1.

R​​​​​​1 = 100 ± 3 R​​​​​​2 = 200 ± 4 Find the resistance in series and parallel combination . My solution : Series connection R​​​​​​S =(100±3)+(200±4) Rs =300±7 Parallel connection R​​​​​​p = (R​​​​​​1R​​​​2)÷(R​1 + R​​​​​​2) R​​​​​​p = {(3/100)(4/200)}÷(300±7) R​​​​​​p = (3/5000)÷(300±7) What to do after this ??? I am not able to get it . Please help me out .

Answer»

R​​​​​​1 = 100 ± 3

R​​​​​​2 = 200 ± 4

Find the resistance in series and parallel combination .

My solution :

Series connection

R​​​​​​S =(100±3)+(200±4)

Rs =300±7

Parallel connection

R​​​​​​p = (R​​​​​​1R​​​​2)÷(R​1 + R​​​​​​2)

R​​​​​​p = {(3/100)(4/200)}÷(300±7)

R​​​​​​p = (3/5000)÷(300±7)

What to do after this ???

I am not able to get it .

Please help me out .



Discussion

No Comment Found

Related InterviewSolutions