CGPET 2019 Solved Paper

According to the question,
Given, capacitance between A and B,CAB=1µF.
∴ Capacitors between point C and D are connected in parallel. So, capacitance. CCD=2µF+2µF=4µF.
∴ Capacitors between points E and F connected in series, so the capacitance between points E and F, is

1

CEF

=

1

6

+

1

12

or

1

CEF

=

2+1

12

=

3

12

or CEF=

12

3

=4µF
Capacitors between points G and D are connected in series, therefore

1

CGD

=

1

8µF

+

1

4µF

∴ =

1+2

8

or

1

CGD

=

3

8

or CGD=

8

3

µF
∴ Capacitance between points E and F, are connected in parallel, so
∴CEF=4µF+4µF=8µF
Capacitance between points G and F, are connected in series, therefore

1

CGF

=

1

1

+

1

8

or=

8+1

8

=

9

8

orCGF=

8

9

µF
Now,
∴CPR=

8

9

+

8

3

=

24+8

9

=

32

9

µFCapacitors are connected in series, then the total capacitance across points A and B,
∴

1

CAB

=

1

C

+

1

32∕9

or 1=

1

C

+

9

32

Here CAB=1µF
or

1

C

=1−

9

32

=

32−9

32

=

23

32

or

1

C

=

23

32

C=

32

23

µF