Series and Parallel RCL Circuits
Fundamentals
(non-resonant conditions)
Series RCL circuit
R
100Ω
V1
125 Vrms
0 Hz
0°
XL
50 Ohms
XC
125 Ohms
X=?
VR = ?
ZT = ?
VC = ?
I=?
VL = ?
θZ = ?
Also determine Real
Power, Apparent Power
and Power Factor
Calculations to determine phasor
values
• Net Reactance (X) = XL – XC = 50Ω - 125Ω =
-75Ω
– This tells us that the overall phase angle will be in
the fourth quadrant (XC > XL). The formula is
written specifically as the leading minus the
lagging.
• ZT = 𝑋 2 + 𝑅 2 = −752 + 1002 =
5625 + 10000 = 15625 = 125Ω
• I=
𝑉𝑇
𝑍𝑇
=
125
125
= 1A
– Since this is a series circuit, all currents are the
same (only one path)
• VR = IR = 1A × 100Ω = 100v
• VC = IXC = 1A × 125Ω = 125v
• VL = IXL = 1A × 50Ω = 50v
• We could double check using the formula VT =
𝑉𝐿 − 𝑉𝐶
2
+ 𝑉𝑅 2 , but since the values of
voltage are the same as the values of
resistance/reactance, this will not be
necessary.
Phasor diagram
y
R = 100Ω
(Adj)
x
X = -75Ω
(Opp)
Calculation for θ
• tan 𝜃 =
𝑂𝑝𝑝
𝐴𝑑𝑗
𝑋
𝑅
= =
𝑂𝑝𝑝
𝐻𝑦𝑝
𝑋
𝑍𝑇
−75
100
∴𝜃=
tan−1 −0.75 = -36.87°
Double check:
• sin 𝜃 =
=
=
−75
125
−1 −75
tan
100
=
−1 −75
sin
125
=
∴𝜃=
sin−1 −0.6 = -36.87° (verified)
Power equations
• Real Power → P =I2R = (1)2(100) = 100W
• Apparent Power → VA = VTI = (125)(1) = 125VA
• Power Factor → PF = cos 𝜃 =
𝑅
𝑍𝑇
∴ PF =
100
125
=
0.8
– This is also equivalent to taking the real power
and dividing it by the apparent power.
Parallel RCL circuit
VA
36 Vrms
0 Hz
0°
R
240Ω
IR = ?
IX = ?
IC = ?
IT = ?
XC
XL
60 Ohms 90 Ohms IL = ?
Zeq = ?
θI = ?
Also determine Real
Power, Apparent Power
and Power Factor
Calculations to determine phasor
values
• Since this is a parallel circuit, all voltages need
to be considered equal. If this were not true,
then everything we learned about the basic
circuits is for naught.
•
•
•
𝑉𝐴
36
𝐼𝑅 = =
= 0.15A = 150mA
𝑅
240
𝑉𝐴 36
𝐼𝑋𝐶 = = = 0.6A = 600mA
𝑋𝐶 60
𝑉𝐴 36
𝐼𝑋𝐿 = = = 0.4A = 400mA
𝑋𝐿 90
• Current due to net reactance (IX) = IC - IL =
0.6A – 0.4A = 0.2A = 200mA
– This tells us that the capacitor will have more
effect on the phase angle. Since this is a parallel
circuit, that angle will be in the first quadrant.
2
2
• IT = 𝐼𝑋 + 𝐼𝑅 =
0.2
2
+ 0.15
2
=
0.04 + 0.0225 = 0.0625 = 0.25A = 250mA
• Zeq =
𝑉𝐴
𝐼𝑇
=
36
0.25
= 144 Ohms
Phasor diagram
y
IX = 200mA
(Opp)
x
IR = 150mA
(Adj)
Calculation for θ
• tan 𝜃 =
𝑂𝑝𝑝
𝐴𝑑𝑗
=
𝐼𝑋
𝐼𝑅
=
0.2
0.15
∴𝜃=
−1 0.2
tan
0.15
=
∴𝜃=
−1 0.15
cos
0.25
=
tan−1 1.333 = 53.13°
Double check:
• cos 𝜃 =
𝐴𝑑𝑗
𝐻𝑦𝑝
=
𝐼𝑅
𝐼𝑇
=
0.15
0.25
cos −1 0.6 = 53.13° (verified)
Power equations
• Real Power → P =I2R = (0.15)2(240) = 5.4W
• Apparent Power → VA = VAIT = (36)(.25) = 9VA
• Power Factor → PF = cos 𝜃 =
𝐼𝑅
𝐼𝑇
∴ PF =
0.15
=
0.25
0.6
– This is also equivalent to taking the real power
and dividing it by the apparent power.
The end….