Question

An engineer wants to build a circuit with total resistance of 3 ohms. Show how he can do this with two resistors, where the resistance of one is 8 ohms greater than the other

Answer 1

Answer:

Step-by-step explanation:

He can do so with parallel combination

1/x+1/(x+8)=1/3

2x+8/x^2 +8x=1/3

6x+24=x^2 +8x

x^2 +2x-24=0

x^2 +6x-4x-24=0

x(x+6) - 4(x+6)

x=4,-6

x=4

Neglecting negative value -6

Another resistance is of 12 ohm

Answer 2

HERE IS YOUR ANSWER

let one of the resistors resistance be X

then the other ones resistance will be X + 8

the equivalent resistance should be 3 ohms

the resistors should be connected in parallel because in parallel combination the equivalent resistance will be less than the individual resistance

therefore

1/equivalent resistance = 1/R1 + 1/R2

1/3 = 1/X + 1/X+8

1/3 = 2X + 8/ X^2 + 8X

X^2 + 8X = 6X + 24

X^2 + 2 X -24 = 0

(X + 6 )( X - 4 ) = 0

so the value of X can be +4 or - 6

as the resistance can not be negative

therefore X = 4ohms

therefore the resistors resistances are 4 ohms and 12 ohms

HOPE MY ANSWER IS CORRECT