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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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Answer» 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 |
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