SAT II Physics (Gary Graff) Episode 1 Part 8 pps

Peterson’s SAT II Success: Physics 170 SERIES AND SERIES PARALLEL CIRCUITS A source of electrical potential sometimes has a single task, such as to operate a flashlight bulb. But just as often the power source is used to perform many tasks through several different pathways. The first example we will discuss is the single task or series circuit; and the second type is called a parallel circuit. SERIES CIRCUITS Series circuits have only one pathway in which current can flow. Every unit of charge flowing in a series circuit must pass through every position of the circuit. When more than one resistor is in a series circuit, the value of each resistor is added together to give the total resistance. RRRRR t =+++ 123 … The total resistance for the circuit shown iss 12Ω ΩΩΩ Ω . R R t t =++ = 246 12 CHAPTER 5 Peterson’s: www.petersons.com 171 PARALLEL CIRCUITS Parallel circuits have two or more pathways through which current can flow. The total value of the current depends on the parallel resistance offered against the current flow. Since current and resis- tance are indirectly related, the highest resistance will allow less current to flow, and the lowest resistance will allow more current to flow. The total resistance of a set of parallel resistors is found in the following manner: 1111 12 3 RRRR t =++ 11 2 1 5 1 10 1 40 1 5 2 1 025 1 825 121 R R R R t t t t =++ + =+++ = = ΩΩ Ω Ω ΩΩΩ Ω Ω . . The value of the final resistance shows another characteristic of parallel resistances. The total resistance of a set of parallel resistors is always equal to or less than the value of the smallest resistor in the parallel set. When an extremely large resistor (500 meg-ohms) is ELECTRIC CIRCUITS Peterson’s SAT II Success: Physics 172 in parallel with an extremely small resistor, the total resistance is equal to the smallest resistor. When a set of parallel resistors are all of equal resistance, we can divide the resistance value of one of them by the total number of resistors in parallel to find the resistance of the parallel system. Example Let’s do a problem where elements of both the series circuit and the parallel circuit are combined. 1.1. 1.1. 1. Find the total resistance for the circuit.Find the total resistance for the circuit. Find the total resistance for the circuit.Find the total resistance for the circuit. Find the total resistance for the circuit. The first step to solve the problem is to find the equivalent resistance for the R 2 – R 3 – R 4 parallel set of resistors: 1111 11 10 1 8 1 4 1 25 4 12 3 RRRR R R R t t t t =++ =++ = = ΩΩΩ Ω Ω . CHAPTER 5 Peterson’s: www.petersons.com 173 Next we find the equivalent resistance for the R 6 – R 7 pair. 111 11 10 1 15 1 1667 6 67 RRR R R R t t t t =+ =+ = = ΩΩ Ω . Finally, we add the two parallel equivalent values to the values of R 1 , R 5, and R 8, yielding: R R t t =++++ = 24369 24 ΩΩΩΩΩ Ω 2.2. 2.2. 2. Use Ohm’s La Use Ohm’s La Use Ohm’s La Use Ohm’s La Use Ohm’s La w to fw to f w to fw to f w to f ind the total curind the total cur ind the total curind the total cur ind the total cur rr rr r ent in the cirent in the cir ent in the cirent in the cir ent in the cir cuit.cuit. cuit.cuit. cuit. VIR I V R I V IA = = = = 24 24 1 Ω 3. Find the voltage change between points B and C.3. Find the voltage change between points B and C. 3. Find the voltage change between points B and C.3. Find the voltage change between points B and C. 3. Find the voltage change between points B and C. This may seem difficult at first, but remember Ohm’s Law. The current passing through R 5 is 1A, and the resistance is 3Ω. V = I R V = (1A)(3Ω) V = 3 V ELECTRIC CIRCUITS Peterson’s SAT II Success: Physics 174 4.4. 4.4. 4. F F F F F ind the vind the v ind the vind the v ind the v oltaolta oltaolta olta gg gg g e ce c e ce c e c hanghang hanghang hang e betwe betw e betwe betw e betw een points een points een points een points een points A and B.A and B. A and B.A and B. A and B. Again, the problem seems more difficult than it really is—it’s another Ohm’s Law problem. The total current passing through the parallel resistors is 1A, and the equivalent resistance is 4Ω. VIR VA VV = = = ()( )14 4 Ω POWER Power in circuits is the rate at which electric energy is used. The power capability of the circuit elements is the product of the voltage and the current. P = VI Through substitution from Ohm’s Law, we see two other ways to calculate power: PIR P V R ==()() 2 2 and . Power is an important quantity in circuits. The voltage source must have enough power to operate the devices in the circuit. Furthermore, the devices in the circuitry will burn out and open if their power capacity is not large enough to perform work at the required rate. Remember, power is the rate at which work is done. The power requirement for a circuit or a circuit element can be calculated if any two of the Ohm’s