Question

2 resistance are expressed as R1=( 4 +- 0.5) ohm and R2= (12+-0.5)ohm. what is absolute error in the net resistance when they are connected in series
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Answer 1

Answer:

The error in the equivalent resistance R of two resistances R1 and R2 maybe calculated as under-

When the resistors are in series-

R= R1+R2

Let the error in R1 be ΔR1 and the error in R2 be ΔR2 and the error in R be ΔR.

By differentiating the above equation-

dR = d(R1) + d(R2)

ΔR = lΔR1l+lΔR2l

Given that

R1=4, ΔR1=+-0.5 ohm

R2=12 ΔR2 = +-0.5 ohm

Therefore

R= R1+R2= 4+12 =16

ΔR = 0.5+0.5= 1 ohm

When the resistors are in parallel,

1/R = 1/R1 + 1/R2

1/R = 1/4+/12

1/R = 1/3

R= 3

By differentiating the above equation-

dR/R^2 = dR1/R1^2 + dR2/R2^2

ΔR/9 = l ΔR1/16 l + lΔR2/144 l

ΔR = 9x (0.5/16 + 0.5/144)

ΔR = 4.5x ( 10/144)

ΔR = 0.3125

ΔR =0.3 ohm

Explanation:

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Answer 2

Explanation:

The error in the equivalent resistance R of two resistances R1 and R2 maybe calculated as under-

When the resistors are in series-

R= R1+R2

Let the error in R1 be ΔR1 and the error in R2 be ΔR2 and the error in R be ΔR.

By differentiating the above equation-

dR = d(R1) + d(R2)

ΔR = lΔR1l+lΔR2l

Given that

R1=4, ΔR1=+-0.5 ohm

R2=12 ΔR2 = +-0.5 ohm

Therefore

R= R1+R2= 4+12 =16

ΔR = 0.5+0.5= 1 ohm

When the resistors are in parallel,

1/R = 1/R1 + 1/R2

1/R = 1/4+/12

1/R = 1/3

R= 3

By differentiating the above equation-

dR/R^2 = dR1/R1^2 + dR2/R2^2

ΔR/9 = l ΔR1/16 l + lΔR2/144 l

ΔR = 9x (0.5/16 + 0.5/144)

ΔR = 4.5x ( 10/144)

ΔR = 0.3125

ΔR =0.3 ohm

Explanation:

please folluw meh, worked hard on the question.