Two resistors of resistances `R_(1) = 150 pm 2` Ohm and `R_(2) = 220 pm 6` Ohm are connected in parallel combination. Calculate the equivalent resistance.
Hint: `(1)/(R') = (1)/(R_(1)) + (1)/(R_(2))`

Explanation

The equivalent resistance of a parallel combination
`R. = (R_(1)R_(2))/(R_(1) + R_(2)) = (150 xx 220)/(150+220) = (33000)/(370) = 89.1` Ohm
We know that, `(1)/(R.) = (1)/(R_(1)) + (1)/(R_(2))`
`(DeltaR.)/((R.)^(2)) (DeltaR_(1))/(R_(1)^(2)) + (DeltaR_(2))/(R_(2)^(2))` `DeltaR. = (R.)^(2) (DeltaR_(1))/(R_(1)^(2)) + (R.)^(2) (DeltaR_(2))/(R_(2)^(2)) = ((R.)/(R_(1)))^(2) Delta R_(1) + ((R.)/(R_(2)))^(2) DeltaR_(2)`
Substituting the value,
`DeltaR. = [(89.1)/(150)]^(2) xx 2 + [(89.1)/(220)]^(2) xx 6 = 0.070 + 0.098 = 0.168` `R. = 89.1 pm 0.168` Ohm