JEE Main 1-Sep-2021 Shift 2 Solved Paper

Given,
R1=(4±0.8)Ω
R2=(4±0.4)Ω
Equivalent resistance when the resistors are connected in parallel isgiven by

1

Req

=

1

R1

+

1

R2

⇒

1

Req

=

1

4

+

1

4

Req=2Ω
Now,

ΔReq

Req2

=

ΔR1

R12

+

ΔR2

R22

Substituting the values in the above equation, we get

ΔReq

4

=

0.8

16

+

0.4

16

ΔReq=0.3Ω
∴ The equivalent resistance in parallel combination is
Req=(2±0.3)Ω.