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`        Two resistors R1 = 600 +- 3 Ohm and R2 = 300 +- 6 Ohm are connected in series combination. Find the percentage error in net resistance.`
8 years ago

17 Points
```										Net resistance R = R1 + R2
(In a series combination, resistances get added)
Max error in measurement of R1 = 3 ohm.
Relative error = Max. error/Measured value
Therefore,
Relative error for R1 = 3/600 = 0.005
Max error in measurement of R2 = 6 ohm.
Relative error for R2 = 6/300 = 0.02

When two quantities are added, the respective errors involved also get added.
R = R1 + R2
Measured value of R = 600 + 300 = 900 ohms
Error in measuring R = 3 + 6 = 9 ohm
Relative Error in measurement of R = 9/900 = 0.01
Percentage error = Relative error * 100
or, Percentage error = 0.01* 100
or, Percentage error in measuring R = 1 %

```
8 years ago
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