Question

When two resistors Ry and Rare connected in parallel, then their parallel combination is given by R,R2 Rp If error in measurement of Ra and R1+R2 R2 are AR, and AR, respectively then error in measurement of their parallel combination, that is ARp is equal to​

Answer 1

Explanation:

solution : equivalent resistance of two resistor R₁ and R₂ is given by, 1/R=1/R₁+ 1/R₂

To find error in equivalent resistance, AR/R² = AR₁/R₁² + AR₂/R₂² 2

first find R,

i.e., 1/R = 1/10 + 1/20 = 3/20

> R= 20/3 Q

now AR/(20/3)² = 0.1/(10)² + (0.4)/(20)²

AR = (20/3)² [ 0.1/100 + 0.4/400]

→AR = 400/9 [ 0.1/100 + 0.1/100 ]

→AR = 400/9 × 0.2/100

→AR = 0.8/9 = 0.08

Therefore the relative error in their equipment resistance is 0.08.

Answer 2

Answer:

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Explanation:

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