An alternating emf E = 440 sin100 t is applited to a circuit containing an inductance of 2 H. If an a.c. ammeter is connected in the circuit, its reading will be:

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Accepted Answer

Correct answer: C. (c) E=440 sin100 πt ,L= 2 H X_L= ωL=100 2 =100 2 Peak current I_0= E_0X_L= 440100 2=2.2 2A AC ammeter reads RMS value therefore reading will I_rms I_ rms= I_0 2=2.2A

$An alternating emf E = 440 sin100 π t is applited to a circuit containing an inductance of \frac{2}{π} H. If an a.c. ammeter is connected in the circuit, its reading will be:$

Solution

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