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

E=440sin⁡100πt,L=2πHXL=ωL=100π2π=1002ΩPeak current I0=E0XL=4401002=2.22AAC ammeter reads RMS value therefore reading will be lrmsIrms=I02=2.2A

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

E=440sin⁡100πt,L=2πHXL=ωL=100π2π=1002ΩPeak current I0=E0XL=4401002=2.22AAC ammeter reads RMS value therefore reading will be lrmsIrms=I02=2.2A

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An alternating emf E = 440 sin 100 $π$ t is applied to a circuit containing an inductance of $\frac{\sqrt{2}}{π} H .$ If an a.c. ammeter is connected in the circuit, its reading will be

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3.11 A

4.4 A

1.55 A

2.2 A

answer is C.

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Detailed Solution

$\begin{array}{l} E = 440 \sin 100 πt, L = \frac{\sqrt{2}}{π} H \\ X_{L} = ωL = 100 π \frac{\sqrt{2}}{π} = 100 \sqrt{2} Ω \end{array}$
Peak current $I_{0} = \frac{E_{0}}{X_{L}} = \frac{440}{100 \sqrt{2}} = 2 .2 \sqrt{2} A$
AC ammeter reads RMS value therefore reading will be l_rms
$I_{rms} = \frac{I_{0}}{\sqrt{2}} = 2 .2 A$