If power factor is 1/2 in a series RL circuit R=100Ω. If AC mains of 50Hz is used then L is :-

Question

If power factor is 1/2 in a series RL circuit R=100Ω. If AC mains of 50Hz is used then L is :- (A) (√3/π ) Henry (B) π Henry (C) (π /√3) Henry (D)  text None of these

Tardigrade Admin · Accepted Answer

cosφ=(R/Z)=(100/Z)=(1/2) Z=200=√R2 + XL2 ⇒ XL=√3× 100Ω=ω L=100π L L=(√3/π )H

Q. If power factor is $1/2$ in a series $R L$ circuit $R = 100Ω.$ If $A C$ mains of $50 Hz$ is used then $L$ is :-

$\frac{3}{π}$ Henry

$π$ Henry

$\frac{π}{3}$ Henry

$None of these$

$cos ϕ = \frac{R}{Z} = \frac{100}{Z} = \frac{1}{2}$
$Z = 200 = R^{2} + X_{L}^{2}$
$\Rightarrow X_{L} = 3 \times 100Ω = ω L = 100 π L$
$L = \frac{3}{π} H$

Q. If power factor is 1/2 in a series RL circuit R=100Ω. If AC mains of 50Hz is used then L is :-

π3​​ Henry

π Henry

3​π​ Henry

None of these