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Measurements and Instrumentation

Q1:

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

A 75.0 μF

B 75.0 ± 0.1 μF

C 75.1 ± 0.1 μF

D 75.002 ± 0.1 μF

ANS:B - 75.0 ± 0.1 μF

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.



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