A resistor of 10 k Ω having tolerance 10% is connected in series with another resistor of 20 k Ω having tolerance 20%. The tolerance of the combination will be(a) 10%(b) 13%(c) 30%(d) 20%
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in place of 10 mark as 6
Propagation of error through mathematical operations:
When adding or subtracting quantities, their errors add.
When any number of quantities are multiplied or divided, their percentage error add.
So, if two resistors are connected in series, then the effective resistance is addition of individual resistance values. Hence according to rule one above, the values of their error will add.
10% of 6k is 600 ohm and 10% of 4k is 400 ohm hence the answer in this case is 1000 ohm, which is interestingly 10% of 10k.
If these resistors are connected in parallel, then the effective resistance is given by (R1*R2)/(R1+R2).
Note that the denominator here is again addition of both resistor values hence it will give an error of 1000 ohm or 10% of 10k.
According to rule 2 above, the numerator will have an error of 20% (10%+10%=20%).
When we divide numerator by denominator, again percentage errors will get added (20% for N+ 10% for D= 30%)
Hence parallel combination will give 30% error.
Edit 1: As pointed out by Phil chalk, the resistance tolerance indeed remains 10% for the parallel combination of resistance, as opposed to 30% calculated by me in the original answer.
The reason being the original expression of parallel resistance combination is 1/(1/R1+1/R2). The expression (R1*R2)/(R1+R2) have redundant terms hence the tolerance I got earlier was wrong.
in place of 10 mark as 6
Propagation of error through mathematical operations:
When adding or subtracting quantities, their errors add.
When any number of quantities are multiplied or divided, their percentage error add.
So, if two resistors are connected in series, then the effective resistance is addition of individual resistance values. Hence according to rule one above, the values of their error will add.
10% of 6k is 600 ohm and 10% of 4k is 400 ohm hence the answer in this case is 1000 ohm, which is interestingly 10% of 10k.
If these resistors are connected in parallel, then the effective resistance is given by (R1*R2)/(R1+R2).
Note that the denominator here is again addition of both resistor values hence it will give an error of 1000 ohm or 10% of 10k.
According to rule 2 above, the numerator will have an error of 20% (10%+10%=20%).
When we divide numerator by denominator, again percentage errors will get added (20% for N+ 10% for D= 30%)
Hence parallel combination will give 30% error.
Edit 1: As pointed out by Phil chalk, the resistance tolerance indeed remains 10% for the parallel combination of resistance, as opposed to 30% calculated by me in the original answer.
The reason being the original expression of parallel resistance combination is 1/(1/R1+1/R2). The expression (R1*R2)/(R1+R2) have redundant terms hence the tolerance I got earlier was wrong.
Answered by
10
Answer:
C) 30%
Explanation:
R1 = 10K & Tolerance is 10%
10K = 10000 Ohms
10% of 10K = 1000 Ohms
Therefore, R1 Min = 10000- 1000 = 9000 = 9K
R1 max = 10000 + 1000 = 11000 = 11.1K
R2 = 20K & Tolerance = 20%
20K = 20000 ohms
R2 Min = 20000–4000 = 16000 = 16K
R2 Max = 20000+4000 = 24000 = 24K
In series the total Resistance = R1+R2
R1 Min+ R2 Min = 9K+16K = 25K
R1 Max + R2 Max = 11+ 24K = 35K
Nominal Value of Series is R1+R2 = 20K +10K = 30K
Overall tolerance = +/- 30%
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