Problem 16: Three equal resistors each of 12Ω can be connected in four different ways. What is equivalent resistance of each combination?

Solution:

Step: 1 Overview

Parallel Combination of resistors.

See overview of problem 7.

Series combination of resistors.

See overview of problem 8.

Step: 2 Calculation

First possible combination can be:

Now equivalent resistance of this combination will be:
R = R₁+R₂+R₃
R = 12+12+12
R = 36Ω

Second Combination can be:

In this combination two resistors are in series so their net resistance will be:
R_series = 12+12
R_series = 24Ω
And third resistance is connected in parallel with this combination, so total resistance will be:
1/R = 1/12 + 1/R_series
1/R = 1/12 + 1/24
1/R = (2+1)/24
1/R = 3/24
1/R = 1/8
R = 8Ω

Third possible combination can be:

In this combination the resistance of parallel combination will be:
1/R_parallel = 1/12 +1/12
1/R_parallel = 2/12
1/R_parallel = 1/6
R_parallel = 6Ω
Series combination with third resistor gives net equivalent resistance:
R = R_parallel + 12
R = 6 + 12
R = 18Ω

Fourth possible combination can be:

In this combination all the resistors are in parallel, so their equivalent resistance will be:
1/R = 1/12 + 1/12 +1/12
1/R = (1+1+1)/12
1/R = 3/12
1/R = 1/4
R = 4Ω

So the possible equivalent resistances of these resistors each of 12Ω, will be 4Ω, 8Ω, 18Ω and 36Ω.