Two resistance when connected in parllel give resultant value of 2 ohm when connected in series the value become 9 ohms. Calculate the value of each resistane
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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.
I hope I was of some help
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.
I hope I was of some help
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