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a. $\dfrac{6}{5}C$

b. $2C$

c. $\dfrac{9}{{11}}C$

d. $\dfrac{{11}}{5}C$

Answer

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Parallel connection of the capacitors

${C_T} = {C_A} + {C_B} + ....$

Series connection of the capacitors

${C_T} = \dfrac{{{C_A} \times {C_B} \times ..}}{{{C_A} + {C_B} + ..}}$

We will start with the different arrangements of the components and analyze whether which are connected in series connection or parallel connection and then they are calculated separately and then the equivalence capacitor is then evaluated.

Staring with the rearrangements of the points $A$ and $B$ given on the circuits. All the capacitors have the same capacitance

${C_1} = {C_2} = {C_3} = {C_4} = {C_5} = {C_6} = C$

From the figure, it can be concluded that the capacitors ${C_1}$ and ${C_3}$ are in the parallel connection between points $P$ , and $A$ hence their equivalent capacitance is given as

${C_{AP}} = {C_1} + {C_3}$

Substitute the values of ${C_1}$and ${C_3}$ hence,

${C_{AP}} = C + C$

$ \Rightarrow {C_{AP}} = 2C$

Similarly, the between points $P$ and $B$ the capacitors ${C_2}$,${C_4}$ and ${C_5}$ are in parallel connection hence their equivalent capacitance is given as

${C_{BP}} = {C_2} + {C_4} + {C_5}$

$ \Rightarrow {C_{BP}} = C + C + C = 3C$

Now on analyzing, it is found that the capacitor with $3C$ and $2C$ are in series hence their equivalent capacitance is given as

$C''' = \dfrac{{3C \times 2C}}{{3C + 2C}}$

$ \Rightarrow C''' = \dfrac{{6C}}{5}$

Hence the final equivalent capacitance between the points $A$ and $B$ can be given by the parallel connection between $C'''$ and $C$ which is given as

${C_{AB}} = \dfrac{{6C}}{5} + C$

$ \Rightarrow {C_{AB}} = \dfrac{{6C + 5C}}{5} = \dfrac{{11}}{5}C$

Hence the final equivalent capacitance between the points $A$ and $B$ can be given as $\dfrac{{11}}{5}C$.

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