A resistor of 500 Ω, an inductance of 0.5 H are in series with an AC voltage source which is given by V = 100 √2 sin (1000t). The power factor of the combination is

Question

A resistor of 500 Ω, an inductance of 0.5 H are in series with an AC voltage source which is given by V = 100 √2 sin (1000t). The power factor of the combination is (A)  (1/√2) (B)  (1/√3) (C) 0.5 (D) 0.6

Tardigrade Admin · Accepted Answer

Voltage V=100 √2 sin (1000 t) and V=V0 sin ω t On comparing ω=1000 Inductive reactance XL=ω L=1000 × 0.5=500 Ω ∴ cos φ=(500/√(500)2+(500)2) =(1/√2)

Q. A resistor of $500 Ω$ , an inductance of $0.5 H$ are in series with an $A C$ voltage source which is given by $V = 1002$ sin $(1000 t)$ . The power factor of the combination is

$\frac{1}{2}$

$\frac{1}{3}$

$0.5$

$0.6$

Voltage $V = 1002 sin (1000 t)$
and $V = V_{0} sin ω t$
On comparing $ω = 1000$
Inductive reactance
$X_{L} = ω L = 1000 \times 0.5 = 500 Ω$
$∴ cos ϕ = \frac{500}{( 500 ) ^{2} + ( 500 ) ^{2}}$
$= \frac{1}{2}$

Q. A resistor of 500Ω, an inductance of 0.5H are in series with an AC voltage source which is given by V=1002​ sin (1000t). The power factor of the combination is

2​1​

3​1​

0.5

0.6