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When two resistor of resistance R1 anf R2 are connected in parallel the net resistance is 3ohm, when connected in series, its value is 16ohm, calculate the values of R1 and R2 |
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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) 3 = R1R2/(R1 + R2) 3(R1 + R2) = R1R2 When they are in series effective resistance Rs = R1 + R2 16 = R1 + R2 ⇒ R2 = 16 - R1 3 x 16 = R1 R2 = R1 (16 - R1) = 16R1 - R12. 48 = 16R1 - R12. R12 - 16R1 + 48 = 0 Solve this quadratic equation. You will get R1 = (16 ± √(256 - 4 x 1 x 48))/2 = (16 ± √64)/2 = (16 ± 8)/2 R1 = 12 ohm or 4 ohm If R1 = 12 ohm, R2 = 4 ohm and vice versa. I hope it HELPED you |
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