Physics Question please solve it

Solution:-

As we know that,

It is given that the resistance of two conductors in parallel is 12 ohm and in series is 50 Ohm.

Given,

r1 = 12 Ohm
r2 = 50 Ohm

Let r1 and r2 be the unknown resistance.

(i).

When the resistances are in series combination.

It is given that:

r1 + r2 = 50 Ohm.

(ii).

When the resistances are in parallel combination.

As we know that,

=>1/r1 + 1/r2

=> r1r2/r1 + r2 = 1/R(say)

where R is given as 12 Ohm.

Therefore, r1 + r2 = 50 Ohm

and r1r2/r1 + r2 = 12 Ohm.

=> r1r2 = 12 × 50

=> r1r2 = 600

Also,

=> ( r1 - r2 )^2 = ( r1 + r2 )^2 - 4r1r2

=> ( r1 - r2 )^2 = 2500 - 4 × 600

=> ( r1 - r2 )^2 = 100

=> ( r1 - r2 )^2 = (10)^2

=> r1 - r2 = 10

From (i) and (iii), 2r1 = 60

Therefore, r1 = 30 Ohm and r2 = 20 Ohm.

abhaysilver1 abhaysilver1 · Answer 2 · 2019-03-18T08:03:49+01:00

Here is your answer

Let r1 and r2 be the resistances of the two conductors.

In series,

Resultant resistance,R= r1+r2

So, r1+ r2= 50ohm ...... (i)

In parallel,

Resultant resistance,

1/R= 1/r1 + 1/r2

[tex]\frac{1}{R}= \frac{r1+r2}{r1×r2}[/tex]

[tex]R= \frac{r1×r2}{r1+r2}[/tex]

[tex]12= \frac{r1×r2}{50} [/tex] - using (i)

r1×r2= 12×50= 600.......(ii)

Now,

(r1-r2)^2= (r1+r2)^2 - 4r1×r2

(r1-r2)^2= 50^2 - 4×600

(r1-r2)^2= 2500-2400= 100

(r1-r2)^2= 10^2

r1-r2= 10 ...... (iii)

Adding i and iii, we get

2r1= 60

r1= 30

So, r2= 50- 30 = 20

Hence,

r1= 30 ohms

and, r2= 20 ohms

HOPE IT IS USEFUL

Physics Question please solve it​

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Physics Question please solve it