For electrical resistors connected in parallel the reciprocal of the combined resistances equals the sum of the reciprocals of the individual resistances. if three resistors are connected in parallel such that the second resistance is 1.40 mega ohm more than the first and the third is 5.60 mega ohm more than the first, find the resistances for a combined resistance of 1.40 mehga ohm
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Let the individual resistances be R1, R2, R3.
let the combined resistance R = 1.4 unit. 1 unit = 1 mega ohm.
1/R = 1/R1 + 1/R2 + 1/R3
1/1.4 = 1/R1 + 1/(R1+1.4) + 1/(R1+5.6)
1/1.4 - 1/(R1+1.4) = (2R1+5.6) / R1 (R1+5.6)
R1 / [1.4* (R1+1.4)] = 2(R1+2.8) / [ R1(R1+5.6) ]
R1*R1*(R1+5.6) = 2.8 (R1+1.4)(R1+2.8)
R1³ +5.6 R1² = 2.8 R1² + 2.8*4.2 R1 + 2.8*2.8*1.4
R1³ + 2.8 R1² - 2.8*4.2 R1 - 2.8*2.8*1.4 = 0
Solve for R1.
R1 is possibly a factor of constant term 2.8*2.8*1.4 i.e.,
R1 may be one of 0.7, 1.4, 2.8, 5.6, 1.4², 2.8², 1.4³, 2.8²*1.4 etc.
Try substituting these values and check if RHS = 0. Then, we find that
R1 is 2.8 mega ohms R2 = 4.2 mega ohm R3 = 8.4 mega ohms
let the combined resistance R = 1.4 unit. 1 unit = 1 mega ohm.
1/R = 1/R1 + 1/R2 + 1/R3
1/1.4 = 1/R1 + 1/(R1+1.4) + 1/(R1+5.6)
1/1.4 - 1/(R1+1.4) = (2R1+5.6) / R1 (R1+5.6)
R1 / [1.4* (R1+1.4)] = 2(R1+2.8) / [ R1(R1+5.6) ]
R1*R1*(R1+5.6) = 2.8 (R1+1.4)(R1+2.8)
R1³ +5.6 R1² = 2.8 R1² + 2.8*4.2 R1 + 2.8*2.8*1.4
R1³ + 2.8 R1² - 2.8*4.2 R1 - 2.8*2.8*1.4 = 0
Solve for R1.
R1 is possibly a factor of constant term 2.8*2.8*1.4 i.e.,
R1 may be one of 0.7, 1.4, 2.8, 5.6, 1.4², 2.8², 1.4³, 2.8²*1.4 etc.
Try substituting these values and check if RHS = 0. Then, we find that
R1 is 2.8 mega ohms R2 = 4.2 mega ohm R3 = 8.4 mega ohms
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