Find the equivalent capacitance of the circui | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The circuit is symmetric. Due to symmetry,the potential at the nodes on the upper branch and the corresponding nodes on the lower branch are equal

Vedclass · Accepted Answer

$C$

Vedclass · Answer

(C) The circuit is symmetric. Due to symmetry,the potential at the nodes on the upper branch and the corresponding nodes on the lower branch are equal. Therefore,no current flows through the vertical capacitors,and they can be removed from the circuit.After removing the vertical capacitors,the circuit simplifies into two parallel branches. Each branch consists of capacitors in series: $C, 2C, 4C, 8C, \dots$The equivalent capacitance of one branch $(C_{branch})$ is given by:$\frac{1}{C_{branch}} = \frac{1}{C} + \frac{1}{2C} + \frac{1}{4C} + \frac{1}{8C} + \dots$$\frac{1}{C_{branch}} = \frac{1}{C} \left( 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots ight)$This is a geometric series with first term $a = 1$ and common ratio $r = \frac{1}{2}$. The sum is $S = \frac{a}{1-r} = \frac{1}{1 - 1/2} = 2$.So,$\frac{1}{C_{branch}} = \frac{1}{C} 	imes 2 = \frac{2}{C} \Rightarrow C_{branch} = \frac{C}{2}$.Since there are two such branches in parallel,the total equivalent capacitance is:$C_{eq} = C_{branch} + C_{branch} = \frac{C}{2} + \frac{C}{2} = C$.

Find the equivalent capacitance of the circuit between points $A$ and $B$.

Similar Questions

Three condensers of capacities $C_{1}, C_{2}, C_{3}$ are connected in series with a source of e.m.f. $V$. The potentials across the three condensers are in the ratio of

The effective capacitance between $X$ and $Y$ is ... $\mu F$.

Find the equivalent capacitance across points $A$ and $B$.

Two capacitors of values $0.6 \ \mu F$ and $0.3 \ \mu F$ are connected in series with a $6 \ V$ source. Find the ratio of energy stored in each capacitor.

Vedclass Test Series

Exam Paper Generator

Online Exam Module