a
3/4µF
b
\( 2 \frac{10}{13} µF\)
c
12µF
d
\( 4 \frac{12}{13} µF\)
e
13µF
Explanation
Correct Option
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Discussions (7)

Deepthinker581
4 years ago
The effective capacitance in parallel =4+3+4=9
The total effective capacitance would now be a series of 4 and 9
For that we have 1/C
1/4 and 1/9
4+9=13
4×9 =36
C=36/13
Which in solve would be 2(10)/3

Boluwa124456
2 years ago
the effective capacitance 4+3+2=9
the remaining capacitance is 9 and 4
9*4 ÷ 9+4
36/13



