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

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For the series combination: R1 + R2 = 10 [1]

For the parallel combination: (R1•R2) / (R1+R2) = 12/5 [2]

Solve [1] R1 = 10 - R2 [3]

Substitute [3] into [2]: (10 - R2) ·R2/[(10- R2) + R2] = 12/5

Simplifying: (10·R2 = R2²)/10 = 12/5,

or

10·R2 - R2² = 24

Or R2² - 10 R2 + 24 = 0 ,

Or (R2 - 6)·(R2 - 4) = 0

If R2 = 6Ω, then R1 = 10 = R2 = 10 = 6 = 4Ω

Alternately,

if R2 = 4Ω, then R1 = 10 = 4 = 6Ω..

Therefore, R1= 4ohm ; R2= 6ohm..

HOPE IT HELPS..

Thanks and Regards..

Answer 2

Explanation:

this is your answer mate

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