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3 parts
iii

31.4 Ω\text{}31.4 \text{ } \Omega\text{}

iv

31.85 Ω\text{}31.85 \text{ } \Omega\text{}

v

18.03 Ω\text{}18.03 \text{ } \Omega\text{}

Step-by-step solution

10 steps

Here are the solutions for the requested calculations:

Part iii· 3 steps

Calculate the inductive reactance XL

The inductive reactance (XLX_L) is the opposition of an inductor to the flow of alternating current. It is given by the formula:

XL=2πfLX_L = 2 \pi f L

where ff is the frequency and LL is the inductance.

  1. Identify the given values.

    Given frequency f=50 Hzf = 50 \text{ Hz}.
    Given inductance L=0.1 HL = 0.1 \text{ H}.
    Given π=3.14\pi = 3.14.

  2. Substitute the values into the formula.

    XL=2×3.14×50 Hz×0.1 HX_L = 2 \times 3.14 \times 50 \text{ Hz} \times 0.1 \text{ H}

  3. Calculate the inductive reactance.

    XL=31.4 ΩX_L = 31.4 \text{ } \Omega

    The unit for reactance is Ohms (Ω\Omega).

    The inductive reactance is 31.4 Ω\boxed{\text{}31.4 \text{ } \Omega\text{}}

Part iv

31.85 Ω\text{}31.85 \text{ } \Omega\text{}

Calculate the capacitive reactance XC

The capacitive reactance (XCX_C) is the opposition of a capacitor to the flow of alternating current. It is given by the formula:

XC=12πfCX_C = \frac{1}{2 \pi f C}

where ff is the frequency and CC is the capacitance.

  1. Identify the given values and convert units.

    Given frequency f=50 Hzf = 50 \text{ Hz}.
    Given capacitance C=100μF=100×10−6 FC = 100 \mu \text{F} = 100 \times 10^{-6} \text{ F}.
    Given π=3.14\pi = 3.14.

  2. Substitute the values into the formula.

    XC=12×3.14×50 Hz×100×10−6 FX_C = \frac{1}{2 \times 3.14 \times 50 \text{ Hz} \times 100 \times 10^{-6} \text{ F}}

  3. Calculate the capacitive reactance.

    XC=131400×10−6X_C = \frac{1}{31400 \times 10^{-6}} XC=10.0314X_C = \frac{1}{0.0314} XC≈31.847 ΩX_C \approx 31.847 \text{ } \Omega

    The capacitive reactance is 31.85 Ω\boxed{\text{}31.85 \text{ } \Omega\text{}} (rounded to two decimal places).

Part v

18.03 Ω\text{}18.03 \text{ } \Omega\text{}

Calculate the total impedance Z

The total impedance (ZZ) in a series RC circuit (resistance and capacitive reactance) is the total opposition to current flow. It is calculated using the Pythagorean theorem for phasors:

Z=R2+XC2Z = \sqrt{R^2 + X_C^2}

where RR is the resistance and XCX_C is the capacitive reactance.

  1. Identify the given values.

    Given resistance R=10 ΩR = 10 \text{ } \Omega.
    Given capacitive reactance XC=15 ΩX_C = 15 \text{ } \Omega.

  2. Substitute the values into the formula.

    Z=(10 Ω)2+(15 Ω)2Z = \sqrt{(10 \text{ } \Omega)^2 + (15 \text{ } \Omega)^2}

  3. Calculate the squares of the resistance and capacitive reactance.

    Z=100 Ω2+225 Ω2Z = \sqrt{100 \text{ } \Omega^2 + 225 \text{ } \Omega^2}

  4. Sum the squared values and take the square root.

    Z=325 Ω2Z = \sqrt{325 \text{ } \Omega^2} Z≈18.027 ΩZ \approx 18.027 \text{ } \Omega

    The total impedance is 18.03 Ω\boxed{\text{}18.03 \text{ } \Omega\text{}} (rounded to two decimal places).

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