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### Transcript of Series and Parallel RCL Circuits Fundamentals Series RCL circuit V1 125 Vrms 0 Hz 0آ° R XC 125...

• Series and Parallel RCL Circuits Fundamentals

(non-resonant conditions)

• Series RCL circuit

V1

125 Vrms

0 Hz

R

100Ω

XC 125 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 = 𝑋 2 + 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 − 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

x

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

R 240Ω

XC 60 Ohms

XL 90 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 = 𝐼𝑋 2 + 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

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….