Topic Menu

Cyber-chess a chess instruction site
Circuits - series and parallel

Q2. Four resistors, each having resistance of 50 , are connected to form a square. A resistance meter is used to measure the resistance between different corners of the square.

Determine the resistance the meter records when connected between the following corners.

This type of question always confuses students, so if it floored you don't worry!.

To start off let's redraw the circuits in a way we can recognize… it makes the analysis MUCH easier! - see the diagrams I have drawn below.

(a) Between A and C

Next, we have to simplify the circuit. We know that resistors in series can be replaced by a single resistor by adding the values, so let's do that!

From your data sheet so redaw the circuits... see below each diagram for the next stage.

Easy to solve – for n identical resistors (R) in parallel a single equivalent resistor is R/n so you can do it 'by inspection' but show some working – otherwise you may not get any marks! RAC = 50

(2)

(b) Between A and B

This can't be done by inspection - you need to use the resistances in parallel equation from your data sheet.

1/R = 1/150 + 1/50 = 4/150

commonest mistake is to forget to find the inverse of this stage...

RAB = 150/4 = 37.5 = 38

(3)

(Total 5 marks)