Electronics and Communication Engineering - Measurements and Instrumentation - Discussion

Discussion Forum : Measurements and Instrumentation - Section 7 (Q.No. 44)
44.
Four capacitors are in parallel. Their values are 36.3 μF, 3.85 μF, 34.002 μF and 850 μF with an uncertainty of digit in the last place. The total capacitance in significant figures is
75.0 μF
75.0 ± 0.1 μF
75.1 ± 0.1 μF
75.002 ± 0.1 μF
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
2 comments Page 1 of 1.

Muhammad Owais said:   4 years ago
If capacitor are connected in parallel then their total capacitance is equal to their sum. So (36.3+3.85+34.002+0.850)uF = 75.002uF.

Note: In question, it should be 850nF not 850uF.

So the least sig no is 3.
Total answer is (75.0+_ 0.1)uF.
(2)

Sandy said:   7 years ago
Please explain the solution in detail.
(2)

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