Series and Parallel RCL Circuits Fundamentals · Series RCL circuit V1 125 Vrms 0 Hz 0° R XC 125...
Transcript of Series and Parallel RCL Circuits Fundamentals · Series RCL circuit V1 125 Vrms 0 Hz 0° R XC 125...
Series and Parallel RCL Circuits Fundamentals
(non-resonant conditions)
Series RCL circuit
V1
125 Vrms
0 Hz
0°
R
100Ω
XC125 Ohms
XL
50 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 = 𝑋2 + 𝑅2 = −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 − 𝑉𝐶2 + 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
x
R = 100Ω(Adj)
X = -75Ω(Opp)
Calculation for θ
• tan 𝜃 = 𝑂𝑝𝑝
𝐴𝑑𝑗 =
𝑋
𝑅 =
−75
100 ∴ 𝜃 = tan−1 −75
100 =
tan−1 −0.75 = -36.87°
Double check:
• sin 𝜃 = 𝑂𝑝𝑝
𝐻𝑦𝑝 =
𝑋
𝑍𝑇 =
−75
125 ∴ 𝜃 = sin−1 −75
125 =
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°
R240Ω
XC60 Ohms
XL90 Ohms
IR = ? IX = ? IC = ? IT = ? 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.
• 𝐼𝑅 = 𝑉𝐴
𝑅 =
36
240 = 0.15A = 150mA
• 𝐼𝑋𝐶 =
𝑉𝐴
𝑋𝐶 =
36
60 = 0.6A = 600mA
• 𝐼𝑋𝐿 =
𝑉𝐴
𝑋𝐿 =
36
90 = 0.4A = 400mA
• 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.
• IT = 𝐼𝑋2 + 𝐼𝑅
2 = 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
x
IR = 150mA(Adj)
IX = 200mA(Opp)
Calculation for θ
• tan 𝜃 = 𝑂𝑝𝑝
𝐴𝑑𝑗 =
𝐼𝑋
𝐼𝑅 =
0.2
0.15 ∴ 𝜃 = tan−1 0.2
0.15 =
tan−1 1.333 = 53.13°
Double check:
• cos 𝜃 = 𝐴𝑑𝑗
𝐻𝑦𝑝 =
𝐼𝑅
𝐼𝑇 =
0.15
0.25 ∴ 𝜃 = cos−1 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….