Question

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

gives 2.4Ω. Then the individual resistance are

(a) each of 5Ω (b) 6Ω and 4Ω (c) 7Ω and 4Ω (d) 8Ω and 2Ω​

Answer 1

When in series, R1+R2=10
when in parallel,1/R1+1/r2=2.4
on solving, 10/R1R2= 2.4
hence,answer is b..6 and 4

Answer 2

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 :

$R=R_1+R_2$

$10=R_1+R_2$................(1)

Equivalent resistance when connected in parallel is :

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

$2.4=\dfrac{1}{R_1}+\dfrac{1}{R_2}$............(2)

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

$R_1=6\ \Omega\\\\R_2=4\ \Omega$

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

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