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  • An alternating emf <img alt="\mathrm{E}=440 \sin 100\: \pi \mathrm{t}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BE%7D%3D440%20%5Csin%20100%5C%3A%20%5Cpi%20%5Cmathrm%7Bt%7D" /

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

Option: 1

4.4 \mathrm{~A}


Option: 2

1.55 \mathrm{~A}


Option: 3

2.2 \mathrm{~A}


Option: 4

3.11 \mathrm{~A}


Answers (1)

best_answer

\mathrm{E=440 \sin (100 \pi t) }

\mathrm{L=\frac{\sqrt{2}}{\pi} \mathrm{H} }

\mathrm{\omega=100 \pi }

\mathrm{X_L=\omega L=100 \sqrt{2} }

\mathrm{E_0=400}
AC ammeter reading is

\mathrm{ I_{rms}=\frac{1\left ( E_{o} \right )}{\sqrt{2}\left (x_{L} \right )}=\frac{I_{o}}{\sqrt{2}}=2.2A }

Hence 3 is correct option
 

Posted by

Deependra Verma

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