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A parallel-plate capacitor of plate separation 2mm is connected in an electric circuit having source voltage 400v. What is the value of the displacement current for 10^-6 s ,if the plate area is 60cm^2?
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A parallel-plate capacitor of plate separation 2mm is connected in an ...
Understanding Displacement Current
Displacement current is a concept introduced by James Clerk Maxwell to account for the changing electric field in situations where the electric field is varying with time, specifically in capacitors. It allows us to apply Ampere's law in scenarios where electric fields change.
Capacitance of the Parallel-Plate Capacitor
- The capacitance (C) of a parallel-plate capacitor can be expressed as:
C = (ε₀ * A) / d
Where:
- ε₀ is the permittivity of free space (approximately 8.85 x 10^-12 F/m).
- A is the plate area in square meters.
- d is the separation between the plates in meters.
- Given:
- Plate separation (d) = 2 mm = 0.002 m
- Plate area (A) = 60 cm² = 0.006 m²
Calculating Capacitance
- Using the formula:
C = (8.85 x 10^-12 F/m * 0.006 m²) / 0.002 m
C ≈ 2.655 x 10^-10 F
Displacement Current Calculation
- The displacement current (I_d) can be calculated using the formula:
I_d = C * (ΔV/Δt)
Where:
- ΔV is the voltage (400 V).
- Δt is the time interval (10^-6 s).
- Substituting the values:
I_d = (2.655 x 10^-10 F) * (400 V / 10^-6 s)
I_d ≈ 0.1062 A
Conclusion
- Therefore, the value of the displacement current for the given parameters is approximately 0.1062 A.
This calculation illustrates the significance of displacement current in understanding the behavior of capacitors in electric circuits.
Displacement current is a concept introduced by James Clerk Maxwell to account for the changing electric field in situations where the electric field is varying with time, specifically in capacitors. It allows us to apply Ampere's law in scenarios where electric fields change.
Capacitance of the Parallel-Plate Capacitor
- The capacitance (C) of a parallel-plate capacitor can be expressed as:
C = (ε₀ * A) / d
Where:
- ε₀ is the permittivity of free space (approximately 8.85 x 10^-12 F/m).
- A is the plate area in square meters.
- d is the separation between the plates in meters.
- Given:
- Plate separation (d) = 2 mm = 0.002 m
- Plate area (A) = 60 cm² = 0.006 m²
Calculating Capacitance
- Using the formula:
C = (8.85 x 10^-12 F/m * 0.006 m²) / 0.002 m
C ≈ 2.655 x 10^-10 F
Displacement Current Calculation
- The displacement current (I_d) can be calculated using the formula:
I_d = C * (ΔV/Δt)
Where:
- ΔV is the voltage (400 V).
- Δt is the time interval (10^-6 s).
- Substituting the values:
I_d = (2.655 x 10^-10 F) * (400 V / 10^-6 s)
I_d ≈ 0.1062 A
Conclusion
- Therefore, the value of the displacement current for the given parameters is approximately 0.1062 A.
This calculation illustrates the significance of displacement current in understanding the behavior of capacitors in electric circuits.
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A parallel-plate capacitor of plate separation 2mm is connected in an electric circuit having source voltage 400v. What is the value of the displacement current for 10^-6 s ,if the plate area is 60cm^2?
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A parallel-plate capacitor of plate separation 2mm is connected in an electric circuit having source voltage 400v. What is the value of the displacement current for 10^-6 s ,if the plate area is 60cm^2? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A parallel-plate capacitor of plate separation 2mm is connected in an electric circuit having source voltage 400v. What is the value of the displacement current for 10^-6 s ,if the plate area is 60cm^2? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel-plate capacitor of plate separation 2mm is connected in an electric circuit having source voltage 400v. What is the value of the displacement current for 10^-6 s ,if the plate area is 60cm^2?.
A parallel-plate capacitor of plate separation 2mm is connected in an electric circuit having source voltage 400v. What is the value of the displacement current for 10^-6 s ,if the plate area is 60cm^2? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A parallel-plate capacitor of plate separation 2mm is connected in an electric circuit having source voltage 400v. What is the value of the displacement current for 10^-6 s ,if the plate area is 60cm^2? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel-plate capacitor of plate separation 2mm is connected in an electric circuit having source voltage 400v. What is the value of the displacement current for 10^-6 s ,if the plate area is 60cm^2?.
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