Electrical circuits are usually not simple, that is, consisting of one load and one source.
They usually consist of multiple loads.
How these are connected determines how they are classified
In series circuits, the component are joined sequentially, whereas parallel circuits are connected as alternate paths
They usually consist of multiple loads.
How these are connected determines how they are classified
In series circuits, the component are joined sequentially, whereas parallel circuits are connected as alternate paths
Resistors in Series
Resistors in Parallel
Lets first explore an analogy to see how voltage , current and resistance plays a part in these circuits.
Using the analogy of a water slide, this will help you understand the comparison of a simple circuit vs a series circuit vs a parallel circuit.
Using the analogy of a water slide, this will help you understand the comparison of a simple circuit vs a series circuit vs a parallel circuit.
Once you have mastered the key concepts, move on to the maths version, where Kirchoff's Laws are used to analyse circuits.
Sample Problem
We are now ready to try a sample problem
Below is a sample problem with a video that explain how to solve it. It is suggested you try the problem beforehand, as this actually aids understanding, even if you are unsure if you are correct.
We are now ready to try a sample problem
Below is a sample problem with a video that explain how to solve it. It is suggested you try the problem beforehand, as this actually aids understanding, even if you are unsure if you are correct.
Check Your Understanding: Electric Circuits
1. Examine the circuit below. The value of the total resistance of this circuit is closest to:
The parallel section is made up of 20 Ω and 25 Ω. For resistors in parallel, 1/R = 1/20 + 1/25 = 0.090, so this section has a resistance of 11.1 Ω. It is in series with the remaining 5 Ω, so the total is 11.1 + 5 = 16.1 Ω, which is closest to 16 Ω.
First combine the two resistors in parallel, 20 Ω and 25 Ω: 1/R = 1/20 + 1/25 = 0.090, so R = 11.1 Ω (always less than the smaller of the two). Then add the remaining 5 Ω in series: 11.1 + 5 = 16.1 Ω, which is closest to 16 Ω. Adding all the resistances (20 + 25 + 5 = 50 Ω) treats them as if they were all in series.
2. A student completes a circuit that includes a mystery box, as shown below. If the voltage between X and Y is 9.0 V, which of the following boxes is in the circuit?
The voltage between X and Y is 9.0 V across 75 Ω, so the current in the circuit is I = 9.0 / 75 = 0.12 A. The supply is 12 V, so the voltage across the box (between P and Q) is 12 - 9.0 = 3.0 V. Its resistance is therefore R = 3.0 / 0.12 = 25 Ω, which is Option 4.
Work out the current first. The voltage between X and Y is 9.0 V across 75 Ω, so I = 9.0 / 75 = 0.12 A, and the same current flows through the box. The supply is 12 V, so the box has 12 - 9.0 = 3.0 V across it. Its resistance is R = 3.0 / 0.12 = 25 Ω, so Option 4 is correct.
- w_intro_to_series_and_parallel.pdf
Intro to series and parallel circuits - a 'rough' worksheet I got from a scanned text. A good start - Circuits from Ingrum
another outsourced worksheet (answers)
Interactive
Use this animation from pHET to set up your own circuits. A good way to apply your skills to to determine the various values using the maths, and then use the ammeter and voltage tools to confirm your calculations
Use this animation from pHET to set up your own circuits. A good way to apply your skills to to determine the various values using the maths, and then use the ammeter and voltage tools to confirm your calculations