In the given figure,the capacitors C_1, C_3, | vedclass

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(B) The given circuit is a Wheatstone bridge network. Let the nodes be defined such that the bridge is formed by $C_1, C_4, C_3, C_5$ with $C_2$ as th

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$4$

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(B) The given circuit is a Wheatstone bridge network. Let the nodes be defined such that the bridge is formed by $C_1, C_4, C_3, C_5$ with $C_2$ as the central capacitor. For a balanced Wheatstone bridge,the ratio of capacitances in the arms must be equal: $\frac{C_1}{C_4} = \frac{C_3}{C_5}$. Given $C_1 = C_3 = C_4 = C_5 = 4\,\mu F$,we have $\frac{4}{4} = \frac{4}{4}$,which is $1 = 1$. Since the bridge is balanced,no charge flows through the central capacitor $C_2$. Therefore,$C_2$ can be removed from the circuit. The circuit simplifies to two series branches in parallel. The upper branch consists of $C_1$ and $C_3$ in series (if we re-draw the bridge,$C_1$ and $C_3$ are in series,and $C_4$ and $C_5$ are in series). Actually,looking at the bridge: $C_1$ and $C_4$ are in series,and $C_3$ and $C_5$ are in series. Equivalent capacitance of upper branch $C_{up} = \frac{C_1 \cdot C_4}{C_1 + C_4} = \frac{4 \cdot 4}{4 + 4} = 2\,\mu F$. Equivalent capacitance of lower branch $C_{low} = \frac{C_3 \cdot C_5}{C_3 + C_5} = \frac{4 \cdot 4}{4 + 4} = 2\,\mu F$. These two branches are in parallel,so $C_{eq} = C_{up} + C_{low} = 2 + 2 = 4\,\mu F$.

In the given figure,the capacitors $C_1, C_3, C_4, C_5$ have a capacitance of $4\,\mu F$ each. If the capacitor $C_2$ has a capacitance of $10\,\mu F$,then the effective capacitance between $A$ and $B$ will be.....$\mu F$.

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