Question

Answer units estion 9 (1 point) C2 HH cl C3 CA C5 n the following circuit, C1=7.1 F, C2=2*7.1 F, C3=2*7.1 F, C4=2*8.1 F, and C5=3*8.1. Calculate the equivalent capacitance in the circuit.

Answer 1

Let the equivalent capacitance in the circuit is C.

Let the equivalent capacitance of capator C₂ and C₃ is C₂₃.

The equivalent capacitance of capacitor C₂ and C₃ is

  $C_{23}=C_{2}+C_{3}\: \: \: \: \: \: \: \left [ \: \because \: C_{2}\: and\: C_{3} \: parallel \: to\: each \: other\: \right ]$

     

or C23 14.2 + 14.2

     $\therefore \: \: \: \: \: C_{23}=28.4\: \: \: \: \: \: F$

The equivalent capacitance between C₄ and C₅ is

   $C_{45}=\frac{C_{4}\times C_{5}}{C_{4}+C_{5}}\: \: \: \: \: \: \: \left [ \: \because \: C_{4}\: and\: C_{5} \: are\: in\: series\: \right ]$

$or\: \: \: \: \: C_{45}=\frac{\left ( 2*8.1 \right )\times \left ( 3*8.1 \right )}{\left ( 2*8.1 \right )+\left ( 3*8.1 \right )}$

  $or\: \: \: \: \: C_{45}=\frac{6*8.1}{5*8.1}$

  $or\: \: \: \: \: C_{45}=\frac{48.6}{40.5}$

  $\therefore \: \: \: \: \: C_{45}=1.2\: \: \: \: F$

Now the equivalent capacitance between C₁ and C₂₃ is

     $C_{123}=\frac{C_{1}\times C_{23}}{C_{1}+C_{23}}$

$or\: \: \: \: \: \: C_{123}=\frac{\left ( 7.1 \right )\times\left ( 28.4 \right ) }{7.1+28.4}$

  $or\: \: \: \: \: \: C_{123}=\frac{201.64 }{35.5}$

  $\therefore \: \: \: \: \: \: C_{123}=5.68\: \: \: \: F$

Now the equivalent capacitance between C₁₂₃ and C₄₅ is

$C=C_{123}+C_{45}$

$\therefore \: \: \: \: \: \: C=6.88\: \: \: \: \: F$

$\therefore$ The equivalent capacitance in the circuit is 6.88 F.