For an L-R circuit, the inductive reactance is equal to the resistance R of the circuit. An emf E=E0 cos (ω t) is applied to the circuit. Then, the power consumed in the circuit is

Question

For an L-R circuit, the inductive reactance is equal to the resistance R of the circuit. An emf E=E0 cos (ω t) is applied to the circuit. Then, the power consumed in the circuit is (A)  (E0/R)  (B)  (E02/4R)  (C)  (4R/E0)  (D)  (R/E0)

Tardigrade Admin · Accepted Answer

For L - R conbination, given, X L = R Impedance Z =√ R 2+ R 2 =√2 R Power consumed =(( V rms / Z ))2 R =(( E 0 / √2/√2 R ))2 R =( E 02/4 R )

Q. For an L-R circuit, the inductive reactance is equal to the resistance R of the circuit. An emf $E = E_{0} cos (ω t)$ is applied to the circuit. Then, the power consumed in the circuit is

$\frac{E _{0}}{R}$

$\frac{E _{0}^{2}}{4 R}$

$\frac{4 R}{E _{0}}$

$\frac{R}{E _{0}}$

For $L - R$ conbination, given,
$X_{L} = R$
Impedance $Z = R^{2} + R^{2}$
$= 2 R$
Power consumed $= (\frac{V _{r m s}}{Z})^{2} R$
$= (\frac{E _{0} / 2}{2 R})^{2} R = \frac{E _{0}^{2}}{4 R}$

Q. For an L-R circuit, the inductive reactance is equal to the resistance R of the circuit. An emf E=E0​cos(ωt) is applied to the circuit. Then, the power consumed in the circuit is

RE0​​

4RE02​​

E0​4R​

E0​R​