An inductor has a resistance R and Inductance L. It is connected to an A.C. source of e.m.f Ev, and angular frequency- ω, then the current Iv in the circuit is

Question

An inductor has a resistance R and Inductance L. It is connected to an A.C. source of e.m.f Ev, and angular frequency- ω, then the current Iv in the circuit is (A) (Ev/ω L) (B) (Ev/R) (C) (Ev/√R2+ω2 L2) (D) √((Ev/R))2+((Ev/ω L))2

Tardigrade Admin · Accepted Answer

The impedance in R-L circuit is Z =√ R 2+ X L 2=√ R 2+(ω L )2 The current, I v =( E V / Z ) =( E V /√ R 2+(ω L )2)

Q. An inductor has a resistance R and Inductance L. It is connected to an A.C. source of $e . m . f E_{v}$ , and angular frequency- $ω$ , then the current $I_{v}$ in the circuit is

$\frac{E _{v}}{ω L}$

$\frac{E _{v}}{R}$

$\frac{E _{v}}{R ^{2} + ω ^{2} L ^{2}}$

$(\frac{E _{v}}{R})^{2} + (\frac{E _{v}}{ω L})^{2}$

The impedance in R-L circuit is
$Z = R^{2} + X_{L}^{2} = R^{2} + (ω L)^{2}$
The current, $I_{v} = \frac{E _{V}}{Z}$
$= \frac{E _{V}}{R ^{2} + ( ω L ) ^{2}}$

Q. An inductor has a resistance R and Inductance L. It is connected to an A.C. source of e.m.fEv​, and angular frequency- ω, then the current Iv​ in the circuit is

ωLEv​​

REv​​

R2+ω2L2​Ev​​

(REv​​)2+(ωLEv​​)2​