An inductor of inductance 2 H and a resistance of 10 Ω are connected in series to an ac source of 110 V, 60 Hz. The current in the circuit will be

Question

An inductor of inductance 2 H and a resistance of 10 Ω are connected in series to an ac source of 110 V, 60 Hz. The current in the circuit will be (A) 0.32 A (B) 0.15 A (C) 0.48 A (D) 0.80 A

Tardigrade Admin · Accepted Answer

XL=ω L=2 π ∪ L=2 π × 60 × 2=240 π Ω, R=10 Ω Z=√R2+XL2=√102+(240 π)2 ≈ 240 π I rms =(V rms /Z)=(110/240 π)=0.15 A

Q. An inductor of inductance $2 H$ and a resistance of $10 Ω$ are connected in series to an ac source of $110 V, 60 Hz$ . The current in the circuit will be

0.32 A

0.15 A

0.48 A

0.80 A

$X_{L} = ω L = 2 π \cup L = 2 π \times 60 \times 2 = 240 π Ω, R = 10Ω$
$Z = R^{2} + X_{L}^{2} = 1 0^{2} + (240 π)^{2} \approx 240 π$
$I_{r m s} = \frac{V _{r m s}}{Z} = \frac{110}{240 π} = 0.15 A$

Q. An inductor of inductance 2H and a resistance of 10Ω are connected in series to an ac source of 110V,60Hz. The current in the circuit will be