Document Sample

ELECTRICAL CIRCUITS The CELL The cell stores chemical energy and transfers it to electrical energy when a circuit is connected. When two or more cells are connected together we call this a Battery. The cells chemical energy is used up pushing a current round a circuit. What is an electric current? An electric current is a flow of microscopic particles called electrons flowing through wires and components. - + In which direction does the current flow? from the Negative terminal to the Positive terminal of a cell. simple circuits Here is a simple electric circuit. It has a cell, a lamp and a switch. cell wires switch lamp To make the circuit, these components are connected together with metal connecting wires. simple circuits When the switch is closed, the lamp lights up. This is because there is a continuous path of metal for the electric current to flow around. If there were any breaks in the circuit, the current could not flow. circuit diagram Scientists usually draw electric circuits using symbols; cell lamp switch wires circuit diagrams In circuit diagrams components are represented by the following symbols; cell battery switch lamp buzzer ammeter voltmeter motor resistor variable resistor types of circuit There are two types of electrical circuits; SERIES CIRCUITS PARALLEL CIRCUITS SERIES CIRCUITS The components are connected end-to-end, one after the other. They make a simple loop for the current to flow round. If one bulb ‘blows’ it breaks the whole circuit and all the bulbs go out. PARALLEL CIRCUITS The components are connected side by side. The current has a choice of routes. If one bulb ‘blows’ there is still be a complete circuit to the other bulb so it stays alight. measuring current Electric current is measured in amps (A) using an ammeter connected in series in the circuit. A measuring current This is how we draw an ammeter in a circuit. A A SERIES CIRCUIT PARALLEL CIRCUIT measuring current SERIES CIRCUIT • current is the same 2A 2A at all points in the circuit. 2A PARALLEL CIRCUIT 2A 2A • current is shared between the 1A components 1A copy the following circuits and fill in the missing ammeter readings. ? 4A ? 3A 3A ? 1A ? 4A 4A 1A ? 1A measuring voltage The ‘electrical push’ which the cell gives to the current is called the voltage. It is measured in volts (V) on a voltmeter V measuring voltage Different cells produce different voltages. The bigger the voltage supplied by the cell, the bigger the current. Unlike an ammeter a voltmeter is connected across the components Scientist usually use the term Potential Difference (pd) when they talk about voltage. measuring voltage This is how we draw a voltmeter in a circuit. V V SERIES CIRCUIT PARALLEL CIRCUIT measuring voltage V V V V series circuit • voltage is shared between the components 3V 1.5V 1.5V parallel circuit • voltage is the same in all parts of the circuit. 3V 3V 3V measuring current & voltage copy the following circuits on the next two slides. complete the missing current and voltage readings. remember the rules for current and voltage in series and parallel circuits. measuring current & voltage a) 6V 4A A V V A measuring current & voltage b) 6V 4A A V A V A answers a) b) 6V 6V 4A 4A 6V 4A 4A 3V 3V 2A 4A 6V 2A Electric Circuits In electricity, the concept of voltage will be like pressure. Water flows from high pressure to low pressure (this is consistent with our previous analogy that Voltage is like height since DP = rgh for fluids) ; electricity flows from high voltage to low voltage. But what flows in electricity? Charges! How do we measure this flow? By Current: current = I = Dq / Dt UNITS: Amp(ere) = Coulomb / second The rate at which electrons move along field lines is called drift speed, typically about 10-4 m/s Electric current defined in terms of the flow of positive charge opposite the electrons is called conventional current Current will always be in the same direction as the local electric field Voltage Sources: batteries and power supplies A battery or power supply supplies voltage. This is analogous to what a pump does in a water system. Question: Does a water pump supply water? If you bought a water pump, and then plugged it in (without any other connections), would water come out of the pump? Question: Does the battery or power supply actually supply the charges that will flow through the circuit? • Charges move from higher to lower potential • For the process to continue, charges that have moved from a higher to lower potential must be raised back to a higher potential again • A battery is able to add charges and raise the charges to higher electric potential Symbol for a battery Voltage Sources: batteries and power supplies Just like a water pump only pushes water (gives energy to the water by raising