any resistors in series:

V_{x} = V \cdot \cfrac{R_{x}}{R_{x} + R_{n1} + R_{n2} + \dots}

V_{1} = 24 V \cdot \cfrac{R_{1}}{R_{1} + R_{2} + R_{3}}

V_{1} = 24 V \cdot \cfrac{10k \Omega}{10k \Omega + 4.7k \Omega + 1.8k \Omega} =14.5454 V

V_{2} = 24 V \cdot \cfrac{R_{2}}{R_{1} + R_{2} + R_{3}}

V_{2} = 24 V \cdot \cfrac{4.7k \Omega}{10k \Omega + 4.7k \Omega + 1.8k \Omega} =6.836 V

V_{3} = 24 V \cdot \cfrac{R_{3}}{R_{1} + R_{2} + R_{3}}

V_{3} = 24 V \cdot \cfrac{1.8k \Omega}{10k \Omega + 4.7k \Omega + 1.8k \Omega} =2.618 V

We can now calculate the power dissipated into every resistors using the power law:

P=\cfrac{V^2}{R}

P_{1} = \cfrac{V_{1}^2}{R_{1}} = \cfrac{14.5454V \cdot 14.5454V}{10k \Omega} = 0.0212W = 21.2mW

P_{2} = \cfrac{V_{2}^2}{R_{2}} = \cfrac{6.836V \cdot 6.836V}{4.7k \Omega} = 0.0099W = 9.9mW

P_{3} = \cfrac{V_{3}^2}{R_{3}} = \cfrac{2.618V \cdot 2.618V}{1.8k \Omega} = 0.0038W = 3.8mW

We have now completed all examples.