SET 1: AC CIRCUIT PROBLEMS Q1: R-L Series Circuit A coil has R = 30\ and L = 0.127H. It is connected to 220V, 50Hz supply. Find: a. Inductive reactance XL b. Total impedance Z c. Current I d. Power factor Q2: R-C Series Circuit A 100\ resistor and 25mu F capacitor are in series across 240V, 60Hz. Find: a. Capacitive reactance XC b. Impedance Z c. Current I d. Phase angle theta. State if leading or lagging Q3: R-L-C Series Circuit R = 50\ , L = 0.2H, C = 50mu F connected to 230V, 50Hz. Find: a. XL and XC b. Net reactance c. Impedance Z d. Current I e. Real power P

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Q1: R-L Series Circuit

Given: R=30 ΩR = 30\ \Omega, L=0.127 HL = 0.127\ H, V=220 VV = 220\ V, f=50 Hzf = 50\ Hz.

a) Inductive reactance XLX_L

  1. Calculate the inductive reactance using the formula XL=2πfLX_L = 2\pi f L.

    XL=2π(50 Hz)(0.127 H)X_L = 2 \pi (50\ \text{Hz}) (0.127\ \text{H}) XL=39.898 ΩX_L = 39.898\ \Omega XL=39.90 Ω\boxed{X_L = 39.90\ \Omega}

    b) Total impedance ZZ

  2. Calculate the total impedance using the formula Z=R2+XL2Z = \sqrt{R^2 + X_L^2}.

    Z=(30 Ω)2+(39.898 Ω)2Z = \sqrt{(30\ \Omega)^2 + (39.898\ \Omega)^2} Z=900 Ω2+1591.85 Ω2Z = \sqrt{900\ \Omega^2 + 1591.85\ \Omega^2} Z=2491.85 Ω2Z = \sqrt{2491.85\ \Omega^2} Z=49.918 ΩZ = 49.918\ \Omega Z=49.92 Ω\boxed{Z = 49.92\ \Omega}

    c) Current II

  3. Calculate the current using Ohm's law for AC circuits, I=VZI = \frac{V}{Z}.

    I=220 V49.918 ΩI = \frac{220\ \text{V}}{49.918\ \Omega} I=4.407 AI = 4.407\ \text{A} I=4.41 A\boxed{I = 4.41\ \text{A}}

    d) Power factor

  4. Calculate the power factor using the formula cos⁡ϕ=RZ\cos \phi = \frac{R}{Z}.

    cos⁡ϕ=30 Ω49.918 Ω\cos \phi = \frac{30\ \Omega}{49.918\ \Omega} cos⁡ϕ=0.6009\cos \phi = 0.6009 Power factor=0.601\boxed{\text{Power factor} = 0.601}

    Q2: R-C Series Circuit

    Given: R=100 ΩR = 100\ \Omega, C=25 μF=25×10−6 FC = 25\ \mu F = 25 \times 10^{-6}\ F, V=240 VV = 240\ V, f=60 Hzf = 60\ Hz.

    a) Capacitive reactance XCX_C

  5. Calculate the capacitive reactance using the formula XC=12πfCX_C = \frac{1}{2\pi f C}.

    XC=12π(60 Hz)(25×10−6 F)X_C = \frac{1}{2 \pi (60\ \text{Hz}) (25 \times 10^{-6}\ \text{F})} XC=10.00942477X_C = \frac{1}{0.00942477} XC=106.103 ΩX_C = 106.103\ \Omega XC=106.10 Ω\boxed{X_C = 106.10\ \Omega}

    b) Impedance ZZ

  6. Calculate the impedance using the formula Z=R2+XC2Z = \sqrt{R^2 + X_C^2}.

    Z=(100 Ω)2+(106.103 Ω)2Z = \sqrt{(100\ \Omega)^2 + (106.103\ \Omega)^2} Z=10000 Ω2+11258.04 Ω2Z = \sqrt{10000\ \Omega^2 + 11258.04\ \Omega^2} Z=21258.04 Ω2Z = \sqrt{21258.04\ \Omega^2} Z=145.801 ΩZ = 145.801\ \Omega Z=145.80 Ω\boxed{Z = 145.80\ \Omega}

    c) Current II

  7. Calculate the current using Ohm's law for AC circuits, I=VZI = \frac{V}{Z}.

    I=240 V145.801 ΩI = \frac{240\ \text{V}}{145.801\ \Omega} I=1.646 AI = 1.646\ \text{A} I=1.65 A\boxed{I = 1.65\ \text{A}}

    d) Phase angle θ\theta. State if leading or lagging.

