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VITEEE 2014 Solved Paper
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© examsnet.com
Question : 14
Total: 120
A circuit has a self-inductance of 1 H and carries a current of 2A. To prevent sparking, when the circuit is switched off, a capacitor which can withstand 400 V is used. The least capacitance of capacitor connected across the switch must be equal to
50 µF
25 µF
100 µF
12.5 µF
Validate
Solution:
Energy stored in capacitor = energy stored in inductance
i.e.
1
2
C
V
2
=
1
2
L
I
2
⇒ C =
L
I
2
V
2
=
1
×
(
2
)
2
(
400
)
2
= 25 µF
© examsnet.com
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