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IIT JEE Advanced 2012 Paper 2
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© examsnet.com
Question : 15
Total: 60
In the given circuit, the AC source has ω = 100 rad/s. Considering the inductor and capacitor to be ideal, the correct choice(s) is(are)
[JEE Adv 2012 P2]
The current through the circuit, I is 0.3 A.
The current through the circuit, I is 0.3
√
2
A.
The voltage across 100 Ω resistor = 10
√
2
V.
The voltage across 50 Ω resistor = 10 V.
Validate
Solution:
From the given circuit, we have
I
capacitance
=
20
100
√
2
(progress ahead of the voltage by + π/4)
I
inductive
=
20
50
√
2
(lagging behind the voltage by -Ï€/4)
Therefore,
I =
√
I
1
2
+
I
2
2
=
√
1
10
~ 0.3
V
100
Ω
=
20
100
√
2
×
100
=
10
√
2
V
.
© examsnet.com
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