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Two resistors of resistances R1=100_ 3ohm and R2=200_ 4ohm are connected (a)in series,(b) in parallel.Find the equivalent resistance of the (a) series combination,(b) parallel combination.Use for (a) the relation R=R1+R2 and for (b) 1/R' = 1/R1+ 1/R2 and ∆R'/R'^2 =∆R1/R1^2+ ∆R2/R2^2. Can anyone please explain this?
if anyone didn't understand u can refer NCERT P.S +1 part-1 pg.no 27 ( example 2.10 )?
if anyone didn't understand u can refer NCERT P.S +1 part-1 pg.no 27 ( example 2.10 )?
Most Upvoted Answer
Two resistors of resistances R1=100_ 3ohm and R2=200_ 4ohm are connect...
Finding Equivalent Resistance of Resistors in Series and Parallel
Series Combination of Resistors
In a series combination of resistors, the resistors are connected end-to-end such that the current flows through them one after the other. The equivalent resistance of the series combination is the sum of the individual resistances.
Formula: R = R1 + R2
Where R is the equivalent resistance, R1 and R2 are the resistances of the individual resistors.
Parallel Combination of Resistors
In a parallel combination of resistors, the resistors are connected such that the current is divided among them. The equivalent resistance of the parallel combination is calculated using the reciprocal of the sum of the reciprocals of individual resistances.
Formula: 1/R' = 1/R1 + 1/R2
Where R' is the equivalent resistance, R1 and R2 are the resistances of the individual resistors.
Calculation of Equivalent Resistance of Resistors in Series and Parallel
Given: R1 = 100_ 3ohm, R2 = 200_ 4ohm
Series Combination:
Using the formula R = R1 + R2, we get:
R = 100_ 3ohm + 200_ 4ohm
R = 100 + 200_ 3 + 4ohm
R = 300_ 7ohm
Therefore, the equivalent resistance of the series combination is 300_ 7ohm.
Parallel Combination:
Using the formula 1/R' = 1/R1 + 1/R2, we get:
1/R' = 1/100_ 3ohm + 1/200_ 4ohm
1/R' = (4 + 3)/(400 + 600)ohm
1/R' = 7/1000ohm
R' = 1000/7_ ohm
Therefore, the equivalent resistance of the parallel combination is 1000/7_ ohm.
Uncertainty in Resistance Calculation
To calculate the uncertainty in resistance, we use the formula:
∆R'/R'^2 = ∆R1/R1^2 + ∆R2/R2^2
Where ∆R' is the uncertainty in the equivalent resistance, ∆R1 and ∆R2 are the uncertainties in the individual resistances.
For the given resistors, assume an uncertainty of ± 1_ ohm for both R1 and R2.
Using the above formula, we get:
∆R'/R'^2 = (∆R1/R1^2) + (∆R2/R2^2)
∆R'/(1000/7)^2 = (1/100^2) + (1/200^2)
∆R' = (1000/7)^2((1/100^2) + (1/200^2))
∆R' = 2.08_ ohm
Therefore, the uncertainty in the equivalent resistance is ± 2.08_ ohm.
Series Combination of Resistors
In a series combination of resistors, the resistors are connected end-to-end such that the current flows through them one after the other. The equivalent resistance of the series combination is the sum of the individual resistances.
Formula: R = R1 + R2
Where R is the equivalent resistance, R1 and R2 are the resistances of the individual resistors.
Parallel Combination of Resistors
In a parallel combination of resistors, the resistors are connected such that the current is divided among them. The equivalent resistance of the parallel combination is calculated using the reciprocal of the sum of the reciprocals of individual resistances.
Formula: 1/R' = 1/R1 + 1/R2
Where R' is the equivalent resistance, R1 and R2 are the resistances of the individual resistors.
Calculation of Equivalent Resistance of Resistors in Series and Parallel
Given: R1 = 100_ 3ohm, R2 = 200_ 4ohm
Series Combination:
Using the formula R = R1 + R2, we get:
R = 100_ 3ohm + 200_ 4ohm
R = 100 + 200_ 3 + 4ohm
R = 300_ 7ohm
Therefore, the equivalent resistance of the series combination is 300_ 7ohm.
Parallel Combination:
Using the formula 1/R' = 1/R1 + 1/R2, we get:
1/R' = 1/100_ 3ohm + 1/200_ 4ohm
1/R' = (4 + 3)/(400 + 600)ohm
1/R' = 7/1000ohm
R' = 1000/7_ ohm
Therefore, the equivalent resistance of the parallel combination is 1000/7_ ohm.
Uncertainty in Resistance Calculation
To calculate the uncertainty in resistance, we use the formula:
∆R'/R'^2 = ∆R1/R1^2 + ∆R2/R2^2
Where ∆R' is the uncertainty in the equivalent resistance, ∆R1 and ∆R2 are the uncertainties in the individual resistances.
For the given resistors, assume an uncertainty of ± 1_ ohm for both R1 and R2.
Using the above formula, we get:
∆R'/R'^2 = (∆R1/R1^2) + (∆R2/R2^2)
∆R'/(1000/7)^2 = (1/100^2) + (1/200^2)
∆R' = (1000/7)^2((1/100^2) + (1/200^2))
∆R' = 2.08_ ohm
Therefore, the uncertainty in the equivalent resistance is ± 2.08_ ohm.
