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In a given AC series circuit containing elements R, L and C & source voltage = 220v, it is known that if L alone is removed or if C alone is removed, phase difference between current & voltage remains 45°. Find iRMS ? (R = 110 $\Omega$)

(1) 2A 

(2) 2.5A 

(3) 1A 

(4) 1.5A

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Electromagnetic induction and Alternating currents, Resonance

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Ans. (1) 2A

Sol. Since $\phi$ remains same,

circuit is in resonance.

$\therefore i_{RMS}={V_{RMS}\over Z}={220\over110}=2A$

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