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Physics-

General

Easy

Question

A network of six identical capacitors, each of value $C$ , is made as shown in the figure.
The equivalent capacitance between the points $A space a n d blank B$ is

$\frac{4 C}{11}$
$\frac{3 C}{4}$
$\frac{3 C}{2}$
$3 C$

The correct answer is: $fraction numerator 4 C over denominator 11 end fraction$

In the given circuit capacitor’s (1) (2) and (3) are connected in series, hence equivalent capacitance is

$fraction numerator 1 over denominator C to the power of ´ ´ end exponent end fraction equals fraction numerator 1 over denominator C end fraction plus fraction numerator 1 over denominator C end fraction plus fraction numerator 1 over denominator C end fraction equals fraction numerator 3 over denominator C end fraction$
$rightwards double arrow C to the power of ´ ´ end exponent equals fraction numerator C over denominator 3 end fraction$
This is connected in parallel with (4).
$therefore blank C to the power of ´ ´ ´ ´ end exponent equals C to the power of ´ ´ end exponent plus C equals fraction numerator C over denominator 3 end fraction plus C equals fraction numerator 4 C over denominator 3 end fraction$
The three capacitor’s (5), $fraction numerator 4 C over denominator 3 end fraction comma$ (6) are now connected in series.
$therefore$ Equivalent capacitance is
$fraction numerator 1 over denominator C blank ´ ´ ´ ´ ´ ´ end fraction equals fraction numerator 1 over denominator C end fraction plus fraction numerator 3 over denominator 4 C end fraction plus fraction numerator 1 over denominator C end fraction$
$fraction numerator 1 over denominator C ´ ´ ´ ´ ´ ´ end fraction equals fraction numerator 11 over denominator 4 C end fraction$
$rightwards double arrow blank C to the power of ´ ´ ´ ´ ´ ´ end exponent equals fraction numerator 4 C over denominator 11 end fraction$

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