the pressure of the water), so the voltage source only pushes the charges (gives energy to the charges by raising the voltage of the charges). Just like a pump needs water coming into it in order to pump water out, so the voltage source needs charges coming into it (into the negative terminal) in order to “pump” them out (of the positive terminal). Voltage Sources: batteries and power supplies Because of the “pumping” nature of voltage sources, we need to have a complete circuit before we have a current. Circuit Elements In this first part of the course we will consider two of the common circuit elements: capacitor resistor The capacitor is an element that stores charge for use later (like a water tower). The resistor is an element that “resists” the flow of electricity. Electrical Resistance & Ohms’ Law The current established is directly proportional to the voltage difference Ohm’s Law: ΔV ∝ I In a plot of ΔV vs I, the slope is called the electrical resistance Resistance Current is somewhat like fluid flow. Recall that it took a pressure difference to make the fluid flow due to the viscosity of the fluid and the size (area and length) of the pipe. So to in electricity, it takes a voltage difference to make electric current flow due to the resistance in the circuit. Resistance By experiment we find that if we increase the voltage, we increase the current: V is proportional to I. The constant of proportionality we call the resistance, R: V = I*R Ohm’s Law UNITS: R = V/I so Ohm = Volt / Amp. The symbol for resistance is Resistance Just as with fluid flow, the amount of resistance does not depend on the voltage (pressure) or the current (volume flow). The formula V=IR relates voltage to current. If you double the voltage, you will double the current, not change the resistance. The same applied to capacitance: the capacitance did not depend on the charge and voltage - the capacitance related the two. As was the case in fluid flow and capacitance, the amount of resistance depends on the materials and shapes of the wires. Resistance The resistance depends on material and geometry (shape). For a wire, we have: R=rL/A where r is called the resistivity (in Ohm-m) and measures how hard it is for current to flow through the material, L is the length of the wire, and A is the cross-sectional area of the wire. The second lab experiment deals with Ohm’s Law and the above equation. Electrical Power The electrical potential energy of a charge is: U = q*V . Power is the change in energy with respect to time: Power = DU / Dt . Putting these two concepts together we have: Power = D(qV) / Dt = V(Dq) / Dt = I*V. Electrical Power Besides this basic equation for power: P = I*V remember we also have Ohm’s Law: V = I*R . Thus we can write the following equations for power: P = I2*R = V2/R = I*V . To see which one gives the most insight, we need to understand what is being held constant. Example When using batteries, the battery keeps the voltage constant. Each D cell battery supplies 1.5 volts, so four D cell batteries in series (one after the other) will supply a constant 6 volts. When used with four D cell batteries, a light bulb is designed to use 5 Watts of power. What is the resistance of the light bulb? Example We know V = 6 volts, and P = 5 Watts; we’re looking for R. We have two equations: P = I*V and V = I*R which together have 4 quantities: P, I, V & R.. We know two of these (P & V), so we should be able to solve for the other two. Example Using the power equation we can solve for I: P = I*V, so 5 Watts = I * (6 volts), or I = 5 Watts / 6 volts = 0.833 amps. Now we can use Ohm’s Law to solve for R: V = I*R, so R = V/I = 6 volts / 0.833 amps = 7.2 W . Example extended If we wanted a higher power light bulb, should we have a bigger resistance or a smaller resistance for the light bulb? We have two relations for power that involve resistance: P=I*V; V=I*R; eliminating V gives: P = I2*R and P=I*V; I=V/R; eliminating I gives: P = V2 / R . In the first case, Power goes up as R goes up; in the second case, Power goes down as R goes up. Which one do we use to answer the above question? Example extended Answer: In this case, the voltage is being held constant due to the nature of the batteries. This means that the current will change as we change the resistance. Thus, the P = V2 / R would be the most straight-forward equation to use. This means that as R goes down, P goes up. (If we had used the P = I2*R formula, as R goes up, I would decrease – so it would not be clear what happened to power.) The answer: for more power, lower the resistance. This will allow more current to flow at the same voltage, and hence allow more power! Kirchhoff’s Laws • Junction Law: at a junction in a circuit, the sum of the current entering the junction will