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Question:

Figure shows a network of capacitors where the numbers indicates capacitances in micro Farad. The value of capacitance C if the equivalent capacitance between point A and B is to be 1µF is :

3223µF

3123µF

3323µF

3423µF

Solution:

Between points E and D: 12µF and 6µF are connected in series, So replace them with a single capacitor C′ ∴C′=12×612+6=4µF
Now this 4µF is connected in parallel with another 4µF capacitor. Equivalent capacitance between E and D CED=4µF+4µF=8µF
Now this 8µF is connected in series with another 1µF capacitor. Equivalent capacitance between R and D CRD=1×81+8=89µF
Between points P and Q: Equivalent capacitance between points P and Q, CPQ=2µF+2µF=4µF
This 4µF is connected in series with another 8µF capacitor. Equivalent capacitance between R and Q CRQ=4×84+8=83µF
Equivalent capacitance between R and B CRB=83µF+89µF=329µF
Equivalent capacitance between A and B CAB=C×329C+329 ∴1µF=329CC+329 ⇒C=3223µF