Math Problem Statement

Given the following information, how would we Find the average power delivered to the 18-Ω resistor?

Given the following information, how would we find the average power delivered to the 18-Ω resistor?

Solution

To find the average power delivered to the 18Ω18 \, \Omega resistor in this AC circuit, we can follow these steps:

Step-by-Step Solution

  1. Determine the Secondary Voltage (V₂):

    • Given the primary voltage V1=400VV_1 = 40 \angle 0^\circ \, \text{V} across the source, and the transformer turn ratio 2:52 : 5.
    • Since the turn ratio of the transformer is 2:52 : 5, the voltage ratio will also be 2:52 : 5.
    • Therefore, V2=52×V1=52×400=1000VV_2 = \frac{5}{2} \times V_1 = \frac{5}{2} \times 40 \angle 0^\circ = 100 \angle 0^\circ \, \text{V}.
  2. Calculate Impedance on the Secondary Side:

    • The load on the secondary side is a combination of 18Ω18 \, \Omega and j45Ωj45 \, \Omega (which is an inductor).
    • The equivalent impedance of the load, ZloadZ_{\text{load}}, is: Zload=18+j45ΩZ_{\text{load}} = 18 + j45 \, \Omega
  3. Find the Current in the Secondary Circuit (I₂):

    • Using Ohm’s law for the secondary side: I2=V2Zload=100018+j45I_2 = \frac{V_2}{Z_{\text{load}}} = \frac{100 \angle 0^\circ}{18 + j45}
    • Calculate the magnitude and phase of ZloadZ_{\text{load}}: Zload=182+452=324+2025=234948.46Ω|Z_{\text{load}}| = \sqrt{18^2 + 45^2} = \sqrt{324 + 2025} = \sqrt{2349} \approx 48.46 \, \Omega θZload=tan1(4518)68.2\theta_{Z_{\text{load}}} = \tan^{-1} \left( \frac{45}{18} \right) \approx 68.2^\circ
    • Thus, Zload=48.4668.2ΩZ_{\text{load}} = 48.46 \angle 68.2^\circ \, \Omega.
    • Now, find I2I_2: I2=100048.4668.2=2.06468.2AI_2 = \frac{100 \angle 0^\circ}{48.46 \angle 68.2^\circ} = 2.064 \angle -68.2^\circ \, \text{A}
  4. Calculate the Voltage Across the 18 Ω\Omega Resistor:

    • The voltage across the resistor VRV_R is I2I_2 times the real part of the impedance, which is 18 Ω\Omega.
    • VR=I2×18V_R = I_2 \times 18. VR=2.06468.2×1837.1568.2VV_R = 2.064 \angle -68.2^\circ \times 18 \approx 37.15 \angle -68.2^\circ \, \text{V}
  5. Find the Average Power Delivered to the 18 Ω\Omega Resistor:

    • The average power PP delivered to the resistor is given by: P=VR2RP = \frac{|V_R|^2}{R}
    • So, P=37.152181379.021876.61WP = \frac{37.15^2}{18} \approx \frac{1379.02}{18} \approx 76.61 \, \text{W}

Answer

The average power delivered to the 18 Ω\Omega resistor is approximately 76.61W76.61 \, \text{W}.


Would you like more detailed calculations for any step? Or do you have additional questions?

Related Questions

  1. How would the average power change if the resistance was doubled to 36 Ω\Omega?
  2. What would be the effect of changing the turn ratio of the transformer on the power delivered to the load?
  3. Can you calculate the reactive power in the circuit, specifically in the inductor?
  4. How would you determine the apparent power delivered by the source?
  5. What is the phase angle difference between the source voltage and current in the primary circuit?

Tip

In AC power calculations, remember that average (real) power is only delivered by resistive components, not reactive (inductive or capacitive) ones.

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Math Problem Analysis

Mathematical Concepts

AC Circuit Analysis
Complex Power
Average Power Calculation

Formulas

Ohm's Law: V = IZ
Average Power in Resistor: P = V^2 / R
Impedance Magnitude: |Z| = √(R^2 + X^2)
Phase Angle: θ = tan^(-1)(X/R)

Theorems

Impedance Transformation
Power Calculation in AC Circuits

Suitable Grade Level

Undergraduate Electrical Engineering