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Q.

An alternating emf E = 440 sin 100πt is applied to a circuit containing an inductance of 2πH. If an a.c. ammeter is connected in the circuit, its reading will be

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a

4.4 A

b

1.55 A

c

2.2 A

d

3.11 A

answer is C.

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Detailed Solution

E=440sin⁡100πt,L=2πHXL=ωL=100π2π=1002Ω
Peak current I0=E0XL=4401002=2.22A
AC ammeter reads RMS value therefore reading will be lrms
Irms=I02=2.2A

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