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Transcript
DC Electricity & Capacitors NCEA AS3.6 Text Chapters 12-13 Kirchhoff’s Laws Current Law: The total current entering a junction in a circuit equals the total current leaving. I1=2A I3=5A I2=3A I1 I 2 I 3 Kirchhoff’s Laws Voltage Law: The total of all the potential differences around a closed loop in a circuit is zero V2 V1 V3 V1 V2 V3 0 Kirchhoff’s Laws For voltage sources: Going from + to – represents a loss of energy so potential difference is negative Going from – to + represents a gain in energy so potential difference is positive +V -V Kirchhoff’s Laws For resistors: Passing in the same direction as the current represents a loss of energy -V=-(IR) Passing in the opposite direction to the current represents a gain in energy +V=+(IR) +V=+IR -V=-IR I I Kirchhoff’s Laws To solve problems, follow a loop around a circuit… a-b 0V b-c +6V c-d -8V (2Ax4Ω) d-a +3V (3Ax1Ω) –V Then add up the p.d’s. 0V+6V-8V+3V-V=0 V=1V! a b 2A V? 6V 1Ω 3A 4Ω d c Kirchhoff’s Laws Harder…. a a-b 0V b-c -0.75A x 4Ω = -3V I1? c-d -I1x 2 = -2I1V d-a +5V Adding: -3 -2I1 +5 = 0 b 4Ω So I1 = 1A Using the current law: d 5V 2Ω 0.75A I2? 1A = 0.75A + I2 So I2 = 0.25A V? e c 8Ω f Kirchhoff’s Laws Continued… d a b-c -0.75A x 4Ω = -3V c-f +0.25A x 8Ω = +2V f-e +V? e-b 0V Adding: -3V + 2V + V? = 0 b So V = 1V 5V I1? 2Ω 0.75A c 4Ω I2? V? e 8Ω f Internal Resistance All components have some resistance, including cells and meters The potential difference (voltage) measured across a cell when no current is being drawn is called the E.M.F (electromotive force) When current flows, the voltage measured across the terminals of the cell will drop, because of the cell’s internal resistance. Internal Resistance V (V) EMF r EMF r A V V EMF Ir I (A) Electrical Meters Galvanometer – very sensitive meter that can be adapted to read either current or voltage Must know 2 things about the galvanometer The internal resistance The current that causes full scale deflection of the needle If.s.d Electrical Meters To make a voltmeter the galvanometer must be connected in series with a large resistor Rs Small current passes through G G r Most current continues through circuit Rs Electrical Meters Example: Need to measure up to 10V with a meter that has 200Ω of internal resistance and a full scale deflection at 5mA. V=IR 10V= 0.005A x (200 + Rs) Rs=1800Ω G 200Ω 5mA 10V Rs Electrical Meters To make an ammeter the galvanometer must be connected in parallel with a very small resistor Rp Small current passes through G G r Rp Most current continues through circuit Electrical Meters Example: Need to measure up to 1A with a meter that has 200Ω of internal resistance and a full scale deflection at 5mA. V (across meter) =IR = 0.005 x 200 = 1V Current through Rp = 1A - 0.005A = 0.995A Rp = V/I = 1V / 0.995A = 1Ω G 200Ω 5mA Rp Capacitors A capacitor is a device that can store electric charge and release it later on. They consist of two metal plates, separated by an insulating material called a dielectric. Capacitors are used for Tuning circuits eg radios Flashing circuits eg camera flashes Controlling alternating current Capacitors There are several types of capacitor: Non-polar Electrolytic Variable Capacitors Capacitors are charged by connecting them to a power supply or battery This causes an accumulation of electrons on one plate (negative) and removal of an equal number of electrons from the other (positive) Capacitors When a capacitor is fully charged: the current flow in the circuit stops both plates have equal but opposite amount of charge on them The voltage across the plates equals the supply voltage An electric field exists between the plates. The strength of this field is given by: E=V/d Capacitance C The value of a capacitor is a measure of how much charge it can store per volt applied across the plates The unit for capacitance is the farad F 1F=1CV-1 (Most capacitors are much less than 1F in size, it’s a big unit!) Q C V Capacitance The capacitance of a capacitor depends on 3 things: The area of the plates which overlap (Cα A ie. The more area overlapping, the bigger the C) The distance separating the plates (C α 1/d ie. The closer the plates the bigger the C) What is used as the dielectric material. Capacitance These are summed up in the capacitor construction formula: A d εo εr = Area of overlapping plates = Distance separating plates = Absolute permittivity of free space (!!) – A constant = 8.84x10-12 Fm-1 =Dielectric constant (no units, it’s a factor) Air 1 Oiled paper 2 Polystyrene 2.5 Glass Water 6 80 Charging Capacitors When the switch is closed, charge begins to build up on the plates. The process continues until the cap has the same voltage across it as the power source Charging capacitors It’s easy at first because the plates are empty, but as they fill the stored charge starts to repel further charge and the rate decreases V Vmax t Charging Capacitors As the rate of charge movement decreases, the current flow in the circuit decreases I Imax t Discharging Capacitors The voltage will be at a max to begin with but will drop to zero as charge leaves the plates. V Vmax t Discharging Capacitors The current will be large at first, but decrease as potential left to “push” decreases I Imax t Charge/Discharge Rates The rate that a cap will charge or discharge depends on two things: The size of the capacitor. The bigger the cap, the more charge it can store, so it will take longer to fill/empty. The initial current in the circuit. This is determined by the resistance of the circuit. More R means smaller current, so longer to fill/empty. Time Constant This is defined as the time it takes the voltage/current to rise to 63% of it’s max value or for it to drop 63% from it’s max value It is calculated using this formula: RC A cap is considered fully charged or discharged after 3 time constants Capacitor in Series Joining capacitors in series: 1 1 1 CT C1 C2 Why? The charge on each cap is the same. The voltage across each cap adds to the supply voltage V=V1+V2 -Q +Q V1 V -Q +Q V2 Q C V 1 V C Q 1 V1 V2 C Q 1 V1 V2 C Q Q 1 1 1 C C1 C 2 Capacitor Networks Joining capacitors in parallel: CT C1 C2 Why? The voltage across each cap is the same as the supply. The total charge stored is the sum of the charge in each cap. Q=Q1+ Q2 V Q1 Q2 Q C V Q1 Q2 C V Q1 Q2 C V V C C1 C2 (NB. This is the opposite way round to resistors) Energy in Capacitors Capacitors store energy as electric potential energy. The amount they store is half of the energy supplied by the battery. (The other half is dissipated as heat in the resistance of the circuit) When a capacitor discharges, the energy is dissipated in the resistance of the circuit as heat, light etc. Energy in Capacitors For a cell: Each coulomb of charge gains V Joules of energy So the energy provided by the cell ΔE=VQ V ΔE= area =VQ Q Energy in Capacitors For a capacitor: V voltage increases gradually as cap charges until it equals the cell voltage So the energy stored by the capacitor ΔE=1/2VQ ΔE= area = ½ VQ Q Energy in Capacitors Some rearrangement and substitution puts this equation into a familiar form… c.f. the formula for kinetic energy E=1/2mv2 c.f. the formula for elastic potential energy E=1/2kx2 1 E p QV 2 but Q C V so 1 2 E p CV 2