Imagine that you have three resistors each of value 30 Ω. How many resistors can you obtained by connecting these three in different ways ? Draw the relevant diagrams.

I can obtain the following resistors by various combinations.

i) When the three resistors are connected in series

{{R}_{eq}} = R1 + R2 + R3

= 30+ 30+ 30 = 90 Ω.

ii) If all the three resistors are connected in parallel

1/ {{R}_{eq}} = 1/r1 + 1/r2 + 1/r3

= 1/30 + 1/30+ 1/30

=1+1+1/30 = 3/30 = 1/10

{{R}_{eq}} = 10 Ω.

iii) If two resistors are connected in series and the third resistor is connected in parallel to them.

{{R}_{eq}} for series = R1 + R2

= 30+ 30 = 60 Ω.

{{R}_{eq}} for 60 Ω, 30 Ω in parallel

1/{{R}_{eq}} = 1/60 + 1/30 = 1+2/30 = 3/60 = 1/20

{{R}_{eq}} = 20 Ω

iv) If two resistors are connected in parallel and third resistor is connected in series.

1/{{R}_{eq}} for parallel connection =1/r1 + 1/r2

= 1/30 + 1/30 = 1+1/30 = 2/30 =1/15

{{R}_{eq}} = 15 Ω

{{R}_{eq}} for series combination of 15 Ω and 30 Ω

{{R}_{eq}} = R1 + R2 = 15+30 = 45 Ω