Question

The resistance of two conductors in series is 40ohm and the resistances become 6.4ohm when connected in parallel find the value of individual conductor
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Answer 1

Answer:

given r1 + r2 = 40 ohms and when connected parallel it is 6.4 ohms ie R = 6.4 ohms

to find the values of r1 and r2

when two resistors are connected in parallel, then as per the standard formula 1/R = 1/r1 + 1/r2

1/R = (r1 + r2)/r1*r2 [ by taking LCM ]

by taking reciprocal we get R = (r1*r2)/(r1 + r2)

where R is the effective resistance of the two connected parallel

given R = 6.4 = (r1*r2)/(r1 + r2) ......1

now replacing r1 + r2 = 40 in the equation 1, we get

6.4 = (r1*r2)/40

therefore r1*r2 = 6.4*40

= 256 ohms

r1 + r2 = 40 .......2

r1*r2 = 256.........3

r1 = 256/r2

substituting value of r1 in equation 1, we get

256/r2 + r2 = 40

256 + r2^2 = 40 r2 [ multiply through by r2 ]

r2^2 - 40r2 + 256 = 0

r2^2 -32 r2 - 8r2 + 256 = 0

r2(r2 - 32) -8(r2 - 32)=0

(r2 - 8)(r2 - 32) = 0

r2 = 8 or r2 = 32

we can say if r2 = 8 ohms then r1 = 32 ohms or vice versa