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Comments ( 4 )

  1. VANAM MANI
    VANAM MANI :
    2 years ago

    Thank you Sajjad Ahamad

  2. Sajjad Ahamad
    Sajjad Ahamad :
    3 years ago

    Tolerance:-
    ∆R1 = 10*5/100 = .5kΩ
    ∆R2 = 5*10/100 = .5kΩ
    ∆Req = .5*.5/.5+.5 = 2.5kΩ [since connected in parallel]
    Equivalent Resistance:-
    R1 = 10kΩ, R2 = 5kΩ
    Req = 10*5/5+10 = 10/3 = 3.33kΩ
    Tolerance limit of parallel network:-
    3.33*2.5kΩ
    8.33%



  3. Brij N.
    Brij N. :
    3 years ago

    tolerance=Req(5/10k +10/5k)
    =3.33k*2.5
    =8.33

  4. Kiran Marathi
    Kiran Marathi :
    4 years ago

    Please share detailed description

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