Law quantities are known. Example A typical power calculation could require you to find the power requirement for a resistor in a circuit. Find the power dissipated by a 4Ω resistor that has a .05A current passing through it. CHAPTER 5 Peterson’s: www.petersons.com 175 Solution PIR PA P = = = 2 05 4 01 (. )( ) . Ω Watt Another type of power question you could be asked would require you to find the total resistance in a circuit, then solve to find either the total power requirement of the circuit or the power requirement for a single element in the circuit. Example Find the power requirement for the series–parallel circuit shown below: ELECTRIC CIRCUITS Peterson’s SAT II Success: Physics 176 Solution To find the total power for the circuit you first must find the total resistance. We find that the total resistance for the circuit is 9Ω. With this information we can now solve for the power required to run the circuit with P V R = 2 . P V R P V P = = = 2 2 6 9 4 () Ω Watts Another question might involve one of the parallel resistors in the diagram above. The current in the circuit would be required to solve the problem. Example Find the power capacity for R 3 in the circuit above. Solution First we find the total current with Ohm’s law. I V R I V IA = = = 6 9 66 Ω . The next step is to calculate the voltage change across the two parallel resistors (R 2 and R 3 ). VIR VA VV = = = ()( ) (. )( ) . 66 5 33 Ω CHAPTER 5 Peterson’s: www.petersons.com 177 With this information the actual current in R 3 can be found by using Ohm’s Law. I V R I V IA = = = 33 30 11 . . Ω Next we calculate the power. P = (V)(I) P = (2.5V)(.11A) P = .275 Watts Note: V R 2 across the parallel resistors does not work because each resistor has a different current through it. CAPACITORS Capacitors (previously mentioned) are parallel plate devices that are capable of storing electric charge and releasing (discharging) it as needed. Their effectiveness is enhanced by placing a non-conductive material, called a dielectric, between the two plates. The energy (in the form of the charge) stored by a capacitor is ECV= 1 2 2 . Capacitors placed in an electric circuit do not allow a steady current to flow. Instead, they build charge between their two plates over a period of time until the potential of the capacitor is almost equal to the applied voltage. When a capacitor is fully charged in a DC circuit, current cannot flow until the capacitor is discharged, because the capacitor acts like an open circuit element. Capacitors used in an electric circuit can be in series or in parallel. The total capacitance (C t ) for capacitors in a parallel circuit is found by adding the values of the capacitors in parallel with one another. You should note that this technique is the opposite of the technique we used to find the value of the resistance of parallel resistors. ELECTRIC CIRCUITS Peterson’s SAT II Success: Physics 178 The six capacitors in the diagram have values of 2, 3, 4, 5, and 6 micro farads, respectively. The total capacitance is found by adding them together. CCCCCC Cfffff Cf t t t =++++ =++++ = 12345 23456 20 µµµµµ µ The total capacitance for capacitors in series with one another is found by the reciprocal method. (Again this is just the opposite of resistors in series). 111111 11 2 1 3 1 4 1 5 1 6 1 5 12345 CCCCCC Cfffff C t t t =++++ =++++ = µµµµµ µ. fffff f Cf t ++++ = . 33 25 2 1667 145 µµµ µ µ CHAPTER 5 Peterson’s: www.petersons.com 179 AMMETERS AND VOLTMETERS Ammeters and voltmeters are used to monitor electric circuits. AmmeterAmmeter AmmeterAmmeter Ammeter s s s s s measure the current flowing in a circuit. They have a low resistance, which keeps them from interfering with the circuit. This is necessary because they are hooked in a series with a circuit being measured, and a large resistance would decrease the current being measured with the meter. VV VV V oltmeteroltmeter oltmeteroltmeter oltmeter s s s s s are used to measure voltage changes in a circuit. They have a high resistance and are used in parallel with the circuit being measured. The high resistance in the voltmeter causes all but the tiniest fraction of current to pass through the circuit being measured, which prevents the voltmeter from interfering with the circuit. MAGNETS AND MAGNETIC FIELDS The properties of naturally occurring magnetic rocks have been known for several thousand years. The Chinese knew that a piece of iron could be magnetized by putting it near lodestone, and sailors have been navigating with magnetic compasses for nearly a thousand years. Some characteristics of magnets that you should know and remember are: 1. Magnets have poles. The north–seeking pole is the north pole of the magnet. The south–seeking pole is the south pole of the magnet. 2. Like poles of magnets repel one another and unlike poles of magnets attract one another. 3. Magnets can induce demagnetized ferrous materials to become magnetized. 