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Parallel plate capacitance made with two square plates of 400mm side. the 14 mm space between the plates is filled with two layers of dielectrics of Er=4,6mm thick and Er=2,8mm thick . neglecting fringing of fields at the edge the capacitence is?
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Parallel plate capacitance made with two square plates of 400mm side. ...
Calculation of Parallel Plate Capacitance with Dielectrics
1. Given Parameters:
- Side length of square plates, a = 400 mm
- Distance between plates, d = 14 mm
- Dielectric constant of first layer, εr1 = 4
- Thickness of first dielectric layer, t1 = 6 mm
- Dielectric constant of second layer, εr2 = 2
- Thickness of second dielectric layer, t2 = 8 mm
2. Calculation of Capacitance with Dielectrics:
- Capacitance of a parallel plate capacitor with dielectrics can be calculated using the formula:
C = (ε0 * εr * A) / d
where:
C = capacitance
ε0 = permittivity of free space = 8.85 x 10^-12 F/m
εr = relative permittivity of the dielectric material
A = area of the plates = a^2
d = distance between plates
3. Calculation for First Dielectric Layer:
- Area A1 = a^2 = (400 mm)^2 = 160,000 mm^2
- Capacitance C1 = (ε0 * εr1 * A1) / t1 = (8.85 x 10^-12 * 4 * 160,000) / 6 x 10^-3 = 94.4 pF
4. Calculation for Second Dielectric Layer:
- Area A2 = a^2 = 160,000 mm^2
- Capacitance C2 = (ε0 * εr2 * A2) / t2 = (8.85 x 10^-12 * 2 * 160,000) / 8 x 10^-3 = 35.4 pF
5. Combined Capacitance:
- The capacitance of the two dielectric layers in series is given by:
1 / C = 1 / C1 + 1 / C2
1 / C = 1 / 94.4 + 1 / 35.4
C = 23.8 pF
Therefore, the capacitance of the parallel plate capacitor with the two dielectric layers is 23.8 pF.
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it depends on the length of the conductor the capacitance of the line is proportional to the length of the transmission line their effect is negligible on the performance of short having a length less than 80 km and low voltage transmission accidents of the transmission line along with the conductances forms the shunted mittens the conductance and the transmission line is because of the leakage over the surface of the conductor considered a line consisting of two conductors and be each of radius are the distance between the conductors being Des shown in the diagram below minus the potential difference between the conductors and via's work QA charge on conductor QB charge on conductor vvab pencil difference between conductor and the Epsilon minus absolute primitivity QA plus QV = 0 so that QA equals QB - equals DBA equals data equals DB equals our substituting these values and voltage equation we get the capacitance between the conductors is cab is referred to as lying to line capacitance if the two conductors are in VR oppositely charge then the potential difference between them is zero then the potential of each conductor is given by one half bath the capacitance between each conductor and point of zero potential and is capacitive CN is called the capacitance to neut or capacitance to ground capacitance cab is the combination of two equal capacity and VN series thus capacitance to neutral is twice the capacitance between the conductors IE CN equals to Cave the absolute primitivity Epsilon is given by Epsilon equals epsilono Epsilon are where epsilano is the permittivity of the free space and Epsilon or is the relative primitivity of the medium prayer capacitance reactants between one conductor and neutral capacitance of the symmetrical three phase line let a balanced system of voltage be applied to a symmetrical three-phase line shown below the phasor diagram of the three phase line with equilateral spacing is shown below take the voltage of conductor to neutral as a reference phaser the potential difference between conductor and we can be written the similarly potential difference between conductors and sea is on adding equations one and two we get also combining equation three and four from equation 6 and 7 the line to neutral capacitance the capacitance of symmetrical three phase line is same as that of the two wire line Related: Capacitance of Transmission Lines?

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Parallel plate capacitance made with two square plates of 400mm side. the 14 mm space between the plates is filled with two layers of dielectrics of Er=4,6mm thick and Er=2,8mm thick . neglecting fringing of fields at the edge the capacitence is?
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Parallel plate capacitance made with two square plates of 400mm side. the 14 mm space between the plates is filled with two layers of dielectrics of Er=4,6mm thick and Er=2,8mm thick . neglecting fringing of fields at the edge the capacitence is? 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 Parallel plate capacitance made with two square plates of 400mm side. the 14 mm space between the plates is filled with two layers of dielectrics of Er=4,6mm thick and Er=2,8mm thick . neglecting fringing of fields at the edge the capacitence is? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Parallel plate capacitance made with two square plates of 400mm side. the 14 mm space between the plates is filled with two layers of dielectrics of Er=4,6mm thick and Er=2,8mm thick . neglecting fringing of fields at the edge the capacitence is?.
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