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Two resistors of resistances R₁ = 300 ± 3 ohm and R2 = 200 ± 2 ohm are connected in parallel. The equivalent resistance of parallel combination with error is [NCERT Pg. 37] (4) [120±2.0] ohm A2B1/2 is (1) [120 1.8] ohm (3) [120 ± 1.6] ohm (2[120 1] ohm?
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Two resistors of resistances R₁ = 300 ± 3 ohm and R2 = 200 ± 2 ohm are...
Solution:
To find the equivalent resistance of the parallel combination, we can use the formula:
1/Req = 1/R1 + 1/R2
Where Req is the equivalent resistance and R1, R2 are the individual resistances.
Given:
R1 = 300 ± 3 ohm
R2 = 200 ± 2 ohm
Let's calculate the minimum and maximum values of R1 and R2:
Minimum value of R1 = 300 - 3 = 297 ohm
Maximum value of R1 = 300 + 3 = 303 ohm
Minimum value of R2 = 200 - 2 = 198 ohm
Maximum value of R2 = 200 + 2 = 202 ohm
Now, let's substitute these values into the formula to find the minimum and maximum values of Req:
1/Req(min) = 1/297 + 1/198
1/Req(max) = 1/303 + 1/202
Calculating these values:
1/Req(min) = 0.003367 + 0.005051 = 0.008418
1/Req(max) = 0.003300 + 0.004950 = 0.008250
Taking the reciprocal of these values to find Req:
Req(min) = 1/0.008418 = 118.763 ohm
Req(max) = 1/0.008250 = 121.212 ohm
So, the minimum and maximum values of the equivalent resistance are 118.763 ohm and 121.212 ohm respectively.
Answer:
The equivalent resistance of the parallel combination with error is [118.763 ± 2.449] ohm.
To find the equivalent resistance of the parallel combination, we can use the formula:
1/Req = 1/R1 + 1/R2
Where Req is the equivalent resistance and R1, R2 are the individual resistances.
Given:
R1 = 300 ± 3 ohm
R2 = 200 ± 2 ohm
Let's calculate the minimum and maximum values of R1 and R2:
Minimum value of R1 = 300 - 3 = 297 ohm
Maximum value of R1 = 300 + 3 = 303 ohm
Minimum value of R2 = 200 - 2 = 198 ohm
Maximum value of R2 = 200 + 2 = 202 ohm
Now, let's substitute these values into the formula to find the minimum and maximum values of Req:
1/Req(min) = 1/297 + 1/198
1/Req(max) = 1/303 + 1/202
Calculating these values:
1/Req(min) = 0.003367 + 0.005051 = 0.008418
1/Req(max) = 0.003300 + 0.004950 = 0.008250
Taking the reciprocal of these values to find Req:
Req(min) = 1/0.008418 = 118.763 ohm
Req(max) = 1/0.008250 = 121.212 ohm
So, the minimum and maximum values of the equivalent resistance are 118.763 ohm and 121.212 ohm respectively.
Answer:
The equivalent resistance of the parallel combination with error is [118.763 ± 2.449] ohm.
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Two resistors of resistances R₁ = 300 ± 3 ohm and R2 = 200 ± 2 ohm are connected in parallel. The equivalent resistance of parallel combination with error is [NCERT Pg. 37] (4) [120±2.0] ohm A2B1/2 is (1) [120 1.8] ohm (3) [120 ± 1.6] ohm (2[120 1] ohm?
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Two resistors of resistances R₁ = 300 ± 3 ohm and R2 = 200 ± 2 ohm are connected in parallel. The equivalent resistance of parallel combination with error is [NCERT Pg. 37] (4) [120±2.0] ohm A2B1/2 is (1) [120 1.8] ohm (3) [120 ± 1.6] ohm (2[120 1] ohm? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two resistors of resistances R₁ = 300 ± 3 ohm and R2 = 200 ± 2 ohm are connected in parallel. The equivalent resistance of parallel combination with error is [NCERT Pg. 37] (4) [120±2.0] ohm A2B1/2 is (1) [120 1.8] ohm (3) [120 ± 1.6] ohm (2[120 1] ohm? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two resistors of resistances R₁ = 300 ± 3 ohm and R2 = 200 ± 2 ohm are connected in parallel. The equivalent resistance of parallel combination with error is [NCERT Pg. 37] (4) [120±2.0] ohm A2B1/2 is (1) [120 1.8] ohm (3) [120 ± 1.6] ohm (2[120 1] ohm?.
Two resistors of resistances R₁ = 300 ± 3 ohm and R2 = 200 ± 2 ohm are connected in parallel. The equivalent resistance of parallel combination with error is [NCERT Pg. 37] (4) [120±2.0] ohm A2B1/2 is (1) [120 1.8] ohm (3) [120 ± 1.6] ohm (2[120 1] ohm? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two resistors of resistances R₁ = 300 ± 3 ohm and R2 = 200 ± 2 ohm are connected in parallel. The equivalent resistance of parallel combination with error is [NCERT Pg. 37] (4) [120±2.0] ohm A2B1/2 is (1) [120 1.8] ohm (3) [120 ± 1.6] ohm (2[120 1] ohm? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two resistors of resistances R₁ = 300 ± 3 ohm and R2 = 200 ± 2 ohm are connected in parallel. The equivalent resistance of parallel combination with error is [NCERT Pg. 37] (4) [120±2.0] ohm A2B1/2 is (1) [120 1.8] ohm (3) [120 ± 1.6] ohm (2[120 1] ohm?.
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Two resistors of resistances R₁ = 300 ± 3 ohm and R2 = 200 ± 2 ohm are connected in parallel. The equivalent resistance of parallel combination with error is [NCERT Pg. 37] (4) [120±2.0] ohm A2B1/2 is (1) [120 1.8] ohm (3) [120 ± 1.6] ohm (2[120 1] ohm?, a detailed solution for Two resistors of resistances R₁ = 300 ± 3 ohm and R2 = 200 ± 2 ohm are connected in parallel. The equivalent resistance of parallel combination with error is [NCERT Pg. 37] (4) [120±2.0] ohm A2B1/2 is (1) [120 1.8] ohm (3) [120 ± 1.6] ohm (2[120 1] ohm? has been provided alongside types of Two resistors of resistances R₁ = 300 ± 3 ohm and R2 = 200 ± 2 ohm are connected in parallel. The equivalent resistance of parallel combination with error is [NCERT Pg. 37] (4) [120±2.0] ohm A2B1/2 is (1) [120 1.8] ohm (3) [120 ± 1.6] ohm (2[120 1] ohm? theory, EduRev gives you an
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