Question

Two resistances when connected in parallel give resultant value of 2 ohm. When connected in series the value become 9ohm. Calculate the value of each resistance

Answer 1

Let R1 and R2 be the two resistances. When they are in parallel, resultant resistance Rp is given by

1/Rp = 1/R1 + 1/R2
Rp = R1R2/(R1 + R2)
2 = R1R2/(R1 + R2)
2(R1 + R2) = R1R2

When they are in series effective resistance

Rs = R1 + R2
9 = R1 + R2 ⇒ R2 = 9 - R1

2 x 9 = R1 R2
= R1 (9 - R1)
= 9R1 - R12.
18 = 9R1 - R12.
R12 - 9R1 + 18 = 0

Solve this quadratic equation. You will get

R1 = (9 ± √(81 - 4 x 1 x 18))/2
= (9 ± √9)/2
= (9 ± 3)/2
R1 = 3 ohm or 6 ohm

If R1 = 3 ohm, R2 = 6 ohm and vice versa.