Four capacitors,each of capacity 3,mu F,are c | vedclass

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(A) Let the capacitance of each capacitor be $C = 3\,\mu F$.$1$. For equivalent capacitance between $A$ and $B$ $(C_{AB})$:When points $A$ and $B$ are

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$4:3$

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(A) Let the capacitance of each capacitor be $C = 3\,\mu F$.$1$. For equivalent capacitance between $A$ and $B$ $(C_{AB})$:When points $A$ and $B$ are connected to a source,the capacitor directly between $A$ and $B$ is in parallel with the series combination of the other three capacitors. $C_{AB} = C + (C/3) = 3 + (3/3) = 3 + 1 = 4\,\mu F$.$2$. For equivalent capacitance between $A$ and $C$ $(C_{AC})$:When points $A$ and $C$ are connected,the circuit forms two parallel branches,each containing two capacitors in series. $C_{AC} = (C/2) + (C/2) = (3/2) + (3/2) = 1.5 + 1.5 = 3\,\mu F$.$3$. The ratio $C_{AB} : C_{AC} = 4 : 3$.

Four capacitors,each of capacity $3\,\mu F$,are connected as shown in the adjoining figure. The ratio of the equivalent capacitance between $A$ and $B$ to that between $A$ and $C$ will be:

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