Electric Circuits
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Electric Circuits

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Electric Circuits Electric Circuits Presentation Transcript

  • +Electric CircuitsChapter 18
  • +18.1 Schematic Diagrams andCircuits
  • + What do you think? • Scientists often use symbols to represent electrical components, such as batteries, bulbs, and wires. On the next slide, you will see the symbols for eight common electrical components that you have seen and discussed previously. • Predict the component shown by looking at each symbol. • Briefly explain why you think each symbol represents that particular electrical component.
  • + What do you think? 1 5 2 6 3 7 4 8
  • + Schematic Diagrams  Schematic diagrams use symbols to represent components.  They show how the parts in an electrical device are arranged.
  • + Electric Circuits  A path through which charges flow  Can have one or more complete paths  Load  An element or group of elements in a circuit that dissipates energy  Ex: A simple circuit consists of a source of potential differences and electrical energy, such as a battery, and a load, such as a bulb or a group of bulbs
  • + Electric Circuits  An electric circuit is a set of components providing a complete, closed-loop path for the movement of electrons.  Called a closed circuit  If the path is broken, the electrons do not flow.  Called an open circuit
  • + Inside a Light bulb  A complete conducting path is established inside the light bulb.  The tip of the bulb (a) is connected to one side of the filament (see the black line).  The threads on the side of the bulb (c) are connected to the other side of the filament (see the white line).
  • + Short Circuits  A shortcircuit bypasses the light bulb or other load.  Itis a closed circuit.  Electrons flow directly from - to + without passing through the bulb.  The current is large and the wire becomes hot.  Short circuits in homes can cause fires.  Fuses or circuit breakers are designed to turn off the electron flow if short circuits occur.
  • + Potential Difference in Circuits A device that increases the PE of the electrons, such as a battery, is a source of emf (electromotive force).  Not really a force, but a PE difference Energy is conserved in electric circuits.  The potential difference ( V) for the battery equals the energy converted into heat as the electrons move through the bulb.  Electrons gain energy (battery) and lose energy (bulb) as they make a complete trip.
  • + Now what do you think? • Draw schematic diagrams showing each of the following circuits: • An open circuit including a battery, open switch, and bulb • A closed circuit including a battery, closed switch, and resistor • A short circuit including a battery, bulb, and closed switch
  • +18.2 Resistors in Series or Parallel
  • + What do you think? • Figure (a) shows a single bulb and battery as seen before. Figures (b) and (c) each show two bulbs connected to the battery. The batteries and bulbs are all identical. Answer the three questions on the next slide and explain your reasoning.
  • + What do you think?• How will the brightness of (b) and (c) compare to each other and how does each compare to (a)? Explain.• How will the brightness of (d) and (e) compare to each other and how does each compare to (a)? Explain.• Compare the total current leaving the battery in each of the three circuits. Explain.
  • + Resistors in Series  Seriesdescribes components of a circuit that provide a single path for the current.  Thesame electrons must pass through both light bulbs so the current in each is the same.
  • + Resistors in Series  Light bulb filaments are resistors  When many resistors are connected in series, the current in each resistor are the same
  • + Resistors in Series  Vbattery= V1 + V2  Conservation of energy  Vbattery= IR1 + IR2  Ohm’s law  Vbattery= I(R1 + R2)  Vbattery= IRequivalent  Requivalent = R1 + R2
  • + Equivalent Resistance  Solving problems with series resistors:  Find the equivalent resistance.  Use Req with Ohm’s law to find V or I.  Use I and R1, R2, etc. to find V1, V2, etc.
  • + Equivalent Resistance  The potential difference across the batter, V, must equal the potential difference across the load.
  • + Classroom Practice Problems  A 6.00V lantern battery is connected to each of the following bulb combinations. Find the equivalent resistance and current in each circuit.  One bulb with a resistance of 7.50  Two bulbs in series, each with a resistance of 7.50  Four bulbs in series, each with a resistance of 7.50
  • + Classroom Practice Problems 1. Start by drawing a picture 2. Take your inventory 3. Decide which equations to use 4. Solve  Answers:  7.5 , 0.800 A  15 , 0.400 A  30 , 0.200 A
  • + Resistors in Parallel Parallel describes components providing separate conducting paths with common connecting points.  The potential difference is the same for parallel components.  Electrons lose the same amount of energy with either path.
