Document Sample

PHY-2049 Current & Circuits February ‘08 A closed circuit Power in DC Circuit In time t, a charge Q is pushed through the resistor by the battery. The amount of work done by the battery is : W VQ Power : W Q V VI t t Power P IV I IR I 2 R E2 P I 2 R IV R The figure below gives the electrical potential V(x) along a copper wire carrying a uniform current, from a point at higher potential (x=0m) to a point at a lower potential (x=3m). The wire has a radius of 2.45 mm. What is the current in the wire? What does the graph tell us?? *The length of the wire is 3 meters. *The potential difference across the wire is 12 m volts. *The wire is uniform. Let’s get rid of the mm radius and convert it to area in square meters: A=pr2 = 3.14159 x 2.452 x 10-6 m2 or copper A=1.9 x 10-5 m 2 12 volts 0 volts Material is Copper so resistivity is (from table) = 1.69 x 10-8 ohm meters We have all we need…. 8 L 1.69 x10 ohm - m 3.0 m R 5 2.67m A 1.9 x10 From Ohm's Law : 6 V 12 10 volts i 3 4.49 ma R 2.67 10 ohms Let’s add resistors ……. SERIES Resistors i i Series Combinations R1 R2 V1 V2 V V1 iR1 V2 iR2 and V V1 V2 iR iR1 iR2 R R1 R2 general : R( series ) Ri i The rod in the figure is made of two materials. The figure is not drawn to scale. Each conductor has a square cross section 3.00 mm on a side. The first material has a resistivity of 4.00 × 10–3 Ω · m and is 25.0 cm long, while the second material has a resistivity of 6.00 × 10–3 Ω · m and is 40.0 cm long. What is the resistance between the ends of the rod? Parallel Combination?? R1, I1 V iR V V V i i1 i2 R1 R2 R R2, I2 so.. 1 1 1 R1 R2 R general 1 1 V R i Ri In Fig. 28-39, find the equivalent resistance What’s This??? between points (a) F and H and [2.5] (b) F and G. [3.13] (a) Find the equivalent resistance between points a and b in Figure P28.6. (b) A potential difference of 34.0 V is applied between points a and b. Calculate the current in each resistor. Power Source in a Circuit The ideal battery does work on charges moving them (inside) from a lower potential to one that is V higher. A REAL Power Source is NOT an ideal battery Internal Resistance V E or Emf is an idealized device that does an amount of work E to move a unit charge from one side to another. By the way …. this is called a circuit! A Physical (Real) Battery Emf i rR Back to which is brighter? Back to Potential Change in potential as one circuits this complete circuit is ZERO! Represents a charge in space Consider a “circuit”. This trip around the circuit is the same as a path through space. THE CHANGE IN POTENTIAL FROM “a” AROUND THE CIRCUIT AND BACK TO “a” is ZERO!! To remember In a real circuit, we can neglect the resistance of the wires compared to the resistors. We can therefore consider a wire in a circuit to be an equipotential – the change in potential over its length is slight compared to that in a resistor A resistor allows current to flow from a high potential to a lower potential. The energy needed to do this is supplied by the battery. W qV NEW LAWS PASSED BY THIS SESSION OF THE FLORIDUH LEGISLATURE. LOOP EQUATION The sum of the voltage drops (or rises) as one completely travels through a circuit loop is zero. Sometimes known as Kirchoff’s loop equation. NODE EQUATION The sum of the currents entering (or leaving) a node in a circuit is ZERO TWO resistors again i R1 R2 V1 V2 V V iR iR1 iR2 or R R1 R2 General for SERIES Resistors R Rj j A single “real” resistor can be modeled as follows: R a b V position ADD ENOUGH RESISTORS, MAKING THEM SMALLER AND YOU MODEL A CONTINUOUS VOLTAGE DROP. We start at a point in the circuit and travel around until we get back to where we started. If the potential rises … well it is a rise. If it falls it is a fall OR a negative rise. We can traverse the circuit adding each rise or drop in potential. The sum of all the rises around the loop is zero. A drop is a negative rise. The sum of all the drops around a circuit is zero. A rise is a negative drop. Your choice … rises or drops. But you must remain consistent. Take a trip around this circuit. Consider voltage DROPS: -E +ir +iR = 0 or E=ir + iR Circuit Reduction i=E/Req Multiple Batteries Reduction Computes i Another Reduction Example 1 1 1 50 1 R 20 30 600 12 R 12 PARALLEL START by assuming a DIRECTION for each Current Let’s write the equations. The Unthinkable …. RC Circuit Initially, no current through the circuit Close switch at (a) and current begins to flow until the capacitor is fully charged. If capacitor is charged and switch is switched to (b) discharge will follow. Close the Switch I need to use E for E Note RC = (Volts/Amp)(Coul/Volt) = Coul/(Coul/sec) = (1/sec) Really Close the Switch Loop Equation q E iR 0 C dq since i dt I need to use E for E dq q R E Note RC = (Volts/Amp)(Coul/Volt) dt C = Coul/(Coul/sec) = (1/sec) or dq q E dt RC R differential This is a equation. To solve we need what is called a particular solution as well as a general solution. We often do this by creative “guessing” and then matching the guess to reality. You may or may not have studied this topic … but you WILL! General Solution q q p Ke at Look at particular solution : dq q E dt RC R When the device is fully charged,dq/dt 0 and q p CE When t 0, q 0 and from solution 0 CE K K -CE dq q E and q CE(1 - e -at ) dt RC R CE (ae at ) CE(1 - e -at ) E / R for t 0 CEa 0 E/R E 1 a RCE RC Time Constant RC Result q=CE(1-e-t/RC) q=CE(1-e-t/RC) and i=(CE/RC) e-t/RC E t / RC i e R Discharging a Capacitor qinitial=CE BIG SURPRISE! (Q=CV) i iR+q/C=0 dq q R 0 dt C solution q q0 e t / RC dq q0 t / RC i e dt RC In Fig. (a), a R = 21, Ohm a resistor is connected to a battery. Figure (b) shows the increase of thermal energy Eth in the resistor as a function of time t. (a)What is the electric potential across the battery? (60) (b) If the resistance is doubled, what is the POWER dissipated by the circuit? (39) (c) Did you put your name on your paper? (1) Looking at the graph, we see that the resistor dissipates 0.5 mJ in one second. Therefore, the POWER =i2R=0.5 mW P 0.5 mW i2 2.38 10 5 amp 2 R 21 Ω i 4.88 10 3 amp 4.88 ma Voltage drop across the reisitor iR or V iR 4.8810-3 amp 21 102mV If the resistance is doubled what is the power dissipated by the circuit? R 42 V 102 mV 3 V 102 10 i 2.43ma R 42 P i R 0.248mJ 2

DOCUMENT INFO

Shared By:

Categories:

Tags:

Stats:

views: | 12 |

posted: | 3/8/2012 |

language: | English |

pages: | 41 |

OTHER DOCS BY ert554898

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.