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VITEEE 2017 Solved Paper
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© examsnet.com
Question : 32
Total: 125
Two capacitors when connected in series have a capacitance of 3 µF, and when connected in parallel have a capacitance of 16 µF. Their individual capacities are
1 µF, 2 µF
6 µF, 2 µF
12 µF,4 µF
3 µF, 16 µF
Validate
Solution:
C
S
=
C
1
C
2
C
1
+
C
2
= 3
C
P
=
C
1
+
C
2
= 16
∴
C
1
C
2
= 48
C
1
−
C
2
=
√
(
C
1
+
C
)
2
)
2
−
4
C
1
C
2
=
√
16
2
−
4
×
48
=
√
64
= 8
C
1
+
C
2
= 16 µF
C
1
−
C
2
= 8µF
⇒
2
C
1
= 24 µF ⇒
C
1
= 12 µF
∴
C
2
=
48
12
= 4 µF
© examsnet.com
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