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A parallel plate capacitor with plate area of 10 cm2 and a plate separation of 6 mm has a voltage 50 Sin 1000t V applied to its plates. Calculate the displacement current if the relative permittivity of the dielctric between the plates is 6?
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A parallel plate capacitor with plate area of 10 cm2 and a plate sepa...
Calculation of Displacement Current in a Parallel Plate Capacitor
Given:
Plate area (A) = 10 cm2
Plate separation (d) = 6 mm
Voltage (V) = 50 Sin 1000t V
Relative permittivity of dielectric (εr) = 6
Formula:
Displacement current (ID) = ε0εr (dV/dt)A
Where, ε0 is the permittivity of free space (8.854 x 10-12 F/m)
Calculation:
Converting plate area to m2: A = 10 cm2 = 10 x 10-4 m2
Converting plate separation to m: d = 6 mm = 6 x 10-3 m
Differentiating voltage with respect to time: dV/dt = 50000 Cos 1000t V/s
Substituting values in the formula:
ID = (8.854 x 10-12 F/m) x 6 x (50000 Cos 1000t) x 10-4 m2
ID = 2.113 x 10-12 Cos 1000t A
Therefore, the displacement current in the parallel plate capacitor is 2.113 x 10-12 Cos 1000t A.
Explanation:
A displacement current is a current that arises due to the time-varying electric field in a region where there is no actual flow of charge. In a capacitor, the electric field between the plates changes with time as the voltage across the plates varies. This changing electric field induces a displacement current in the dielectric between the plates. The displacement current is directly proportional to the rate of change of voltage and the permittivity of the dielectric. The formula for displacement current in a parallel plate capacitor is derived from Ampere's law and the concept of continuity of current.
Given:
Plate area (A) = 10 cm2
Plate separation (d) = 6 mm
Voltage (V) = 50 Sin 1000t V
Relative permittivity of dielectric (εr) = 6
Formula:
Displacement current (ID) = ε0εr (dV/dt)A
Where, ε0 is the permittivity of free space (8.854 x 10-12 F/m)
Calculation:
Converting plate area to m2: A = 10 cm2 = 10 x 10-4 m2
Converting plate separation to m: d = 6 mm = 6 x 10-3 m
Differentiating voltage with respect to time: dV/dt = 50000 Cos 1000t V/s
Substituting values in the formula:
ID = (8.854 x 10-12 F/m) x 6 x (50000 Cos 1000t) x 10-4 m2
ID = 2.113 x 10-12 Cos 1000t A
Therefore, the displacement current in the parallel plate capacitor is 2.113 x 10-12 Cos 1000t A.
Explanation:
A displacement current is a current that arises due to the time-varying electric field in a region where there is no actual flow of charge. In a capacitor, the electric field between the plates changes with time as the voltage across the plates varies. This changing electric field induces a displacement current in the dielectric between the plates. The displacement current is directly proportional to the rate of change of voltage and the permittivity of the dielectric. The formula for displacement current in a parallel plate capacitor is derived from Ampere's law and the concept of continuity of current.
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A parallel plate capacitor with plate area of 10 cm2 and a plate separation of 6 mm has a voltage 50 Sin 1000t V applied to its plates. Calculate the displacement current if the relative permittivity of the dielctric between the plates is 6?
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A parallel plate capacitor with plate area of 10 cm2 and a plate separation of 6 mm has a voltage 50 Sin 1000t V applied to its plates. Calculate the displacement current if the relative permittivity of the dielctric between the plates is 6? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A parallel plate capacitor with plate area of 10 cm2 and a plate separation of 6 mm has a voltage 50 Sin 1000t V applied to its plates. Calculate the displacement current if the relative permittivity of the dielctric between the plates is 6? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel plate capacitor with plate area of 10 cm2 and a plate separation of 6 mm has a voltage 50 Sin 1000t V applied to its plates. Calculate the displacement current if the relative permittivity of the dielctric between the plates is 6?.
A parallel plate capacitor with plate area of 10 cm2 and a plate separation of 6 mm has a voltage 50 Sin 1000t V applied to its plates. Calculate the displacement current if the relative permittivity of the dielctric between the plates is 6? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A parallel plate capacitor with plate area of 10 cm2 and a plate separation of 6 mm has a voltage 50 Sin 1000t V applied to its plates. Calculate the displacement current if the relative permittivity of the dielctric between the plates is 6? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel plate capacitor with plate area of 10 cm2 and a plate separation of 6 mm has a voltage 50 Sin 1000t V applied to its plates. Calculate the displacement current if the relative permittivity of the dielctric between the plates is 6?.
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