Question

when two resistors are connected in series the effective resistance is 80 ohm when they are connected in parallel the effective resistance is 20 ohms . what are the values of the two resistances

Answer 1

Let R1 and R2 be the two resistances. When they are in parallel, resultant resistance Rp is given by

1/Rp = 1/R1 + 1/R2
Rp = R1R2/(R1 + R2)
2 = R1R2/(R1 + R2)
2(R1 + R2) = R1R2

When they are in series effective resistance 

Rs = R1 + R2
9 = R1 + R2 ⇒ R2 = 9 - R1

2 x 9 = R1 R2
          = R1 (9 - R1)
          = 9R1 - R12.
18 = 9R1 - R12.
R12 - 9R1 + 18 = 0

Solve this quadratic equation. You will get 

R1 = (9 ± √(81 - 4 x 1 x 18))/2
      = (9 ± √9)/2
      =  (9 ± 3)/2
R1 = 3 ohm or 6 ohm

If R1 = 3 ohm, R2 = 6 ohm and vice versa.

Answer 2

The values of two resistances are 40 ohm and 40 ohm.

Explanation:

Given:

Two resistors are connected in series the effective resistance is 80 ohm.

Two resistors are connected in parallel the effective resistance is 20 ohms.

To Find:

The values of two resistances.
Formula Used:

Let the two resistances are $R_{1} \ and\ R _{2}$.

In series connection:

The effective resistance,$R_{s} =R_{1} +R_{2}$.  ------ Formula no.01.

In parallel connection:

The effective resistance $R_{p} =\frac{R_{1}.R_{2} }{R_{1} +R_{2} }$    --------- formula no.02.

Solution:

As given, two resistors are connected in series the effective resistance is 80 ohm.

$R_{s} =80 \ ohm.$

Applying formula no.01.

$80 =R_{1} +R_{2}$   --------- equation no.01.

As given,two resistors are connected in parallel the effective resistance is 20 ohms.

$R_{p} =20 \ ohm$

Applying formula no.02.

$20 =\frac{R_{1}.R_{2} }{R_{1} +R_{2} }$    -----  equation no.02.

Putting the value of $R_{2}$ form equation no.01 in equation no.02. we get.

$20 =\frac{R_{1}(80-R_{1} ) }{R_{1} +(80-R_{1}) }$

$20R_{1} +1600- 20R_{1} = 80 R_{1}-R_{1}^{2}$

$R_{1}^{2} - 80R_{1} +1600 = 0$

$R_{1}^{2} - 40R_{1} -40R_{1} +1600 = 0\\R_{1}(R_{1}-40)-40(R_{1}-40)=0\\(R_{1}-40)(R_{1}-40)=0$

$R_{1}-40=0$

$R_{1}=40$

Putting value of $R_{1}$ in equation no.01, we get.

$80 =40 +R_{2}\\R_{2}=40$

Thus,the values of two resistances are 40 ohm and 40 ohm.

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