Question

Three resistors each of 2 Ω are connected together so that their total resistance is 3 Ω. Draw a diagram to show this arrangement and check it by calculation. 

Answer 1

Neither three 2 Ω resistors in series nor all three in parallel can be considered feasible because the resultant resistance would be less than 6 Ω. In the latter scenario, the resultant resistance would have been less than 2 Ω. Therefore, a good configuration has two resistors in parallel and one resistor in series.

`1/"R"_"p" = 1/"R"_1 + 1/"R"_2`

`1/"R"_"p" = 1/2 + 1/2`

= `2/2`

= 1 Ω

∴ R_p = 1 Now R_p and R₃ are in series

∴ R = R_p + R₃

R = 1 + 2

R = 3 Ω

∴ Two resistors are connected in parallel with one another in a series configuration.

Answer 2

The given diagram of the arrangement is shown in figure in which two resistors are in parallel and then this combination is in series with one resistor.

The equivalent resistance of parallel combination:

R' = `(2 xx 2)/(2 + 2) = 4/4 = 1 Omega`

Then total resistance of series combinations of R’ = 1Ω with 2Ω resistors is R = 1 + 2 = 3Ω.