Question

two resistance (150+-3)ohm and (200+-4)are connected in series.find the resultant resistance with error limit.​

Answer 1

Given:

Two resistances with absolute errors

• $\sf{R_1\pm\Delta R_1=150\pm 3\;\;\Omega}$
• $\sf{R_2\pm\Delta R_2=200\pm4\;\;\Omega}$

are connected in Series.

To find:

• Resultant resistance with error limit, $\sf{R\pm\Delta R=?}$

Knowledge required:

• Error of sum of quantities

When two quantities are added, the absolute error in the final result is the sum of the absolute error in individual quantities.

$\purple{\spadesuit}\;\;\;\boxed{\sf{\Delta Z = \Delta A + \Delta B}}$

[ Where $\Delta$ A and $\Delta$ B are the absolute errors in two quantities A and B; and ΔZ is the absolute error in result Z. (Such that Quantity Z = A + B) ]

Solution:

Using the rule for error of a sum calculating absolute error in the resultant $\sf{\Delta R=?}$

$\implies\sf{\Delta R = \Delta R_1 + \Delta R_2}$

$\implies\sf{\Delta R = 3 + 4}$

$\implies\sf{\Delta R = 7\;\;\Omega}$

AND

$\implies\sf{R=R_1+R_2}$

$\implies\sf{R=150+200}$

$\implies\sf{R=350\;\;\Omega}$

Hence,

Calculating Resultant Resistance with error limit

$\implies\boxed{\boxed{\large{\purple{\sf{R\pm\Delta R=350\pm 7\;\;\Omega}}}}}\;\;\;\purple{\spadesuit}$

Answer 2

Given:

• R1±Δ R1= 150±3 Ω

• R2±Δ R2=200±4 Ω
• Connected in series

Here,

• Δ R1= 3 Ω
• Δ R2= 4 Ω

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Need to find:

• R±Δ R =?

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In series connection we know,

Net Resistance =R1+R2

Thus

⟹ΔR=ΔR 1+ΔR2

⟹ΔR=3+4 Ω

$\pink{\boxed{\implies \Delta R=7 \Omega}}$

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Also

R=R1+R2

⟹R=150+200

$\pink{\boxed{\implies R=350 \Omega}}$

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Hence

• $\bf{\red{R\pm \Delta R = 350 \pm 7\Omega}}$

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MORE TO KNOW:

☆What Is absolute error?

• The difference between the measured value of a quantity and its actual value is called the absolute error.

• The absolute error of the sum or difference of a number of quantities is less than or equal to the sum of their absolute errors.