Power in Series R-L Circuit:

Example:  A 200?    resistor and a 50?    XL are placed in series along with a voltage source and the total current flow are 2 amps, as display in Figure.

Find:  

1. pf

2. applied voltage, V

3. P

4. Q

5. S

 Figure: Series R-L Circuit

Solution:

1.      pf= cosθ                     θ -arctan(X_L/R)

= cos(arctan(X_L /R))

= cos(arctan(50 /200))

= cos(14º)

pf= 0.097

2. V=IZ   Z= √R² +X_L²

= I√R²+X_L²

= 2√200²+50²

= 2√42,500

= (2)(206.16)

V=412.3 volts

3. P =      EI cosθ

= (412.3)(2)(0.97)

P = 799.86 watts

4. Q = EI sinθ

= (412.3)(2)(0.242)

Q = 199.6 VAR

5. S =   EI

= (412)(2)

S = 824.6 VA