In the arrangement of capacitors shown in Figure, if each capacitor is $9$ pF, the effective capacitance between A and B is

Question

In the arrangement of capacitors shown in Figure- each capacitor is 9 pF- the effective capacitance between A and B is 10pF15pF20pF5pF

Toppr Admin · Accepted Answer

Capacitor in parallelc3-c1-18pf-c-x2032-and in series-xA0-1c-1c-x2032-1c2-xA0- -xA0- -xA0- -xA0-118-19-xA0- -xA0- 16c-x2032-6pfso equivalent-xA0- - - 9 - 6 - pf-xA0-xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0-15pf

In the arrangement of capacitors shown in Figure, if each capacitor is $9$ pF, the effective capacitance between A and B is

$10$ pF
$15$ pF
$20$ pF
$5 p F$

capacitor in parallel

$c_{3} + c_{1} = 18 p f = c^{'}$

and in series

$\frac{1}{c} = \frac{1}{c^{'}} + \frac{1}{c_{2}}$

$\frac{1}{18} + \frac{1}{9}$

= $\frac{1}{6}$

$c^{'} = 6 p f$

so equivalent = ( 9 + 6 ) pf

$= 15 p f$

In the arrangement of capacitors shown in Figure, if each capacitor is 9 pF, the effective capacitance between A and B is 10pF15pF20pF5pF

capacitor in parallelc3+c1=18pf=c′and in series 1c=1c′+1c2 118+19= 16c′=6pfso equivalent = ( 9 + 6 ) pf =15pf

In the arrangement of capacitors shown in Figure, if each capacitor is $9$ pF, the effective capacitance between A and B is

$10$ pF
$15$ pF
$20$ pF
$5 p F$

capacitor in parallel

$c_{3} + c_{1} = 18 p f = c^{'}$

and in series

$\frac{1}{c} = \frac{1}{c^{'}} + \frac{1}{c_{2}}$

$\frac{1}{18} + \frac{1}{9}$

= $\frac{1}{6}$

$c^{'} = 6 p f$

so equivalent = ( 9 + 6 ) pf

$= 15 p f$