Question

a.
8. A coil having a resistance of 6 12 and an inductance
of 0.03 H is connected across a 50 V, 60 Hz, supply.
Calculate:
Inductive reactance.
b. Impedance.
Current.
d. Phase angle between the voltage and current.
Power factor.
f. Volt-amperes.
Active power.
Reactive power.
c.
تعرفه نه نه​

Answer 1

Answer:

Given:

• Resistance = 6.12 ohms
• Inductance = 0.03 H
• Potential Difference = 50 V
• Frequency = 60 Hz

To calculate:

a ) Inductive Reactance = wL = 2π × f × L

⇒ Inductive Reactance = 2 × 3.14 × 60 Hz × 0.03 H = 11.30 ohms

b ) Impedance ( Z ) = √ ( R² + (wL)² )

⇒ Impedance = √ ( 6.12 )² + ( 11.3 )²

⇒ Impedance = √ ( 37.45 ) + ( 127.69)

⇒ Impedance = √ 165.14 = 12.85 ohms

c ) Current = Potential Difference / Impedance

⇒ Current = 50 V / 12.85 ohms

⇒ Current = 3.89 A

d ) Cos A = R / Z where, A is the phase angle.

⇒ Cos A = 6.12 / 12.85 ≈ 0.5

⇒ A = arccos ( 0.5 ) = 60 degrees = π/3

e ) Power factor = Cos A = 0.5

f  ) Reactive Power = I² ( wL )

⇒ Reactive Power = 3.89 × 3.89 × 11.3 ≈ 171 W

⇒ Active Power = I² R = 3.89 × 3.89 × 6.12 = 92.6 W