Two resistors of resistances R_{1}=100±3 ohm and R_{2}=200±4 ohm are connected

(a) In series, (b) in parallel. Find the equivalent resistance of the (a) series combination, (b) parallel combination. Use for (a) the relation R=R_{1}+R_{2 }and for (b) 1/R=1/R_{1}+1/R_{2} and ΔR_{1}/R_{1} ^{2}+ΔR_{2}/R_{2} ^{2} ?

CASE 1: Series Combination

For series combination equivalent resistance R = R_{1 }+ R_{2}

= R = 100 ± 3 + 200 ± 4 = R = 300 ± 7 ohm

As for addition or substraction the errors are simply added together.

CASE 2: Parallel combination

For parallel combination equivalent resistance is given by

1/R = 1/R_{1} + 1/R_{2} = 1/R = (R_{1} + R_{2})/(R_{1}x R2)

= R = (R_{1}x R2)/(R_{1}+ R_{2})

=Thus, R = (100 x 200)/(100 + 200)

= R = 20000/300 = 200/3

= R = 66.67 ohm

Now to calculate error we use

∆R/R^{2} = ∆R_{1}/R_{1}^{2}^{}+ ∆R_{2}/R_{2}^{2}

Putting values we get

∆R/(66.67)^{2 }= 3/(100)^{2 }+ 4/(200)^{2}

= ∆R/(66.67)^{2}= 3/10000 + 4/40000

= ∆R/(66.67)^{2} = 3/10000 + 1/10000

= ∆R/(66.67)^{2} = 4/10000

= ∆R = (4/10000) x (66.67)^{2}

= ∆R = 4/10000 x 4444.889

= ∆R = 17779.56/10000

= ∆R = 1.7779 = ∆R = 1.78 Ohm

Thus, total equivalent resistance in parallel combination

= R ± ∆R

= 66.67 ± 1.78 Ohm

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