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(D) The circuit consists of three parts in series: $1$. Two $1\,\mu F$ capacitors in parallel at the left,which have an equivalent capacitance of $C_1

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(D) The circuit consists of three parts in series: $1$. Two $1\,\mu F$ capacitors in parallel at the left,which have an equivalent capacitance of $C_1 = 1 + 1 = 2\,\mu F$. $2$. $A$ single $1\,\mu F$ capacitor in the middle,$C_2 = 1\,\mu F$. $3$. Two $1\,\mu F$ capacitors in parallel at the right,which have an equivalent capacitance of $C_3 = 1 + 1 = 2\,\mu F$. These three equivalent capacitors are in series. The total equivalent capacitance $C_{AB}$ is given by: $\frac{1}{C_{AB}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{2} + \frac{1}{1} + \frac{1}{2} = 0.5 + 1 + 0.5 = 2\,\mu F^{-1}$. Therefore,$C_{AB} = \frac{1}{2} = 0.5\,\mu F$. Note: The provided option $(d)$ was $0.5$ in the original text,which is the correct value.

The equivalent capacitance between $A$ and $B$ is.......$\mu \,F$

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