  • + Resistors in Parallel Ibattery = I1 + I2  Conservation of charge Vbattery V1 V2 Req R1 R2  Ohm’s law V V V VReq VR1 VR2 Vbattery= V1 = V2  Potential energy loss is the same across all parallel resistors. Because Vbattery= V1 = V2, the equation above reduces as follows: 1 1 1 Req R1 R2
  • + Equivalent Resistance  Solving problems with parallel resistors:  Find the equivalent resistance.  Use Req with Ohm’s law to find V or Itotal.  Use V to find I1, I2, etc.
  • + Equivalent Resistance  Thesum of currents in parallel resistors equals the total current  The Req for a parallel arrangement of resistors must always be the smallest resistance in the group of resistors
  • + Classroom Practice Problems  A 9.0V battery is connected to 4 resistors as shown 2 below. Find the 4 equivalent 5 resistance for the 7 circuit and the total current in the circuit. 9V
  • + Classroom Practice Problems Given: V= 9V R 1= 2 R 2= 4 R 3= 5 R 4= 7 Req= ?? I= ?? Equations: 1 1 1 1 V=IReq .... Req R1 R2 R3
  • + Classroom Practice Problems 1 1 1 1 1 Req 2 4 5 7 0.5 0.25 0.2 0.14 1.09 1 1 1 1 1 1 1 0.915 Req 1.09 V 9V I 9.84 A Req 0.915
  • + Classroom Practice Problems  Findthe equivalent resistance, the total current drawn by the circuit, and the current in each resistor for a 9.00 V battery connected to:  One 30.0 resistor  Three 30.0 resistors connected in parallel  Answers:  30.0 , 0.300 A, 0.300 A  10.0 , 0.900 A, 0.300 A
  • + Summary
  • + Wiring Lights  The series circuit shows a bulb burned out.  What will happen to the other bulbs?  Would this also happen in the parallel circuit?  Assuming the bulbs are identical:  Which circuit will draw more current?  In which circuit are the bulbs brighter?
  • + Now what do you think?• How will the brightness of (b) and (c) compare to each other and how does each compare to (a)? Explain.• How will the brightness of (d) and (e) compare to each other and how does each compare to (a)? Explain.• Compare the total current leaving the battery in each of the three circuits. Explain.
  • +18.3 Complex ResistorCombinations
  • + What do you think? • Household circuits typically have many outlets and permanent fixtures such as hanging light fixtures on each circuit. • Are these wired in series or in parallel? • Why do you believe one of these methods has an advantage over the other method? • What disadvantages would the other method of wiring have for household circuits?
  • + Complex Resistor Calculations  Forcomplex resistors, you need to follow a few steps to be successful: 1. Combine series in parallel 2. Combine parallel sets 3. Combine series set 4. Finish problem from there
  • + Complex Resistor Calculations 1. 6+2= 8 2. 1 1 0.12 0.25 0.37 8 4 1  Tofind the equivalent 3. 2.70 0.37 resistance for the circuit shown above, follow the steps shown to the 4. 3 + 9 + 2.70 + 1= 12.7 right:
  • + Complex Resistor Calculations  Req for 6.0 and 2.0  Answer: 8.0  Req for 8.0 and 4.0  Answer: 2.7  Tofind the equivalent  Req for 3.0 and 6.0 and resistance for the circuit 2.7 and 1.0 shown above, follow the  Answer: 12.7 steps shown to the right:  So,the resistance of all 6 resistors is equivalent to a single 12.7 resistor.
  • + Complex Resistor Calculations  Findthe total current in the equivalent circuit.  Answer: 0.71 A  This is the current through the 1.0 , 6.0 (on the left), and  Forthe 2.0 resistor, find 3.0 loads the current and the  Findthe total potential potential difference. drop across the parallel  To solve this problem, use the combination of three step-by-step approach resistors. shown.  Answer: 1.9 V  Continued on the next slide
  • + Complex Resistor Calculations  Findthe current through the combined 6.0 and 2.0 resistor.  Answer: 0.24 A  Find the potential difference across the 2.0 resistor.  Answer: 0.48 V
  • + Classroom Practice Problems  For the circuit shown, find the:  Equivalent resistance  Current through the 3.0 resistor  Potential difference across the 6.0 resistor  Answers:  6.6 , 1.8 A, 6.5 V
  • + Now what do you think? • Household circuits typically have many outlets and permanent fixtures such as hanging light fixtures on each circuit. • Are these wired in series or in parallel? • Why do you believe one of these methods has an advantage over the other method? • What disadvantages would the other method of wiring have for household circuits?