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Two resistors have resistances R1=(24+-0.5)ohm and R2=(8+-0.3 ) ohm .Calculate the absolute error and the percentage relative error in calculating the combination of two resistances when they are in parallel ? please answer fast?
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Two resistors have resistances R1=(24+-0.5)ohm and R2=(8+-0.3 ) ohm .C...
Calculating Absolute and Percentage Relative Error in Parallel Resistors
Given:
Resistance of R1 = (24 - 0.5) ohm
Resistance of R2 = (8 - 0.3) ohm
To Find:
Absolute error and percentage relative error in calculating the combination of two resistances when they are in parallel
Solution:
Step 1: Calculate the combined resistance of the two resistors when in parallel.
The formula for calculating the total resistance (Rp) of two resistors in parallel is:
1/Rp = 1/R1 + 1/R2
Substituting the given values:
1/Rp = 1/(24 - 0.5) + 1/(8 - 0.3)
1/Rp = 0.0417 + 0.125
1/Rp = 0.1667
Rp = 6 ohm
Therefore, the combined resistance of the two resistors when in parallel is 6 ohm.
Step 2: Calculate the absolute error in the combined resistance value.
The absolute error is the difference between the measured value and the actual value.
Absolute error = |Actual value - Measured value|
Actual value = 6 ohm (calculated in step 1)
Measured value = (24 - 0.5) || (8 - 0.3)
Measured value = 23.5 || 7.7
Measured resistance in parallel = (23.5 x 7.7) / (23.5 + 7.7)
Measured resistance in parallel = 6.05 ohm
Absolute error = |6 - 6.05|
Absolute error = 0.05 ohm
Therefore, the absolute error in the combined resistance value is 0.05 ohm.
Step 3: Calculate the percentage relative error in the combined resistance value.
The percentage relative error is the absolute error divided by the actual value, multiplied by 100.
Percentage relative error = (Absolute error / Actual value) x 100
Percentage relative error = (0.05 / 6) x 100
Percentage relative error = 0.83%
Therefore, the percentage relative error in the combined resistance value is 0.83%.
Conclusion:
The absolute error in the combined resistance value is 0.05 ohm and the percentage relative error is 0.83%.
Given:
Resistance of R1 = (24 - 0.5) ohm
Resistance of R2 = (8 - 0.3) ohm
To Find:
Absolute error and percentage relative error in calculating the combination of two resistances when they are in parallel
Solution:
Step 1: Calculate the combined resistance of the two resistors when in parallel.
The formula for calculating the total resistance (Rp) of two resistors in parallel is:
1/Rp = 1/R1 + 1/R2
Substituting the given values:
1/Rp = 1/(24 - 0.5) + 1/(8 - 0.3)
1/Rp = 0.0417 + 0.125
1/Rp = 0.1667
Rp = 6 ohm
Therefore, the combined resistance of the two resistors when in parallel is 6 ohm.
Step 2: Calculate the absolute error in the combined resistance value.
The absolute error is the difference between the measured value and the actual value.
Absolute error = |Actual value - Measured value|
Actual value = 6 ohm (calculated in step 1)
Measured value = (24 - 0.5) || (8 - 0.3)
Measured value = 23.5 || 7.7
Measured resistance in parallel = (23.5 x 7.7) / (23.5 + 7.7)
Measured resistance in parallel = 6.05 ohm
Absolute error = |6 - 6.05|
Absolute error = 0.05 ohm
Therefore, the absolute error in the combined resistance value is 0.05 ohm.
Step 3: Calculate the percentage relative error in the combined resistance value.
The percentage relative error is the absolute error divided by the actual value, multiplied by 100.
Percentage relative error = (Absolute error / Actual value) x 100
Percentage relative error = (0.05 / 6) x 100
Percentage relative error = 0.83%
Therefore, the percentage relative error in the combined resistance value is 0.83%.
Conclusion:
The absolute error in the combined resistance value is 0.05 ohm and the percentage relative error is 0.83%.
Community Answer
Two resistors have resistances R1=(24+-0.5)ohm and R2=(8+-0.3 ) ohm .C...
Hey there,,,, Here is Ur ans,,,, =======>
Relative error for parallel combination of resistors is _____
dR/R^2= dR1/R1^2+dR2/R2^2
= 0.5/(24)^2+0.3/8^2
= 0.5/ 24*24 + 0.3/ 64
= 0.5/576+0.3/ 64
= 32 + 172.8/ 36,864
=204.8/ 36,864=0.0055
=0.006_________________________
Percentage error =Relative error *100
= 0.006* 100 = 0.6
Relative error for parallel combination of resistors is _____
dR/R^2= dR1/R1^2+dR2/R2^2
= 0.5/(24)^2+0.3/8^2
= 0.5/ 24*24 + 0.3/ 64
= 0.5/576+0.3/ 64
= 32 + 172.8/ 36,864
=204.8/ 36,864=0.0055
=0.006_________________________
Percentage error =Relative error *100
= 0.006* 100 = 0.6
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Two resistors have resistances R1=(24+-0.5)ohm and R2=(8+-0.3 ) ohm .Calculate the absolute error and the percentage relative error in calculating the combination of two resistances when they are in parallel ? please answer fast?
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Two resistors have resistances R1=(24+-0.5)ohm and R2=(8+-0.3 ) ohm .Calculate the absolute error and the percentage relative error in calculating the combination of two resistances when they are in parallel ? please answer fast? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Two resistors have resistances R1=(24+-0.5)ohm and R2=(8+-0.3 ) ohm .Calculate the absolute error and the percentage relative error in calculating the combination of two resistances when they are in parallel ? please answer fast? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two resistors have resistances R1=(24+-0.5)ohm and R2=(8+-0.3 ) ohm .Calculate the absolute error and the percentage relative error in calculating the combination of two resistances when they are in parallel ? please answer fast?.
Two resistors have resistances R1=(24+-0.5)ohm and R2=(8+-0.3 ) ohm .Calculate the absolute error and the percentage relative error in calculating the combination of two resistances when they are in parallel ? please answer fast? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Two resistors have resistances R1=(24+-0.5)ohm and R2=(8+-0.3 ) ohm .Calculate the absolute error and the percentage relative error in calculating the combination of two resistances when they are in parallel ? please answer fast? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two resistors have resistances R1=(24+-0.5)ohm and R2=(8+-0.3 ) ohm .Calculate the absolute error and the percentage relative error in calculating the combination of two resistances when they are in parallel ? please answer fast?.
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