equal the sum of the current leaving. ΣI= ΣI in out • Loop Law: the sum of the potential drops around any closed loop must add to ΣV= 0 loop Connecting Resistors There are two basic ways of connecting two resistors: series and parallel. In series, we connect resistors together like railroad cars; this is just like we have for capacitors: + - + - high V low V R1 R2 Formula for Series: To see how resistors combine to give an effective resistance when in series, we can look either at R1 I V = I*R, or at + V1 V R2 Vbat 2 R = rL/A . - Formula for Series Using V = I*R, we see that in series the current must move through both resistors. (Think of water flowing down two water falls in series.) Thus Itotal = I1 = I2 . Also, the voltage drop across the two resistors add to give the total voltage drop: (The total height that the water fell is the addition of the two heights of the falls.) Vtotal = (V1 + V2). Thus, Reff = Vtotal / Itotal = (V1 + V2)/Itotal = V1/I1 + V2/I2 = R1 + R2. Formula for Series Using R = rL/A , we see that we have to go over both lengths, so the lengths should add. The distances are in the numerator, and so the values should add. This is just like in R = V/I (from V = IR) where the V’s add and are in the numerator! Note: this is the opposite of capacitors when connected in series! Recall that C = Q/V, where V is in the denominator! Formula for Parallel Resistors The result for the effective resistance for a parallel connection is different, but we can start from the same two places: (Think of water in a river that splits with some water flowing over one fall and the rest falling over the other but all the water ending up joining back together again.) V=I*R, or R = rL/A . Itotal + Vbat I1 R1 I2 R2 - Formula for Parallel Resistors V=I*R, or R = rL/A For parallel, both resistors are across the same voltage, so Vtotal = V1 = V2 . The current can go through either resistor, so: Itotal = (I1 + I2 ) . Since the I’s are in the denominator, we have: R = Vtotal/Itotal = Vtotal/(I1+I2); or 1/Reff = (I1+I2)/Vtotal = I1/V1 + I2/V2 = 1/R1 + 1/R2. Formula for Parallel Resistors If we start from R = rL/A , we can see that parallel resistors are equivalent to one resistor with more Area. But A is in the denominator (just like I was in the previous slide), so we need to add the inverses: 1/Reff = 1/R1 + 1/R2 . Terminal Voltage Terminal voltage, VT , is the potential difference between the terminals of a battery. Ideal voltage, VB , is determined by the chemistry of the battery. Internal resistance, ir : some charge will be los due to the random thermal motion of the battery Terminal voltage will be: VT = VB - ir For recharging a battery: VT = VB + ir Current Division When current enters a junction, Kirchhoff’s first law tells you the sum of the current entering must equal the sum of the current leaving. Example: 8 = I + 4I + 5I I = 0.8 A 1/RP = 1/20 +1/5 +1/4 = ½ RP = 2Ω V = IRP = 8 • 2 = 16V 16 = I1(20) I1 = 0.8A 16 = I2(5) I2 = 3.2A 16 = I3(4) I3 = 4A Simple Circuits • A simple circuit is a connection of batteries and resistors that meets 2 criteria 1. All batteries are in series 2. The equivalent resistance of the entire circuit can be obtained by repeated use of just the series and parallel equivalent resistance formulas VB = 60 – 18 = 42V 1/RP = 1/12 + 1/6 = ¼ RP = 4Ω Requiv = 4 + 8 + 6 + 3 = 21Ω 42 – I(21) = 0 I = 2A VT60 = 60 – 2 • 1 = 58V VT18 = 18 – 2 • 2 = 22V 2 = I + 2I I = 0.67A Capacitance We define capacitance as the amount of charge stored per volt: C = Qstored / DV. UNITS: Farad = Coulomb / Volt Just as the capacity of a water tower depends on the size and shape, so the capacitance of a capacitor depends on its size and shape. Just as a big water tower can contain more water per foot (or per unit pressure), so a big capacitor can store more charge per volt. Parallel Plate Capacitor For a parallel plate capacitor, we can pull charge from one plate (leaving a -Q on that plate) and deposit it on the other plate (leaving a +Q on that plate). Because of the charge separation, we have a voltage difference between the plates, DV. The harder we pull (the more voltage across the two plates), the more charge we pull: C = Q /DV. Note that C is NOT CHANGED by either Q or DV; C relates Q and DV! Capacitance • What happens when a water tower is over- filled? It can break due to the pressure of the water pushing on the walls. • What happens when an electric capacitor is “over-filled” or equivalently a higher voltage is placed across the capacitor than the listed maximum voltage? It will “break” by having the charge “escape”. This escaping charge is like lightning - a spark that usually destroys the capacitor. V or DV ? When we deal with height, h, we usually refer to the change in height, Dh, between the base and