R2ΔR=R12ΔR1+R22ΔR2
ΔR=(R)2[R12ΔR1+R22ΔR2]
=(2)2[420.8+420.4]
=[0.2+0.1]=0.3
Rnet =R+ΔR=2±0.3
JEE Main 2021 — Physics Electromagnetism
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 :
Held on 1 Sept 2021 · Verified 6 Jul 2026.
(4±0.4)Ω
(2±0.4)Ω
(4±0.3)Ω
(2±0.3)Ω
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