Question

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/R12+ΔR2/R​

Answer 1

Answer:

which chapter it belongs from

Answer 2

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(a) Series Combination

⭕️Resistance =R=R1+R2

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

(b) Parallel combination

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

 ⇒R = (R1x R2)/(R1+ R2)

 ∴ R = (100 x 200)/(100 + 200)

 ⇒ R = 20000/300 = 200/3

⇒ R = 66.67 ohm

⭕️Now to calculate error:-

 ∴ ∆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

∴ total equivalent resistance in parallel combination will be :-

 ⇒ R ± ∆R

 ⇒ 66.67 ± 1.78 Ohm!!!!!!!!!!!!

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