Question

Two resistances of when connected in parallel give resultant value of 2 Ohms; when connected in series the value become 9 ohms. calculate the value of each resistance

Answer 1

Hello

Let the resistances be r1 and r2

The given equivalent resistances are

For parallel = 2 ohm
For series = 9 ohms

So, from ohms law

The speed resistance will be

R(eq) = r1 +r2
Or,
r1 + r2 = 9 ohms

Now
For parallel connection

$\frac{1}{r(eq)} = \frac{1}{r1} + \frac{1}{r2} \\ \\ \frac{1}{2} = \frac{1}{r1} + \frac{1}{r2} \\ \frac{1}{2} = \frac{r1 + r2}{r1r2} \\ r1r2 = 2(9) = 18 \\ r1 = \frac{18}{r2}$
Now


R1 = 18/r2

And
R1 = 9 - r2

So,
Equating both
$9 - r2 = \frac{18}{r2} \\ \frac{18}{r2} + r2 = 9 \\ \\ = 18 + {(r2)}^{2} = 9r2 \\ \\ consider \: r2 \: as \: x \\ {x}^{2} - 9x + 18 = 0 \\ {x}^{2} - 6x - 3x + 18 = 0 \\ x(x - 6) - 3(x - 6) \\ (x - 6)(x - 3)$
Now

equating with Zeroes we get the value of x as 6 and 3

So,
Now ,
From equation 1
R1+R2 =9

Of R2 =6 then R1 will be 3
And if R2 =3 then R1 will be 6

So ,
The value of the two resistances is equal to
6 ohm and 3 ohm


Hope this will be helping you ✌️

Answer 2

Answer:

Here is the solution