Question

Two resistances combines in series order
provide 50 ohm resultant resistance and when
it combines in parallel order provides 8 ohm
resultant resistance. Then the value of each
resistance
​

Answer 1

Let the two resistances be R1 and R2

Given,

In series Combination of resistors, equivalent resistance = 50ohm

That is,

Rs = R1 + R2

50ohm= R1 + R2

R2 = 50-R1....(i)

Now,

In parallel combination of resistances, equivalent resistance = 8ohm

1/Rp = 1/R1 + 1/R2

1/8 = (R2+R1)/R1R2

1/8 = 50/R1R2

So, R1R2 = 50×8 = 400ohm ......(ii)

Putting the value of eq(i) in eq(ii) we get,

R1(50-R1) = 400

50R1 -R1² = 400

0 = R1² -50R1+400

0 = R1² -40R1-10R1+400

0 = R1(R1-40)-10(R1-40)

0 = (R1-10)(R1-40)

So,

R1-10 = 0

R1 = 10ohm

and R2 = 50-10 = 40ohm

The correct option is B. 10ohm and 40ohm

Cheers!