Two resistors of resistances R₁=100±3 ohm and R₂=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=R₁+R₂ and for (b) 1/R=1/R₁+1/R₂ and ΔR₁/R₁ ²+ΔR₂/R₂ ² ?

CASE 1: Series Combination

For series combination equivalent resistance R = R₁ + R₂

= 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/R₁ + 1/R₂ = 1/R = (R₁ + R₂)/(R₁x R2)

= R = (R₁x R2)/(R₁+ R₂)

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

= R = 20000/300 = 200/3

= R = 66.67 ohm

Now to calculate error we use

∆R/R² = ∆R₁/R₁²+ ∆R₂/R₂²

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

CASE 1: Series Combination For series combination equivalent resistance R = R1 + R2 = 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/R1 + 1/R2 = 1/R = (R1 + R2)/(R1x R2) = R = (R1x R2)/(R1+ R2) =Thus, R = (100 x 200)/(100 + 200) = R = 20000/300 = 200/3 = R = 66.67 ohm Now to calculate error we use ∆R/R2 = ∆R1/R12+ ∆R2/R22 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

thanx a lot it was very helpful for me☺☺😊

Hope it help

Hope it may help

Answer

This is the answer of this question

Why is ΔR/R² = ΔR1/R1²+ ΔR2/R2² in this case?

CASE 1: Series Combination For series combination equivalent resistance R = R1 + R2 = 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/R1 + 1/R2 = 1/R = (R1 + R2)/(R1x R2) = R = (R1x R2)/(R1+ R2) =Thus, R = (100 x 200)/(100 + 200) = R = 20000/300 = 200/3 = R = 66.67 ohm Now to calculate error we use ∆R/R2 = ∆R1/R12+ ∆R2/R22 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

Why is it ∆R/R^2=∆R1/R1^2+∆R2/R2^2?

yeah its in the TB bro

Why in this case the error is not added in equivalent resistance

CASE 1: Series Combination For series combination equivalent resistance R = R1 + R2 = 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/R1 + 1/R2 = 1/R = (R1 + R2)/(R1x R2) = R = (R1x R2)/(R1+ R2) =Thus, R = (100 x 200)/(100 + 200) = R = 20000/300 = 200/3 = R = 66.67 ohm Now to calculate error we use ∆R/R2 = ∆R1/R12+ ∆R2/R22 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

what the f.................

I'm getting an error of 4.888889 if done by percentage error method

Can anyone please tell me why I am wrong?

CASE 1: Series Combination

For series combination equivalent resistance R = R1?+ R2

= 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/R1?+ 1/R2?= 1/R = (R1?+ R2)/(R1x R2)

= R = (R1x R2)/(R1+ R2)

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

= R = 20000/300 = 200/3

= R = 66.67 ohm

Now to calculate error we use

?R/R2?= ?R1/R12+ ?R2/R22

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