Question

Two resistors are connected in series gives an equivalent resistance of 10 Ω. When

connected in parallel, gives 2.4 Ω. Then the individual resistance are
​

Answer 1

Answer:

The individual resistance are 6 ohms and 4 ohms.

Explanation:

Given that, two resistors are connected in series gives an equivalent resistance of 10Ω. When connected in parallel, gives 2.4Ω. We need to find the individual resistance.

Equivalent resistance in series combination is :

R= R1 + R2................(1)

10 = R1 + R2

Equivalent resistance when connected in parallel is :

1/R = 1/R1 + 1/R2

2.4 = 1/R1 + 1/R2 ............(2)

On solving (1) and (2) we get :

R1 = 6ohm

R2 = 4ohm

So, the individual resistance are 6 ohms and 4 ohms. Hence, this is the required solution

Answer 2

Answer:

6, 4

Explanation:

let the resistors have resistances R1 and R2.

When connected in series,

Equivalent resistance, R' = R1 + R2

R1 + R2 = 10 ohm

When connected in parallel,

Equivalent resistance, R' = R1 × R2/R1 + R2 = 2.4 ohm

R1 × R2/ 10 = 2.4

R1× R2 = 24

we can write,

R1 =10-R2

so,. ( 10 - R2 ) R2 = 24

let R2 = x

(10 - x)x = 24

10x - x^2 = 24

x^2 - 10x + 24 = 0

x^2 - 6x - 4x + 24 =0

x( x - 6 ) - 4( x - 6 ) = 0

x = 6 , 4

So, R1 and R2 = 6 , 4.

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