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The values of two resistors are R1 = (6 ±0.3)kΩ and R2 = (10±0.0.2)kΩ. The percentage error in the equivalent resistors when they are connected in series is :

The values of two resistors are R1 = (6±0.3)kΩ and R2 = (10±0.0.2)kΩ. The percentage error in the equivalent resistors when they are connected in series is :

Grade:11

2 Answers

Khimraj
3007 Points
5 years ago
the equivalent resistor in series will be
(6±0.3)kΩ + (10±0.2)kΩ = (16+0.5)kΩ
Hope it clears.....................
Arun
25750 Points
5 years ago
Dear student
 
equivalent resistance when connected in series = R1 + R2
 
now error in resistance = \DeltaR1 / R1 + \DeltaR2 / R2 = 0.3 / 6 + 0.2 /10
 
 =  0.05 + 0.02 = 0.07
 
Now percentage error = 0.07 * 100 = 7%

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