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A parallel-plate capacitor with plate area of 5 cm2 and plate separation of 3 mm has a voltage 50 sin (103 t) V applied to its plates. If εr = 2, the displacement current is
- a)148 cos(1010 t) nA
- b)261 cos(1010 t) μA
- c)261 cos( 1010 t) nA
- d)148 cos(1010 t) μA
Correct answer is option 'A'. Can you explain this answer?
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Question:
A parallel-plate capacitor with plate area of 5 cm2 and plate separation of 3 mm has a voltage 50 sin (103 t) V applied to its plates. If r= 2, the displacement current is
a) 148 cos(1010 t) nA
b) 261 cos(1010 t)A
c) 261 cos( 1010 t) nA
d) 148 cos(1010 t)A
Answer:
Displacement current is given by the formula:
I = ε₀(dv/dt)
where I is the displacement current, ε₀ is the permittivity of free space, and dv/dt is the rate of change of voltage with respect to time.
Given:
Area of the plates (A) = 5 cm² = 5 × 10⁻⁴ m²
Plate separation (d) = 3 mm = 3 × 10⁻³ m
Voltage (v) = 50 sin (103 t) V
Relative permittivity (r) = 2
To calculate the displacement current, we need to first find the capacitance of the parallel-plate capacitor using the formula:
C = ε₀(rA/d)
Substituting the given values, we get:
C = (8.85 × 10⁻¹² × 2 × 5 × 10⁻⁴)/(3 × 10⁻³)
= 29.5 pF
Now, we can calculate the rate of change of voltage with respect to time using the derivative of the given voltage:
dv/dt = 50 × 10³ cos (103 t)
Substituting the values, we get:
I = ε₀(dv/dt)
= 8.85 × 10⁻¹² × 2 × 29.5 × 10⁻¹² × 50 × 10³ cos (103 t)
= 148 cos (1010 t) nA
Therefore, the correct answer is option a) 148 cos (1010 t) nA.
A parallel-plate capacitor with plate area of 5 cm2 and plate separation of 3 mm has a voltage 50 sin (103 t) V applied to its plates. If r= 2, the displacement current is
a) 148 cos(1010 t) nA
b) 261 cos(1010 t)A
c) 261 cos( 1010 t) nA
d) 148 cos(1010 t)A
Answer:
Displacement current is given by the formula:
I = ε₀(dv/dt)
where I is the displacement current, ε₀ is the permittivity of free space, and dv/dt is the rate of change of voltage with respect to time.
Given:
Area of the plates (A) = 5 cm² = 5 × 10⁻⁴ m²
Plate separation (d) = 3 mm = 3 × 10⁻³ m
Voltage (v) = 50 sin (103 t) V
Relative permittivity (r) = 2
To calculate the displacement current, we need to first find the capacitance of the parallel-plate capacitor using the formula:
C = ε₀(rA/d)
Substituting the given values, we get:
C = (8.85 × 10⁻¹² × 2 × 5 × 10⁻⁴)/(3 × 10⁻³)
= 29.5 pF
Now, we can calculate the rate of change of voltage with respect to time using the derivative of the given voltage:
dv/dt = 50 × 10³ cos (103 t)
Substituting the values, we get:
I = ε₀(dv/dt)
= 8.85 × 10⁻¹² × 2 × 29.5 × 10⁻¹² × 50 × 10³ cos (103 t)
= 148 cos (1010 t) nA
Therefore, the correct answer is option a) 148 cos (1010 t) nA.
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A parallel-plate capacitor with plate area of 5 cm2and plate separation of 3 mm has a voltage 50 sin (103 t) V applied to its plates. Ifεr= 2, the displacement current isa)148 cos(1010 t) nAb)261 cos(1010 t)μAc)261 cos( 1010 t) nAd)148 cos(1010 t)μACorrect answer is option 'A'. Can you explain this answer?
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A parallel-plate capacitor with plate area of 5 cm2and plate separation of 3 mm has a voltage 50 sin (103 t) V applied to its plates. Ifεr= 2, the displacement current isa)148 cos(1010 t) nAb)261 cos(1010 t)μAc)261 cos( 1010 t) nAd)148 cos(1010 t)μACorrect answer is option 'A'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A parallel-plate capacitor with plate area of 5 cm2and plate separation of 3 mm has a voltage 50 sin (103 t) V applied to its plates. Ifεr= 2, the displacement current isa)148 cos(1010 t) nAb)261 cos(1010 t)μAc)261 cos( 1010 t) nAd)148 cos(1010 t)μACorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel-plate capacitor with plate area of 5 cm2and plate separation of 3 mm has a voltage 50 sin (103 t) V applied to its plates. Ifεr= 2, the displacement current isa)148 cos(1010 t) nAb)261 cos(1010 t)μAc)261 cos( 1010 t) nAd)148 cos(1010 t)μACorrect answer is option 'A'. Can you explain this answer?.
A parallel-plate capacitor with plate area of 5 cm2and plate separation of 3 mm has a voltage 50 sin (103 t) V applied to its plates. Ifεr= 2, the displacement current isa)148 cos(1010 t) nAb)261 cos(1010 t)μAc)261 cos( 1010 t) nAd)148 cos(1010 t)μACorrect answer is option 'A'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A parallel-plate capacitor with plate area of 5 cm2and plate separation of 3 mm has a voltage 50 sin (103 t) V applied to its plates. Ifεr= 2, the displacement current isa)148 cos(1010 t) nAb)261 cos(1010 t)μAc)261 cos( 1010 t) nAd)148 cos(1010 t)μACorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel-plate capacitor with plate area of 5 cm2and plate separation of 3 mm has a voltage 50 sin (103 t) V applied to its plates. Ifεr= 2, the displacement current isa)148 cos(1010 t) nAb)261 cos(1010 t)μAc)261 cos( 1010 t) nAd)148 cos(1010 t)μACorrect answer is option 'A'. Can you explain this answer?.
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