If emf E=4 cos 1000 t volt is applied to an L-R circuit of inductance 3 mH and resistance 4 Ω, the amplitude of current in the circuit is

Question

If emf E=4 cos 1000 t volt is applied to an L-R circuit of inductance 3 mH and resistance 4 Ω, the amplitude of current in the circuit is (A) (4/√7)A (B) 1.0 A (C) (4/7)A (D) 0.8 A

Tardigrade Admin · Accepted Answer

Impedence, Z=√R2+XL2 Here, R=4 Ω, XL=L ω =3 × 10-3 × 1000 Ω Then, =3 Ω Z=√(4)2+(3)2 or Z=√16+9=√25 or Z=5 Ω Hence, current, I0 =(E0/Z) =(4/5)=0.8 A

Q. If emf $E = 4 cos 1000 t$ volt is applied to an $L - R$ circuit of inductance $3 m H$ and resistance $4 Ω$ , the amplitude of current in the circuit is

$\frac{4}{7} A$

$1.0 A$

$\frac{4}{7} A$

$0.8 A$

Impedence, $Z = R^{2} + X_{L}^{2}$
Here, $R = 4Ω, X_{L} = L ω$
$= 3 \times 1 0^{- 3} \times 1000 Ω$
Then, $= 3 Ω$
$Z = (4)^{2} + (3)^{2}$
or $Z = 16 + 9 = 25$
or $Z = 5 Ω$
Hence, current, $I_{0} = \frac{E _{0}}{Z}$
$= \frac{4}{5} = 0.8 A$

Q. If emf E=4cos1000t volt is applied to an L−R circuit of inductance 3mH and resistance 4Ω, the amplitude of current in the circuit is

7​4​A

1.0A

74​A

0.8A

Impedence, Z=R2+XL2​​ Here, R=4Ω,XL​=Lω =3×10−3×1000Ω Then, =3Ω Z=(4)2+(3)2​ or Z=16+9​=25​ or Z=5Ω Hence, current, I0​=ZE0​​ =54​=0.8A

Q. If emf $E = 4 cos 1000 t$ volt is applied to an $L - R$ circuit of inductance $3 m H$ and resistance $4 Ω$ , the amplitude of current in the circuit is

$\frac{4}{7} A$

$1.0 A$

$\frac{4}{7} A$

$0.8 A$

Impedence, $Z = R^{2} + X_{L}^{2}$
Here, $R = 4Ω, X_{L} = L ω$
$= 3 \times 1 0^{- 3} \times 1000 Ω$
Then, $= 3 Ω$
$Z = (4)^{2} + (3)^{2}$
or $Z = 16 + 9 = 25$
or $Z = 5 Ω$
Hence, current, $I_{0} = \frac{E _{0}}{Z}$
$= \frac{4}{5} = 0.8 A$