Question

What is d in this ques two resistance 100±3 ohm and 200±4 ohm are connected in series find the equivalent resistance with error limits

Answer 1

Answer:

Explanation:

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