Download RC Circuit

Series RLC Circuit
PHY 213 and PHY 201
Important points to consider:
Sketch the phasor (vector) diagram
A circuit will appear, to the power supply, inductive if XL > XC
or capacitive if XC > XL
Use subscripts
The Problem:
A series ac circuit driven by a 120 V, 60.0 Hz supply consists of the
following components. Solve for all unknowns.
R = 1500 
L = 2.12 H
C = 2.00 F
R
C
L
X L  2 fL  2 (60.0 Hz ) (2.12 H )  799 
1
1
XC 

 1.33k 
6
2 fC 2 (60.0 Hz )(2.00 [10 F ]
 X   X L  X C  799   1.33k    531 
+y
XL
XC
+y
R

R
+x
Z
X
+x
 X is capacitive because XC > XL.
+y
XL
XC
+y
R

R
+x
Z
X
+x
Z  Rx    X  y  1500  x  (799   1.33k ) y
Z  1500  x  531  y
Solving the triangle:
Z  1.59k  @  19.5
o
-19.5 degrees is the phase angle between the current in the
circuit and the potential difference (voltage) applied to the
circuit.
V
120 V
3
I 
 75.5 [10 ] A  75.5 mA
Z 1.59k 
3
VR  I R  75.5 [10 ] A (1500)  113 V
3
VL  I X L  75.5 [10 ] A (799 )  60.3 V
3
VC  I X C  75.5 [10 ] A (1.33k )  100 V
Of special interest is the fact that the sum of the voltages is
greater than the applied voltage.
 V  113 V  60.3 V  100 V  273 V  120 V
This is because these are RMS voltages and not instantaneous
voltages. That is, these voltages do not occur at the same time.
Consider the graph below. It represents the RMS voltages across
the three components in a series RLC circuit. If we consider the
individual RMS voltages at the time represented by the black
vertical line, we will see that these RMS voltages sum to 120 V.
Note that the peak voltage will be 170 V.
200
150
100
-100
-150
-200
61
56
51
46
41
36
31
26
21
16
11
6
0
-50
1
50
Apparent power is the power “apparently” consumed by the
circuit. That is, current times voltage with no regard to
reactance.
3
Papp.  I V  75.5 [10 ] A (120 V )  9.06 Watt
However, “real” power is always the power consumed by the
resistance. This is the power that we pay for on our electric bill.
PRe al  I 2 R  (75.5 [103 ] A) 2 (1500 )  8.55 W
Power factor gives us an indication of the amount of reactive
current flowing in the circuit.
PRe al 8.55 W
power factor  pf 

 0.944
Papp. 9.06 W
Or:
pf  cos  cos19.5  0.944
o