An inductor with a resistance of 30Ω and inductance of 0.8H is connected to a 240V 50Hz supply. The document finds (a) the inductive reactance of 251.2Ω, (b) the impedance of 253Ω, (c) the current of 0.95A, and (d) the phase angle between the current and voltage of 83.2°. A phasor diagram is drawn showing the current lags the voltage by 83.2°.

1. Example Question 1
An inductor has a resistance of 30Ω and an inductance of
0.8H. If itʼs connected
across a 240V 50Hz supply, find:
(a) Its inductive reactance
(b) Its impedance
(c) The current flowing through the inductor
(d) The phase angle between the current and the applied
voltage
(e) Also, draw the phasor diagram for the current and voltage.

2. Z = impedance in ohms
R = resistance in ohms
XL = reactance in ohms

3. Known values:
L = 0.8H
R = 30Ω
V = 240V
f = 50Hz

4. (a) Equation XL = 2πfL = 6.28 x 50 x 0.8
Answer XL = 251.2 Ω

5. (b) Values:
R = 30Ω
XL = 251.2 Ω

6. Z = √(R² + XL²)
Z = √(30² )+(251.2²)
Z = √900 + 63101.44
Z = √64001.44
Answer Z = 253 Ω

7. (c) Values;
Z = 253 Ω
V = 240 V

8. V
I =
Z (Ohmʼs law)
240
I = 253
Answer I = 0.95 A

9. (d) Values:
Z = 253 Ω
R = 50 Ω

10. ∅ = cos¯¹
R
Z

11. ∅ = cos¯¹
30
253

12. ∅ = cos¯¹ 0.118577

13. Answer ∅ = 83.2°

14. Applied
voltage (V)
Current (I)

15. Applied
voltage (V)
∅=
Current (I)

16. Applied
voltage (V)
∅ = 83.2⁰
Current (I)