An alternating emf E =440 sin 100 π t is applited to a circuit containing an inductance of (√2/π) H. If an a.c. ammeter is connected in the circuit, its reading will be :

Question

An alternating emf E =440 sin 100 π t is applited to a circuit containing an inductance of (√2/π) H. If an a.c. ammeter is connected in the circuit, its reading will be : (A) 4.4 A (B) 1.55 A (C) 2.2 A (D) 3.11 A

Tardigrade Admin · Accepted Answer

E =440 operatornameSin 100 π t, L=-(√2/π) H X L =ω L =100 π (√2/π)=100 √2 Ω text Peak current I 0=( E 0/ X L )=(440/100 √2)=2.2 √2 A AC ammeter reads RMS value therefore reading will be I rms I rms =( I 0/√2)=2.2 A

Q. An alternating emf $E = 440 sin 100 π t$ 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 :

$4.4 A$

$1.55 A$

$2.2 A$

$3.11 A$

$E = 440 Sin 100 π t, L = - \frac{2}{π} H$
$X_{L} = ω L = 100 π \frac{2}{π} = 1002 Ω$
$Peak current I_{0} = \frac{E _{0}}{X _{L}} = \frac{440}{100 2} = 2.2 2 A$
$A C$ ammeter reads RMS value therefore reading will be $I_{r m s}$
$I_{r m s} = \frac{I _{0}}{2} = 2.2 A$

Q. An alternating emf E=440sin100πt is applited to a circuit containing an inductance of π2​​H. If an a.c. ammeter is connected in the circuit, its reading will be :

4.4A

1.55A

2.2A

3.11A