the top. Sometimes we do refer to the height as measured from some reference point. It is usually clear from the context whether h refers to an actual h or a Dh. With voltage, the same thing applies. We often just use V to really mean DV. You should be able to determine whether we really mean V or DV when we say V. Parallel Plate Capacitor For this parallel plate capacitor, the capacitance is related to charge and voltage (C = Q/V), but the actual capacitance depends on the size and shape: C plate ∝ A / d A is the area of each plate, d is the distance between the plates Energy Storage If a capacitor stores charge and carries voltage, it also stores the energy it took to separate the charge. The formula for this is: Estored = (1/2)QV = (1/2)CV2 , where in the second equation we have used the relation: C = Q/V . Energy Storage Note that previously we had: U = q*V , and now for a capacitor we have: Ucap = ½ QV = ½ CV2 = ½ Q2/C Energy Storage The reason is that in charging a capacitor, the first bit of charge is transferred while there is very little voltage on the capacitor (recall that the charge separation creates the voltage!). Only the last bit of charge is moved across the full voltage. Thus, on average, the full charge moves across only half the voltage! Hooking Capacitors Together Instead of making and storing all sizes of capacitors, we can make and store just certain values of capacitors. When we need a non-standard size capacitor, we can make it by hooking two or more standard size capacitors together to make an effective capacitor of the value we need. Two basic ways There are two basic ways of connecting two capacitors: series and parallel. In series, we connect capacitors together like railroad cars; using parallel plate capacitors it would look like this: + - + - high V low V C1 C2 Series If we include a battery as the voltage source, the series circuit would look like this: C1 + Vbat C2 Note that there is only one way around the circuit, and you have to jump BOTH capacitors in making the circuit - no choice! Parallel In a parallel hook-up, there is a branch point that allows you to complete the circuit by jumping over either one capacitor or the other: you have a choice! High V C1 Low V C2 Parallel Circuit If we include a battery, the parallel circuit would look like this: + + + Vbat C1 C2 Formula for Series: To see how capacitors combine to give an effective capacitance when in series, we can look either at C = Q/V, or at C plate = A /d Formula for Series Using C = Q/V, we see that in series the charge moved from capacitor 2’s negative plate must be moved through the battery to capacitor 1’s positive plate. C1 + +Q Vbat C2 - -Q ( +Qtotal) Formula for Series But the positive charge on the left plate of C1 will attract a negative charge on the right plate, and the negative charge on the bottom plate of C2 will attract a positive charge on the top plate - just what is needed to give the negative charge on the right plate of C1. Thus Qtotal = Q1 = Q2 . C1 (+Q1 ) + +Q1 -Q 1 +Q2 Vbat C2 - -Q2 ( +Qtotal) Formula for Series Also, the voltage drop across the two capacitors add to give the total voltage drop: Vtotal = (V1 + V2). Thus, Ceff = Qtotal / Vtotal = Qtotal / (V1 + V2), or (with Qtotal = Q1 = Q2) [1/Ceff] = (V1 + V2) / Qtotal = V1/Q1 + V2/Q2 = 1/C1 + 1/C2 = 1/Ceffective Parallel Circuit For parallel, both plates are across the same voltage, so Vtotal = V1 = V2 . The charge can accumulate on either plate, so: Qtotal = (Q1 + Q2). Since the Q’s are in the numerator, we have: Ceff = C1 + C2. + +Q1 +Q2 Vbat C1 -Q1 C2 -Q2 +Q1 +Qtotal = (Q1+Q2) +Q2 Review of Formulas Electric Current I= ΔQ/Δt Resistor Voltage Drop V = IR Resistivity R = rL/A Electric Power P = IV Resistance Power P = I2R = V2/R Junction Law SI = SI in out Loop Law SΔV = 0 loop Series Resistors Rs = R1 + R2 +… Parallel Resistors 1/RP = 1/R1 + 1/R2 +… Terminal Voltage VT = VB ± IR Capacitance C = Q/V Parallel Plate Capacitor C plate ∝ A / d Capacitors in Series 1/Cs = 1/C1 + 1/C2 +… Capacitors in Parallel CP = C1 + C2 +… Energy in Capacitor U = ½ Q2 /C = ½ CV2 = ½ QV

DOCUMENT INFO

Shared By:

Categories:

Tags:

Stats:

views: | 11 |

posted: | 8/19/2012 |

language: | Unknown |

pages: | 74 |

OTHER DOCS BY yurtgc548

Docstoc is the premier online destination to start and grow small businesses. It hosts the best quality and widest selection of professional documents (over 20 million) and resources including expert videos, articles and productivity tools to make every small business better.

Search or Browse for any specific document or resource you need for your business. Or explore our curated resources for Starting a Business, Growing a Business or for Professional Development.

Feel free to Contact Us with any questions you might have.