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Lecture 07 Current & Circuits Upcoming Stuff Watch for more Web-Assignments Quiz on Friday (This weeks material) Examination #2 one week from Friday: March 11th. Last time We discussed current, resistance, power and the elements of the microscopic theory of resistivity. Today we start electric circuits after a short diversion. The Spectrum of Conductors ENGINEERED MATERIALS!!! Semiconductors Materials VERY IMPORTANT MATERIAL SILICON Silicon Crystal Lattice Electron Freed! Add an impurity Special Impurities Effect of Impurities on r Importance P N P N P N P Diode current voltage breakdown Who Cares?? Silicon Transistor CHIPS History Thinking Chips? Back to the Mundane Fun and Frolic With Electric Circuits Power Source in a Circuit V 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 Battery Emf i rR 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. 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 V1 V iR iR1 iR2 R2 V2 V or R R1 R2 General for SERIES Resistors R Rj j A single resistor can be modeled as follows: R a b V position ADD ENOUGH RESISTORS, MAKING THEM SMALLER AND YOU MODEL A CONTINUOUS VOLTAGE DROP. Take a trip around this circuit. Consider voltage DROPS: -E +ir +iR = 0 or E=ir + iR Might as well repeat the parallel thing! i i1 i2 V R1 i1 R2 i2 V V V R R1 R2 1 1 1 R R1 R2 General : 1 1 R j Rj Circuit Reduction i=E/Req Multiple Batteries Reduction Computes i Another Example 1 1 1 50 1 R 20 30 600 12 R 12 PARALLEL NOTICE ASSUMED DIRECTION OF TRAVEL Voltage Drops: -E1 –i1R1 + i2R2 + E2 +i1R1 = 0 From “a” -i3R1 + E2 – E2 –i2R2 =0 NODE I3 +i2 = i1 In the figure, all the resistors have a resistance of 4.0 and all the (ideal) batteries have an emf of 4.0 V. What is the current through resistor R? 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 Loop Equation I need to use E for E Note RC = (Volts/Amp)(Coul/Volt) = Coul/(Coul/sec) = (1/sec) q E iR 0 C dq since i dt dq q R E dt C or dq q E dt RC R This is a differential 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 thenb matching the guess to reality. You may oir 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 q0 t / RC dq 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 i 2.38 105 amp2 R 21 Ω i 4.88 10 3 amp 4.88 ma Voltage drop across the reisitor iR or 2 V iR 4.88 10-3 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 2 P i R 0.248mJ