Question

A 12 ohm resistor and a 0.21 henry inductor are connected in series to an ac source operating at 20 volts, 50 cycle/ second. The phase angle between the current and the sources voltage is

Answer 1

XL = 2π50 x 0.21 = 66
Z = √12*12 + (XL)^2 = √144+(66)^2 =√ 4500 =30√5.
Now, cos(angle)=R/Z = 12/30√5 = 2/5√5
hence, angle= 80

Answer 2

Answer:

The phase angle between the current and the sources voltage is 79.69°.

Explanation:

Given that,

Resistance = 12 ohm

Inductance = 0.21 Henry

Voltage = 20 volts

frequency = 50 cycle/sec

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=2\pi f$

Put the value into the formula

$\omega=2\times\pi\times50$

$\omega=314.15\ rad/sec$

We need to calculate the phase angle between the current and the sources voltage

Using formula of phase

$\phi=tan^{-1}\dfrac{\omega L}{R}$

Put the value into the formula

$\phi=tan^{-1}(\dfrac{314.15\times0.21}{12})$

$\phi=79.69^{\circ}$

Hence, The phase angle between the current and the sources voltage is 79.69°.