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A real inductor has resistance because it is composed of a coil of wire with nonzero resistivity. To measure the inductance of a coil, a student places it across a 12.0-V battery and measures a current of 0.72 A. The student then connects the coil to a 30.0-V (rms) 60 Hz generator and measures an rms current of 0.86 A. What is the inductance of the coil?
- a)0.081 H
- b)0.51 H
- c)34.9 H
- d)30.6 H
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A real inductor has resistance because it is composed of a coil of wir...
The coil has an inductance and resistance. We can visualize the coil therefore as an inductor and resistor in series. When we connect the DC battery across the coil, we have therefore and LR circuit. The steady state current is just i0 = V/R – same as above. So
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A real inductor has resistance because it is composed of a coil of wir...
To find the inductance of the coil, we can use the formula:
L = (Vrms / Irms) * (1 / (2πf))
Where:
L is the inductance of the coil
Vrms is the rms voltage across the coil (30.0 V in this case)
Irms is the rms current through the coil (0.86 A in this case)
f is the frequency of the generator (60 Hz in this case)
Let's calculate the inductance using the given values:
L = (30.0 V / 0.86 A) * (1 / (2π * 60 Hz))
L = 34.9 H
So, according to the calculations, the inductance of the coil is 34.9 H.
However, in the question, it is mentioned that the coil has some resistance due to the nonzero resistivity of the wire. This means that the inductance measured above also includes the effect of the resistance. To find the true inductance of the coil, we need to subtract the resistance component.
The resistance of the coil can be calculated using Ohm's law:
R = V / I
R = 12.0 V / 0.72 A
R = 16.7 Ω
Now, we can calculate the true inductance by considering the resistance:
Ltrue = √(L^2 - R^2)
Ltrue = √(34.9 H^2 - 16.7 Ω^2)
Ltrue = √(1217.01 H^2 - 278.89 Ω^2)
Ltrue = √938.12 H^2
Ltrue = 30.64 H
So, the true inductance of the coil is approximately 30.6 H.
Therefore, the correct answer is option 'D', not option 'A'.
L = (Vrms / Irms) * (1 / (2πf))
Where:
L is the inductance of the coil
Vrms is the rms voltage across the coil (30.0 V in this case)
Irms is the rms current through the coil (0.86 A in this case)
f is the frequency of the generator (60 Hz in this case)
Let's calculate the inductance using the given values:
L = (30.0 V / 0.86 A) * (1 / (2π * 60 Hz))
L = 34.9 H
So, according to the calculations, the inductance of the coil is 34.9 H.
However, in the question, it is mentioned that the coil has some resistance due to the nonzero resistivity of the wire. This means that the inductance measured above also includes the effect of the resistance. To find the true inductance of the coil, we need to subtract the resistance component.
The resistance of the coil can be calculated using Ohm's law:
R = V / I
R = 12.0 V / 0.72 A
R = 16.7 Ω
Now, we can calculate the true inductance by considering the resistance:
Ltrue = √(L^2 - R^2)
Ltrue = √(34.9 H^2 - 16.7 Ω^2)
Ltrue = √(1217.01 H^2 - 278.89 Ω^2)
Ltrue = √938.12 H^2
Ltrue = 30.64 H
So, the true inductance of the coil is approximately 30.6 H.
Therefore, the correct answer is option 'D', not option 'A'.
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A real inductor has resistance because it is composed of a coil of wire with nonzero resistivity. To measure the inductance of a coil, a student places it across a 12.0-V battery and measures a current of 0.72 A. The student then connects the coil to a 30.0-V (rms) 60 Hz generator and measures an rms current of 0.86 A. What is the inductance of the coil?a)0.081 Hb)0.51 Hc)34.9 Hd)30.6 HCorrect answer is option 'A'. Can you explain this answer?
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A real inductor has resistance because it is composed of a coil of wire with nonzero resistivity. To measure the inductance of a coil, a student places it across a 12.0-V battery and measures a current of 0.72 A. The student then connects the coil to a 30.0-V (rms) 60 Hz generator and measures an rms current of 0.86 A. What is the inductance of the coil?a)0.081 Hb)0.51 Hc)34.9 Hd)30.6 HCorrect answer is option 'A'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about A real inductor has resistance because it is composed of a coil of wire with nonzero resistivity. To measure the inductance of a coil, a student places it across a 12.0-V battery and measures a current of 0.72 A. The student then connects the coil to a 30.0-V (rms) 60 Hz generator and measures an rms current of 0.86 A. What is the inductance of the coil?a)0.081 Hb)0.51 Hc)34.9 Hd)30.6 HCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A real inductor has resistance because it is composed of a coil of wire with nonzero resistivity. To measure the inductance of a coil, a student places it across a 12.0-V battery and measures a current of 0.72 A. The student then connects the coil to a 30.0-V (rms) 60 Hz generator and measures an rms current of 0.86 A. What is the inductance of the coil?a)0.081 Hb)0.51 Hc)34.9 Hd)30.6 HCorrect answer is option 'A'. Can you explain this answer?.
A real inductor has resistance because it is composed of a coil of wire with nonzero resistivity. To measure the inductance of a coil, a student places it across a 12.0-V battery and measures a current of 0.72 A. The student then connects the coil to a 30.0-V (rms) 60 Hz generator and measures an rms current of 0.86 A. What is the inductance of the coil?a)0.081 Hb)0.51 Hc)34.9 Hd)30.6 HCorrect answer is option 'A'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about A real inductor has resistance because it is composed of a coil of wire with nonzero resistivity. To measure the inductance of a coil, a student places it across a 12.0-V battery and measures a current of 0.72 A. The student then connects the coil to a 30.0-V (rms) 60 Hz generator and measures an rms current of 0.86 A. What is the inductance of the coil?a)0.081 Hb)0.51 Hc)34.9 Hd)30.6 HCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A real inductor has resistance because it is composed of a coil of wire with nonzero resistivity. To measure the inductance of a coil, a student places it across a 12.0-V battery and measures a current of 0.72 A. The student then connects the coil to a 30.0-V (rms) 60 Hz generator and measures an rms current of 0.86 A. What is the inductance of the coil?a)0.081 Hb)0.51 Hc)34.9 Hd)30.6 HCorrect answer is option 'A'. Can you explain this answer?.
Solutions for A real inductor has resistance because it is composed of a coil of wire with nonzero resistivity. To measure the inductance of a coil, a student places it across a 12.0-V battery and measures a current of 0.72 A. The student then connects the coil to a 30.0-V (rms) 60 Hz generator and measures an rms current of 0.86 A. What is the inductance of the coil?a)0.081 Hb)0.51 Hc)34.9 Hd)30.6 HCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for SSC.
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