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A circuit is shown in figure for which `C_(1)=(3pm 0.011)muF, C_(2)=(5pm 0.01)muF and C_(3)=(1pm 0.01)muF.` If C is the equivalent capacitance across AB, then C is given by A. `(0.9 pm 0.114)muF`B. `(0.9 pm 0.01)muF`C. `(0.9 mu 0.023)muF`D. `(0.9 mu 0.09) muF`

Answer» Correct Answer - C
The capacitor `C_(2)` is shorted, so it is not playing any role in circuit and can be removed. The 3 capacitors each of `C_(1)` are connected in parallel and this is connected to `C_(3)` in series.
`C_(eq)=(3C_(1)C_(3))/(3C_(1)+C_(3))=C`
`=(3xx3xx1)/(3xx3+1)=0.9muF`
So,`" "(DeltaC)/(C)=(3DeltaC_(1))/(C_(1))+(DeltaC_(3))/(C_(3))+(3DeltaC_(1)+DeltaC_(3))/(3C_(1)+C_(3))`
[For computaion of errors worst has to be taken]
`rArr" "(DeltaC)/(0.9)=(3xx0.011)/(3)+(0.01)/(1)+((0.033+0.01))/(10)`
`rArr" "DeltaC=pm 0.023muF`


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