Question

two resistances when connected in parallel give effective resistance 20ohms and when connected in series equivalent resistance is 90ohm calculate the value of two resistances​

Answer 1

Answer:

30 Ω and 60 Ω.

Explanation:

Let first resistance be a and second be b

In series there is sum of both

So  a + b = 90 .

  b = 90 - a  ... ( i )

In parallel there is sum of reciprocal of both

$\large \dfrac{1}{a}+\dfrac{1}{b} =\dfrac{1}{20} \\\\\\\large \dfrac{b + a}{ab} =\dfrac{1}{20}\\\\\\\large \dfrac{90\times20}{a} = b\\\\\\\large \dfrac{1800}{a} =b \ ...(ii))$

From ( i ) and ( ii ) we have

90 - a = 9 / 2 a

$\large 90-a=\dfrac{1800}{a} \\\\\\\large 90a - {a}^{2}= 1800\\\\\\\large a^{2}-90a-1800=0\\\\\\\large \text{On solving this we get a = 30 or a = 60}$

Now putting a = 30 in ( i )

a + b = 90

b = 90 - 30

b = 60

when we put a = 60

we get b = 30

First resistance = 30 Ω

Second resistance = 60 Ω

Thus we get answer .

Answer 2

Answer :-

$R_1 = 60\Omega$

$R_2 = 30 \Omega$

Given:-

Effective parallel resistance = 20 $\Omega$

Effective series resistance = 90 $\Omega$

To find :-

The value of two resistance.

Solution:-

Let the two resistance be $R_1 and R_2$ respectively.

A/Q

$R_S = 90\Omega$

$R_p = 20\Omega$

Now in series arrangement, the equivalent resistance is given by :-

$\boxed{\sf{ R_s = R_1+R_2......R_n}}$

And in parallel arrangement the equivalent resistance is given by :-

$\dfrac{1}{R_p}= \dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_n}$

Now,

$R_s = R_1 + R_2$ ----eq. 1

$\dfrac{1}{R_p}= \dfrac {1}{R_1}+\dfrac{1}{R_2}$

$\implies$ $R_1 + R_2 = 90$

$\implies$ $\dfrac{1}{20}= \dfrac {1}{R_1}+\dfrac{1}{R_2}$

$\implies$ $\dfrac {1}{20}= \dfrac {R_1+R_2}{R_1.R_2}$

$\implies$ $\dfrac {1}{20}= \dfrac{90}{R_1.R_2}$

$\implies$ $R1_.R_2= 1800$

By using suitable identity :-

$(R_1-R_2)^2 = (R_1 + R_2 )^2 -4.R_1.R_2$

$\implies$ $(R_2-R_2)^2 = (90)^2 -4.1800$

$\implies$ $(R1_-R_2)^2 = 8100 - 7200$

$\implies$ $(R_1-R_2)^2 = 900$

$\implies$ $(R1_-R_2) = \sqrt{900}$

$\implies$ $(R_1 - R_2 ) = 30$----eq.2

Now,

$R_1 + R_2 = 90-----eq.1\\R_1 -R_2 = 30-----eq.2$

adding equation 1. and 2.

$R_1 +R_1 + R_2 -R_2 = 120$

$2R_1 = 120$

$R_1 = 60 \Omega$

Put the value of $R_1$ in eq. 1.

$R_2 = 30 \Omega$