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

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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

    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

    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

    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….