# 1-4 - 1 DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING...

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Unformatted text preview: 1 DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING ELECTRICAL PLANT 342 LECTURES 1-4 ENERGY STORAGE CIRCUIT ELEMENTS There are two types: Capacitors stores electrostatic energy in its electric field Inductor stores electromagnetic energy in its magnetic field CAPACITOR A capacitor consists of two metal plates separated by a dielectric material. When the two plates are held at different potentials, a charge is built up on the plates. The balance is maintained with an equal opposite charges. The unit of capacitance is the FARAD and is defined as: C = Q / V Farad The farad is a very large unit and the practical unit is the micro-farad ( uF ), nano-farad ( nF ) or pico-farad ( pF ). The capacity of a parallel-plate capacitor depends on its dimensions and the insulating material that separates the plates, called the dielectric. C = ( ε o ε r A ) F d where ε o is the absolute permittivity = 8.854 x 10 –12 F/m and ε r = relative permittivity Dielectric Properties of Selected Materials Material Relative Permittivity ε r Dielectric Strength (kV/cm) Vacuum 1.00000 ∞ Air 1.00054 8 Paper 3.5 140 Polystyrene 2.6 250 2 Teflon 2.1 600 Titanium Dioxide 100 60 CAPACITORS IN PARALLEL When capacitors are connected in parallel across a voltage source, each capacitor will be charged until the voltage difference across each capacitor is the same as the supply voltage. Let the charge of the capacitor be Q 1 and Q 2 respectively. then Q 1 = C 1 V C Q 2 = C 2 V C Using KCL at the node, total charge Q T = Q 1 + Q 2 = C 1 V + C 2 V Q T / V = C 1 + C 2 hence C p = C 1 + C 2 CAPACITORS IN SERIES When capacitors are connected in series the same charge flows through each capacitor. Voltage across capacitors C 1 and C 2 are: 3 V 1 = Q/C 1 V 2 = Q/C 2 Using KVL, V = V 1 + V 2 Substituting Q/C S = Q/C 1 + Q/C 2 Or 1/C S = 1/C 1 + 1/C 2 CHARGING OF CAPACITOR When a voltage is applied across the plates of an uncharged capacitor, it’s potential difference increases exponentially from zero to the supply voltage in t seconds ( t = ∞ ). The charging current is given by: i ( t ) = d q / d t Amperes Current = rate of change of charge Substitute q = C v coulombs i ( t ) = C d v / d t v ( t ) = 1/C ∫ i d t If the initial voltage across the capacitor is v then the voltage after t seconds is given by: v ( t ) = 1/C ∫ i d t + v volts ENERGY STORED IN A CAPACITOR Since Power = rate of doing work (energy) Energy is the integral of power Energy W = ∫ v i d t = ∫ v [C d v / d t ] d t W = ½ C v 2 watts THE INDUCTOR 4 A current through a wire will produces a magnetic flux....
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1-4 - 1 DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING...

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