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

Question

An infinite number of identical capacitors each of capacitance 1 μ F are connected as shown in figure. Then the equivalent capacitance between A and B is (A) 1 μ F (B) 2 μ F (C) (1/2) μ F (D) ∞

Tardigrade Admin · Accepted Answer

This combination forms a G.P. S=1+(1/2)+(1/4)+(1/8) ldots ldots . Sum of infinite G.P. S=(a/1-r) Here a= first term =1 and r= common ratio =(1/2) ⇒ S=(1/1-(1)2)=2 ⇒ Ce q=2 μ F

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

$1 μ F$

$2 μ F$

$\frac{1}{2} μ F$

$\infty$

This combination forms a $G . P . S = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} \dots\dots .$
Sum of infinite $G . P . S = \frac{a}{1 - r}$
Here $a =$ first term $= 1$ and $r =$ common ratio $= \frac{1}{2}$
$\Rightarrow S = \frac{1}{1 - \frac{1}{2}} = 2$
$\Rightarrow C_{e q} = 2 μ F$

Q. An infinite number of identical capacitors each of capacitance 1μF are connected as shown in figure. Then the equivalent capacitance between A and B is

1μF

2μF

21​μF

∞