For the circuit shown in figure, the ammeter A2 reads 1.6 A and ammeter A3 reads 0.4 A. Then

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ω=4LC

f=2πLC

the ammeter A1 reads 1.2 A

the ammeter A1 reads 2A

answer is C.

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Detailed Solution

The current of 1.6 A lags emf in phase by π/2. The current of 0.4 A leads emf in phase byπ/2. So, these two currents are 1800 out of phase with each other. ∴Net current, I1=1.6-0.4A=1.2A1.6=EVXL and 0.4 = EVXC⇒XCXL=4⇒1ωCωL=4⇒ω=12LC

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