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Determine the average power delivered to the circuit consisting of an impedance Z = 5+j8 when the current flowing through the circuit is I = 5∠30⁰.
- a)61.5
- b)62.5
- c)63.5
- d)64.5
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Determine the average power delivered to the circuit consisting of an ...
The expression of the average power delivered to the circuit is Pavg = Im2 R/2. Given Im = 5, R = 5. So the average power delivered to the circuit = 52×5/2 = 62.5W.
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Determine the average power delivered to the circuit consisting of an ...
Understanding the Circuit Parameters
To determine the average power delivered to the circuit, we begin with the given parameters:
- Impedance (Z) = 5 + j8 ohms
- Current (I) = 5∠30° A
Convert Current to Rectangular Form
First, convert the current from polar to rectangular form:
- I = 5 * (cos(30°) + j*sin(30°))
- I = 5 * (0.866 + j*0.5)
- I = 4.33 + j2.5 A
Calculate Voltage (V)
Using Ohm's Law (V = I * Z):
- V = (4.33 + j2.5) * (5 + j8)
Calculating this product, we get:
- V = (4.33*5 - 2.5*8) + j(4.33*8 + 2.5*5)
- V = (21.65 - 20) + j(34.64 + 12.5)
- V = 1.65 + j47.14 V
Determine Average Power (P)
The average power (P) can be calculated using the formula:
- P = I_rms^2 * R
Where R is the real part of the impedance (5 ohms) and I_rms is the root mean square of the current.
Since I = 5A, I_rms = I / √2 = 5 / √2 = 3.54 A. Thus:
- P = (3.54^2) * 5
- P = 12.54 * 5 = 62.7 W
However, since we use the average power formula directly with the given current:
- P = 5^2 * (Real part of Z) / 2 = 25 * 5 / 2 = 62.5 W
Conclusion
The average power delivered to the circuit is approximately 62.5 W, which corresponds to option 'B'.
To determine the average power delivered to the circuit, we begin with the given parameters:
- Impedance (Z) = 5 + j8 ohms
- Current (I) = 5∠30° A
Convert Current to Rectangular Form
First, convert the current from polar to rectangular form:
- I = 5 * (cos(30°) + j*sin(30°))
- I = 5 * (0.866 + j*0.5)
- I = 4.33 + j2.5 A
Calculate Voltage (V)
Using Ohm's Law (V = I * Z):
- V = (4.33 + j2.5) * (5 + j8)
Calculating this product, we get:
- V = (4.33*5 - 2.5*8) + j(4.33*8 + 2.5*5)
- V = (21.65 - 20) + j(34.64 + 12.5)
- V = 1.65 + j47.14 V
Determine Average Power (P)
The average power (P) can be calculated using the formula:
- P = I_rms^2 * R
Where R is the real part of the impedance (5 ohms) and I_rms is the root mean square of the current.
Since I = 5A, I_rms = I / √2 = 5 / √2 = 3.54 A. Thus:
- P = (3.54^2) * 5
- P = 12.54 * 5 = 62.7 W
However, since we use the average power formula directly with the given current:
- P = 5^2 * (Real part of Z) / 2 = 25 * 5 / 2 = 62.5 W
Conclusion
The average power delivered to the circuit is approximately 62.5 W, which corresponds to option 'B'.
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Question Description
Determine the average power delivered to the circuit consisting of an impedance Z = 5+j8 when the current flowing through the circuit is I = 5∠30.a)61.5b)62.5c)63.5d)64.5Correct answer is option 'B'. Can you explain this answer? for Electrical Engineering (EE) 2025 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about Determine the average power delivered to the circuit consisting of an impedance Z = 5+j8 when the current flowing through the circuit is I = 5∠30.a)61.5b)62.5c)63.5d)64.5Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Determine the average power delivered to the circuit consisting of an impedance Z = 5+j8 when the current flowing through the circuit is I = 5∠30.a)61.5b)62.5c)63.5d)64.5Correct answer is option 'B'. Can you explain this answer?.
Determine the average power delivered to the circuit consisting of an impedance Z = 5+j8 when the current flowing through the circuit is I = 5∠30.a)61.5b)62.5c)63.5d)64.5Correct answer is option 'B'. Can you explain this answer? for Electrical Engineering (EE) 2025 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about Determine the average power delivered to the circuit consisting of an impedance Z = 5+j8 when the current flowing through the circuit is I = 5∠30.a)61.5b)62.5c)63.5d)64.5Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Determine the average power delivered to the circuit consisting of an impedance Z = 5+j8 when the current flowing through the circuit is I = 5∠30.a)61.5b)62.5c)63.5d)64.5Correct answer is option 'B'. Can you explain this answer?.
Solutions for Determine the average power delivered to the circuit consisting of an impedance Z = 5+j8 when the current flowing through the circuit is I = 5∠30.a)61.5b)62.5c)63.5d)64.5Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
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