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# 4.1.6 RL Example 2

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### 4.1.6 RL Example 2

1. 1. Example Question 2 Referring to the circuit diagram: a resistor of 50 Ω is connected in series with an 0.2H inductor that has a resistance of 4 Ω. This circuit is powered from a 50Hz supply. A voltage drop of 25V is measured across the resistor. Find:
2. 2. (a) the current in the circuit (b) the inductive reactance of the inductor (c) the applied voltage (d) the phase angle between the current and the applied voltage
3. 3. Values; R1 = 50Ω VR1 = 25V L = 0.2H f = 50Hz RL = 4Ω
4. 4. I = 25 = 0.5 A 50 Answer I = O.5A
5. 5. XL = 2 π f L XL = 6.28 x 50 x 0.2 = 62.8 Ω Answer XL = 62.8Ω
6. 6. L
7. 7. L
8. 8. Req = R1 + RL = 50 + 4 Req = 54 Ω
9. 9. Therefore, VReq = I X Req = 0.5 x 54 VReq = 27V
10. 10. So, VL = IL x XL = 0.5 x 62.8 VL = 31.4V
11. 11. 27V 31.4V L
12. 12. 31.4 27
13. 13. V = √(27² +31.4²) V = √729 + 985.96 = √1714.96 = 41.4 V Answer V = 41.4V
14. 14. Adjacent cos Φ = Hypotenuses
15. 15. Φ = Cos-¹ 27 = 49.3⁰ 41.4 Answer Φ = 49.3°