Class 11 Exam > Class 11 Questions > Two resistors have resistances R1=(24+-0.5)oh...
Start Learning for Free
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?
Most Upvoted Answer
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
Attention Class 11 Students!
To make sure you are not studying endlessly, EduRev has designed Class 11 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 11.
|
Explore Courses for Class 11 exam
|
|
Similar Class 11 Doubts
Top Courses for Class 11View all
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?
Question Description
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?.
Solutions 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? in English & in Hindi are available as part of our courses for Class 11.
Download more important topics, notes, lectures and mock test series for Class 11 Exam by signing up for free.
Here you can find the meaning of 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? defined & explained in the simplest way possible. Besides giving the explanation of
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?, a detailed solution 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? has been provided alongside types of 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? theory, EduRev gives you an
ample number of questions to practice 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? tests, examples and also practice Class 11 tests.
|
|
Explore Courses for Class 11 exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup