Two resistors R 1 = ( 4 ± 0 . 8 )   Ω and R 2 = ( 4 ± 0 . 4 )   Ω are…

Two resistors R1=(4±0.8) Ω and R2=(4±0.4) Ω are connected in parallel. The equivalent resistance of their parallel combination will be :
  1. (4±0.4) Ω
  2. (2±0.4) Ω
  3. (4±0.3) Ω
  4. (2±0.3) Ω

Solution

ΔRR2=ΔR1R12+ΔR2R22
ΔR=R2ΔR1R12+ΔR2R22
=220.842+0.442
=[0.2+0.1]=0.3
Rnet =R+ΔR=2±0.3

Asked in: JEE Main 2021 (01 Sep Shift 2)

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