4. Temporary magnets cease acting like magnets as soon as the permanent magnet is removed. 5. Permanent magnets retain their magnetism for a long time. MAGNETS AND MAGNETIC FIELDS [...]... magnet) 3 Your palm faces in the direction the force is applied to the particle This is the direction the particle will move The charged particle has a charge q on it, and its velocity is v The definition of current is a number of moving charges Thus, we can say F = B⊥ IL for a current-carrying wire 18 4 Peterson’s SAT II Success: Physics MAGNETS AND MAGNETIC FIELDS The current in a wire is made of a... single charged particle, the charge on the particle and the velocity with which it is moving can be equated to the current and the length of the wire This leads to the following equation, which describes the force exerted on a moving particle in a magnetic field F = B⊥ qv Unless the particle leaves the magnetic field, the direction of the force on the particle keeps changing as the path of the particle changes... emanates from the wire The magnetic field from the primary side coils cuts through the coils of the secondary side, inducing a current into the secondary side of the transformer (see below) 18 8 Peterson’s SAT II Success: Physics MAGNETS AND MAGNETIC FIELDS Transformers that reduce the applied voltage are called stepdown transformers Those that increase the applied voltage are called step-up transformers... of the secondary side of the transformer? Peterson’s: www.petersons.com 18 9 CHAPTER 5 Solution Vs N s = Vp N p Vs = Vs = (Vp )(N s ) Np (80 00V )(900 turns) 30, 000 turns Vs = 240V 0 19 0 Peterson’s SAT II Success: Physics CHAPTER SUMMARY CHAPTER SUMMARY • Charged particles are called ions • Electrostatic charge is caused by the transfer of electrons from one object to another object • Objects with a... magnetic field (B) can be determined Peterson’s: www.petersons.com 18 1 CHAPTER 5 Example A 5m length of wire carries a 6A current within a magnetic field The wire is at right angles to the field and experiences a force of 45N What is the strength of the magnetic field? Solution F = B⊥ IL F B⊥ = ( I )( L ) 45N B⊥ = (6A)(.5m) N = 15 tesla (T ) B⊥ = 15 A• M A current-carrying wire is a source of a magnetic field... We also know that the length of the wire (L), the strength of the magnetic field (B), and the rate at which the wire is moved in the magnetic field (V) affect the induced current as well 18 6 Peterson’s SAT II Success: Physics MAGNETS AND MAGNETIC FIELDS The following represents the relationship between the induced voltage and the moving magnetic field in which a wire is placed (The wire may be moved... of a magnetic field too This is the reason the wire has a force exerted on it when it is in a magnetic field Let’s place a compass beside a wire hooked to a battery with an open switch 18 2 Peterson’s SAT II Success: Physics MAGNETS AND MAGNETIC FIELDS Notice that with the switch open the compass points correctly to the north When the switch is closed the compass points in a different direction A force... similar magnetic pole is a manifestation of the lines of force not crossing one another The attraction between unlike poles also can be explained in a manner similar to the electric field 18 0 Peterson’s SAT II Success: Physics MAGNETS AND MAGNETIC FIELDS Some magnets exert large repulsive or attractive forces on other magnets regardless of how physically close the two magnets are to one another The way... a magnetic field • Transformers change high voltage to low voltage by stepping down the voltage • Transformers change low voltage to high voltage by stepping up the voltage 19 2 Peterson’s SAT II Success: Physics Chapter 6 MODERN PHYSICS ... within a magnetic field The right-hand rule can be applied to charged particles within a magnetic field Unlike the current in a wire that followed its pathway in the wire, a charged particle can have its pathway changed by the force applied to it Reviewing the right-hand rule: 1 Point the thumb of your right hand in the direction the particle is moving 2 Point your fingers in the direction of the field . resistors: 11 11 11 10 1 8 1 4 1 25 4 12 3 RRRR R R R t t t t =++ =++ = = ΩΩΩ Ω Ω . CHAPTER 5 Peterson’s: www.petersons.com 17 3 Next we find the equivalent resistance for the R 6 – R 7 pair. 11 1 11 10 1 15 1 1667 6 67 RRR R R R t t t t =+ =+ = = ΩΩ Ω . Finally,. of a set of parallel resistors is found in the following manner: 11 11 12 3 RRRR t =++ 11 2 1 5 1 10 1 40 1 5 2 1 025 1 82 5 12 1 R R R R t t t t =++ + =+++ = = ΩΩ Ω Ω ΩΩΩ Ω Ω . . The value of. them together. CCCCCC Cfffff Cf t t t =++++ =++++ = 12 345 23456 20 µµµµµ µ The total capacitance for capacitors in series with one another is found by the reciprocal method. (Again this is just the opposite of resistors in series). 11 111 1 11 2 1 3 1 4 1 5 1 6 1 5 12 345 CCCCCC Cfffff C t t t =++++ =++++ = µµµµµ µ.