Question

18. In Fig. 4.72. calculate. (i) resultant capacitance between
A and B, (ii) potential difference across 4uF capacitor.
3uF
4uF
1uF 2uF
6uF
HHH
12uF
41
6 V
Fig. 4.72​

Answer 1

Answer:

PLEASE change the topic

Answer 2

Explanation:

The equivalent capacitance for RS is C

RS

=4+

6+12

6×12

=4+4=8μF

Now, C

PS

=

1+8

1×8

=8/9μF and C

PQ

=

8+4

8×4

=8/3μF

Now C

PQ

and C

RS

are in parallel and then it is series with C.

Thus, C

AB

=

C+32/9

C(32/9)

=1

or C+32/9=32C/9

or 9C+32=32C or 23C=32

so, C=32/23=1.4μF