What is the equivalent capacitance of the sys | vedclass

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Question

The capacitors in the right hand square are in series. So they can be replaced by an equivalent capacitor of capacitance of$:$

$1 /(1 / C+1 / C)=C

Vedclass · Accepted Answer

$1.6\, C$

Vedclass · Answer

The capacitors in the right hand square are in series. So they can be replaced by an equivalent capacitor of capacitance of$:$

$1 /(1 / C+1 / C)=C / 2$

The next step is to replace $\mathrm{C}$ and $\mathrm{C} / 2$ by a capacitor having an equivalent capacitance of $\mathrm{C}$ and $\mathrm{C} / 2$ in parallel.

$C+C / 2=3 C / 2$

Then, $\mathrm{C}$ and $3 \mathrm{C} / 2$ are in series. So, they can be replaced by a capacitor having an equivalent capacitance of

$1 /(1 / C+2 / 3 C)=3 C / 5$

Then since in parallel they can be replaced by an equivalent capacitance of

$C+$ $3 C / 5=8 C / 5=1.6 C$

So the equivalent capacitance is $1.6 \mathrm{C}$.

What is the equivalent capacitance of the system of capacitors between $A$ and $B$ :-

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