  8. Calculate the phase angle using tan⁡θ=−XCR\tan \theta = \frac{-X_C}{R}.

    tan⁡θ=−106.103 Ω100 Ω\tan \theta = \frac{-106.103\ \Omega}{100\ \Omega} tan⁡θ=−1.06103\tan \theta = -1.06103 θ=arctan⁡(−1.06103)\theta = \arctan(-1.06103) θ=−46.69∘\theta = -46.69^\circ

    In an R-C series circuit, the current leads the voltage.

    θ=−46.69∘,Current leads voltage\boxed{\theta = -46.69^\circ, \text{Current leads voltage}}

    Q3: R-L-C Series Circuit

    Given: R=50 ΩR = 50\ \Omega, L=0.2 HL = 0.2\ H, C=50 μF=50×10−6 FC = 50\ \mu F = 50 \times 10^{-6}\ F, V=230 VV = 230\ V, f=50 Hzf = 50\ Hz.

    a) XLX_L and XCX_C

  9. Calculate the inductive reactance XL=2πfLX_L = 2\pi f L.

    XL=2π(50 Hz)(0.2 H)X_L = 2 \pi (50\ \text{Hz}) (0.2\ \text{H}) XL=62.832 ΩX_L = 62.832\ \Omega

  10. Calculate the capacitive reactance XC=12πfCX_C = \frac{1}{2\pi f C}.

    XC=12π(50 Hz)(50×10−6 F)X_C = \frac{1}{2 \pi (50\ \text{Hz}) (50 \times 10^{-6}\ \text{F})} XC=10.01570796X_C = \frac{1}{0.01570796} XC=63.662 ΩX_C = 63.662\ \Omega XL=62.83 Ω,XC=63.66 Ω\boxed{X_L = 62.83\ \Omega, X_C = 63.66\ \Omega}

    b) Net reactance

  11. Calculate the net reactance Xnet=XL−XCX_{net} = X_L - X_C.

    Xnet=62.832 Ω−63.662 ΩX_{net} = 62.832\ \Omega - 63.662\ \Omega Xnet=−0.83 ΩX_{net} = -0.83\ \Omega Xnet=−0.83 Ω\boxed{X_{net} = -0.83\ \Omega}

    c) Impedance ZZ

  12. Calculate the impedance using the formula Z=R2+(XL−XC)2Z = \sqrt{R^2 + (X_L - X_C)^2}.

    Z=(50 Ω)2+(−0.83 Ω)2Z = \sqrt{(50\ \Omega)^2 + (-0.83\ \Omega)^2} Z=2500 Ω2+0.6889 Ω2Z = \sqrt{2500\ \Omega^2 + 0.6889\ \Omega^2} Z=2500.6889 Ω2Z = \sqrt{2500.6889\ \Omega^2} Z=50.0069 ΩZ = 50.0069\ \Omega Z=50.01 Ω\boxed{Z = 50.01\ \Omega}

    d) Current II

  13. Calculate the current using Ohm's law for AC circuits, I=VZI = \frac{V}{Z}.

    I=230 V50.0069 ΩI = \frac{230\ \text{V}}{50.0069\ \Omega} I=4.599 AI = 4.599\ \text{A} I=4.60 A\boxed{I = 4.60\ \text{A}}

    e) Real power PP

  14. Calculate the real power using the formula P=I2RP = I^2 R.

    P=(4.599 A)2(50 Ω)P = (4.599\ \text{A})^2 (50\ \Omega) P=(21.1508 A2)(50 Ω)P = (21.1508\ \text{A}^2) (50\ \Omega) P=1057.54 WP = 1057.54\ \text{W} P=1057.54 W\boxed{P = 1057.54\ \text{W}}
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