Let the resistance be R1 and R2 when two resistances are connected in series
RS = R1 + R2
= 9
R1 + R2 = 9 ……….(1)
When two resistance are connected in parallel
Using (1) equation (2) becomes
\(\frac{9}{R_1R_2}\)
R1R2 = 18 ...... (3)
(R1 – R2)2 = (R1 + R2)2 – 4R1R2
= (9)2 – 4 × 18
= 81 – 72 = 9
∴ (R1 – R2 ) = √9 = 3 ………(4)
From (1)
R1 + R2 = 9
2R1 =12
∴ R1 =\(\frac{12}{2}\) = 6 ohm
From (1)
R2 = 9 – R1
= 9 – 6 = 3Ω
The values of resistances are
R1 = 6 ohm
R2 = 3 ohm
