Given, ‌C1=(3±0.01l)µF ‌C2=(5±0.01)µF ‌C3=(1±0.01)µF=CQ The given circuit diagram is shown as
It can be redrawn as
As we know that, Parallel equivalent capacitance CP be ∴‌CP‌‌=Ca+Cb+Cc ∴‌‌CP‌‌=3C1‌ i.e. in series with ‌C3 ‌‌=(9±0.033)Ω ∴ Equivalent capacitance, C‌eq ‌=‌
CP×C3
CP+C3
C‌eq ‌‌‌=‌
9×1
9+1
∴‌‌C‌eq ‌‌‌=‌
9
10
=0.9 ‌ Now, net error ‌‌‌=3∆C1−∆C3 ‌‌=3×0.011−0.01 ‌‌=0.033−0.01=0.023 ∴ Equivalent capacitance with error, Ceq=(0.9±0.023)