Question

The series combination of resistance R and inductance L is connected to an alternating source of e. m. f. e = 311 sin (100
π
t
). If the value of wattless current is 0.5 A and the impedance of the circuit is 311
Ω
, the power factor will be - ​

Answer 1

Answer:

Power factor = $\frac{1}{\sqrt{2} }$

Formulae used :

1. $\boxed{power factor = cos(\phi)}$  where Φ is phase difference between V and I

    2. $\boxed{sin(\phi) = \frac{I'}{Irms}}$ where I' is wattless current

    3. $\boxed{\frac{Eo}{Io}=Impedence }$

    4.$\boxed{E=(Eo)sin(\omega t)}$

Solution :

From (4) , and since $E=311sin(100\pi t)$,

$Eo=311----(1)$

Since $Io=(Eo)(impedence)---(2)$

$Io=1---(3)$               [FROM (1) and (2)]

Using Formula 2;

$sin(\phi) = \frac{1/2}{1/\sqrt{2} }$

$\phi =\frac{\pi}{4}$

∴ $\boxed{\boxed{power factor=cos(\phi) =\frac{1}{\sqrt{2}}}}$

Hope this answer helped you :)

Answer 2

Thus the value of power factor is cos (Φ) = 1 / √2

Explanation:

Power factor = cos (Φ)    Φ is the phase difference.

Sin (Φ) = I' / I (rms) Where I' is wattless current.

Eo / Io = I (impedenec)

E = Eo sin (ωt)

Now using the formula of power factor:

Sin (Φ)  = 1 /2 ÷ 1 / √2

Φ = π / 4

Power factor = cos (Φ) = 1 / √2

Thus the value of power factor is cos (Φ) = 1 / √2