Electronics and Communication Engineering - Measurements and Instrumentation - Discussion
Discussion Forum : Measurements and Instrumentation - Section 4 (Q.No. 11)
11.
The current through a resistance R is shown in figure. The computed value of power is
Answer: Option
Explanation:
% error in current is 2% and % error in resistance is 0.2%.
Since P = I2R, % error in power = 2 x 2% + 0.2% = 4.2%.
Discussion:
3 comments Page 1 of 1.
Vijay Kumar said:
5 years ago
Please explain clearly.
Rajiv Nautiyal said:
9 years ago
%error in current is 1% not 2% and in resistance, it is 0.2%.
So 1X2% + 0.2% = 2.2%.
And 2.2% of 400 = 8.8 W.
So 1X2% + 0.2% = 2.2%.
And 2.2% of 400 = 8.8 W.
Pardeep Kumar said:
9 years ago
Please explain it in clear.
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