TS EAMCET 5-Aug-2021 Shift 2 Question Paper

Given,
‌C1=(3±0.01l)µF
‌C2=(5±0.01)µF
‌C3=(1±0.01)µF=CQ
The given circuit diagram is shown asIt can be redrawn asAs 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)