An infinite number of identical capacitors ea | vedclass

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(B) The circuit consists of parallel branches.In the $1^{st}$ branch,there is $1$ capacitor of $1 \mu F$. Its equivalent capacitance is $C_1 = 1 \mu F

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(B) The circuit consists of parallel branches.In the $1^{st}$ branch,there is $1$ capacitor of $1 \mu F$. Its equivalent capacitance is $C_1 = 1 \mu F$.In the $2^{nd}$ branch,there are $2$ capacitors in series,each of $1 \mu F$. Its equivalent capacitance is $C_2 = \frac{1 \mu F}{2} = 0.5 \mu F$.In the $3^{rd}$ branch,there are $4$ capacitors in series,each of $1 \mu F$. Its equivalent capacitance is $C_3 = \frac{1 \mu F}{4} = 0.25 \mu F$.In the $4^{th}$ branch,there are $8$ capacitors in series,each of $1 \mu F$. Its equivalent capacitance is $C_4 = \frac{1 \mu F}{8} = 0.125 \mu F$.Since all these branches are connected in parallel,the total equivalent capacitance $C_{eq}$ is the sum of the capacitances of each branch:$C_{eq} = C_1 + C_2 + C_3 + C_4 + \dots$$C_{eq} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$This is an infinite geometric progression with first term $a = 1$ and common ratio $r = \frac{1}{2}$.The sum of an infinite geometric series is given by $S = \frac{a}{1 - r}$.$C_{eq} = \frac{1}{1 - 1/2} = \frac{1}{1/2} = 2 \mu F$.

An infinite number of identical capacitors each of capacitance $1 \mu F$ are connected as shown in the figure. Then,the equivalent capacitance between $A$ and $B$ is .......... $\mu F$.

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