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Two resistors of resistances R1=100_ 3ohm and R2=200_ 4ohm are connect...
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Two resistors of resistances R1=100_ 3ohm and R2=200_ 4ohm are connected (a)in series,(b) in parallel.Find the equivalent resistance of the (a) series combination,(b) parallel combination.Use for (a) the relation R=R1+R2 and for (b) 1/R' = 1/R1+ 1/R2 and ∆R'/R'^2 =∆R1/R1^2+ ∆R2/R2^2. Can anyone please explain this? if anyone didn't understand u can refer NCERT P.S +1 part-1 pg.no 27 ( example 2.10 )?
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Two resistors of resistances R1=100_ 3ohm and R2=200_ 4ohm are connected (a)in series,(b) in parallel.Find the equivalent resistance of the (a) series combination,(b) parallel combination.Use for (a) the relation R=R1+R2 and for (b) 1/R' = 1/R1+ 1/R2 and ∆R'/R'^2 =∆R1/R1^2+ ∆R2/R2^2. Can anyone please explain this? if anyone didn't understand u can refer NCERT P.S +1 part-1 pg.no 27 ( example 2.10 )? 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=100_ 3ohm and R2=200_ 4ohm are connected (a)in series,(b) in parallel.Find the equivalent resistance of the (a) series combination,(b) parallel combination.Use for (a) the relation R=R1+R2 and for (b) 1/R' = 1/R1+ 1/R2 and ∆R'/R'^2 =∆R1/R1^2+ ∆R2/R2^2. Can anyone please explain this? if anyone didn't understand u can refer NCERT P.S +1 part-1 pg.no 27 ( example 2.10 )? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two resistors of resistances R1=100_ 3ohm and R2=200_ 4ohm are connected (a)in series,(b) in parallel.Find the equivalent resistance of the (a) series combination,(b) parallel combination.Use for (a) the relation R=R1+R2 and for (b) 1/R' = 1/R1+ 1/R2 and ∆R'/R'^2 =∆R1/R1^2+ ∆R2/R2^2. Can anyone please explain this? if anyone didn't understand u can refer NCERT P.S +1 part-1 pg.no 27 ( example 2.10 )?.
Two resistors of resistances R1=100_ 3ohm and R2=200_ 4ohm are connected (a)in series,(b) in parallel.Find the equivalent resistance of the (a) series combination,(b) parallel combination.Use for (a) the relation R=R1+R2 and for (b) 1/R' = 1/R1+ 1/R2 and ∆R'/R'^2 =∆R1/R1^2+ ∆R2/R2^2. Can anyone please explain this? if anyone didn't understand u can refer NCERT P.S +1 part-1 pg.no 27 ( example 2.10 )? 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=100_ 3ohm and R2=200_ 4ohm are connected (a)in series,(b) in parallel.Find the equivalent resistance of the (a) series combination,(b) parallel combination.Use for (a) the relation R=R1+R2 and for (b) 1/R' = 1/R1+ 1/R2 and ∆R'/R'^2 =∆R1/R1^2+ ∆R2/R2^2. Can anyone please explain this? if anyone didn't understand u can refer NCERT P.S +1 part-1 pg.no 27 ( example 2.10 )? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two resistors of resistances R1=100_ 3ohm and R2=200_ 4ohm are connected (a)in series,(b) in parallel.Find the equivalent resistance of the (a) series combination,(b) parallel combination.Use for (a) the relation R=R1+R2 and for (b) 1/R' = 1/R1+ 1/R2 and ∆R'/R'^2 =∆R1/R1^2+ ∆R2/R2^2. Can anyone please explain this? if anyone didn't understand u can refer NCERT P.S +1 part-1 pg.no 27 ( example 2.10 )?.
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Two resistors of resistances R1=100_ 3ohm and R2=200_ 4ohm are connected (a)in series,(b) in parallel.Find the equivalent resistance of the (a) series combination,(b) parallel combination.Use for (a) the relation R=R1+R2 and for (b) 1/R' = 1/R1+ 1/R2 and ∆R'/R'^2 =∆R1/R1^2+ ∆R2/R2^2. Can anyone please explain this? if anyone didn't understand u can refer NCERT P.S +1 part-1 pg.no 27 ( example 2.10 )?, a detailed solution for Two resistors of resistances R1=100_ 3ohm and R2=200_ 4ohm are connected (a)in series,(b) in parallel.Find the equivalent resistance of the (a) series combination,(b) parallel combination.Use for (a) the relation R=R1+R2 and for (b) 1/R' = 1/R1+ 1/R2 and ∆R'/R'^2 =∆R1/R1^2+ ∆R2/R2^2. Can anyone please explain this? if anyone didn't understand u can refer NCERT P.S +1 part-1 pg.no 27 ( example 2.10 )? has been provided alongside types of Two resistors of resistances R1=100_ 3ohm and R2=200_ 4ohm are connected (a)in series,(b) in parallel.Find the equivalent resistance of the (a) series combination,(b) parallel combination.Use for (a) the relation R=R1+R2 and for (b) 1/R' = 1/R1+ 1/R2 and ∆R'/R'^2 =∆R1/R1^2+ ∆R2/R2^2. Can anyone please explain this? if anyone didn't understand u can refer NCERT P.S +1 part-1 pg.no 27 ( example 2.10 )? theory, EduRev gives you an
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