In the given network of capacitors as shown in the figure, given that C1=C2=C3=400pF and C4=C5=C6=200pF. The effective capacitance of the circuit between X and Y is.
A
810pF
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B
205pF
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C
600pF
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D
410pF
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Solution
The correct option is D410pF Let us use the technique of succesive reduction.
Here, in thefigure C′=(C3+C4)(C2)C3+C4+C2
⇒C′=(400+200)×(400)(400+200+400)=(600)(400)1000
⇒C′=240pF
Now, applying the formula and redrawing the circuit,
C′′=(400)(440)840=209.5pF
Now, C′′ and C6 are in parallel. So, the equivalent capacitance,