Question

The resistance of two conductors in parallel is 12 ohm and in series is 50 Ohm then find value of resistance of each conductor.​

Answer 1

Let the resistance of each conductor be R1and R2 .

In parallel connection,

Rp =R1 × R2 / R1 + R2 ----------1

in series connection ,

Rs = R1 + R2 ------------2

multiple equation 1 and 2

Rp × Rs = R1 × R2

R1 × R2 = 12 × 50 = 600

=> R1 = 600 / R2

substituting in equation 2

=> R²2 -50R2 + 600 = 0

=> (R2 - 20) (R2 - 30) = 0

=> R2 = 20 or 30 ohm

The value of each resistance is 30 and 20 ohms.

Answer 2

Solution:- It is given that the resistance of two conductors in parallel is 12 ohm and in series is 50 Ohm.

Let r1 and r2 be the unknown resistance.

(i).

When the resistances are in series combination.

r1 + r2 = 50 Ohm.

(ii).

When the resistances are in parallel combination.

=>1/r1 + 1/r2

=> r1r2/r1 + r2 = 1/R(say)

where R is given as 12 Ohm.

Therefore, r1 + r2 = 50 Ohm

and r1r2/r1 + r2 = 12 Ohm.

=> r1r2 = 12 × 50

=> r1r2 = 600

Also,

=> ( r1 - r2 )^2 = ( r1 + r2 )^2 - 4r1r2

=> ( r1 - r2 )^2 = 2500 - 4 × 600

=> ( r1 - r2 )^2 = 100

=> ( r1 - r2 )^2 = (10)^2

=> r1 - r2 = 10

From (i) and (iii), 2r1 = 60

Therefore, r1 = 30 Ohm and r2 = 20 Ohm.