Can You Light the Bulb?

AP PHYSCS 2 Can You Light the Bulb? UNT 4 DC circuits and RC circuits. CHAPTER 16 DC CRCUTS 1. Draw wires and make the bulb light. 2. Modify your drawing and use ONE wire only! Complete circuits To check whether any circuit you build is complete, trace the path of an imaginary positive charge moving from the positive terminal of the battery to the negative terminal. The path must pass along conducting material at every location. 1

Observation experiment outcomes 2

The initial charge separation between the oppositely charged spheres allowed charge to flow between the spheres. Observation experiment outcomes Observation experiment outcomes The electric potential energy of the spheres was converted into some other form of energy. The presence of a charge conduction pathway allowed charge to flow. We need to learn how to keep these processes happening continuously. Alessandro Giuseppe Antonio Anastasio Volta (1745-1827) Alessandro Giuseppe Antonio Anastasio Volta (1745-1827) After results reported by Luigi Galvani Volta decided to experiment with chemical baths. He experimented with multiple metals and multiple acids. Copper Sulfuric acid Zinc Volta observed that if he touched both metals with his tongue, there was a tingly sensation: An electric current. Sulfuric acid Zinc 3

Alessandro Giuseppe Antonio Anastasio Volta (1745-1827) Alessandro Giuseppe Antonio Anastasio Volta (1745-1827) Two types of metal in the acid bath generates two opposite charges in the two metal rods. The charge separation leads to a potential difference between the rods. Making the process continuous To achieve a steady flow of electric charge, we need a device that can maintain a steady potential difference. Alessandro Giuseppe Antonio Anastasio Volta (1745-1827) Volta is the father of the voltaic pile, an early term to describe what we know today as a battery!!! Fluid flow and charge flow Fluid flow and charge flow You connect a hose between the two containers such that the two ends of the hose are at different pressures. Water starts flowing until the pressures are the same. 4

Making the process continuous Pumping water from B back to A maintains a pressure difference between the ends of the hose and results in a continuous flow from A to B. Voltage causes electric current to flow. Through a series of chemical reactions, a battery pushes electrons back to the negative terminal. t does work on electrons, moving them from the positive to the negative terminal within the battery. Batteries are a source of voltage Voltage is a property of the battery itself. The voltage across the terminals of a battery is the work done by the battery per unit charge moved inside the battery from one terminal to the other. Whereas current varies depending on the circuit that it is in, voltage does not. DV = W q Units: Volts = J/C The voltage of a battery indicates the amount of work done by the battery per coulomb of charge to move charges from one terminal to the other. f a battery has a voltage of 9 Volts, it means that 9 Joules of energy are added to the circuit for every Coulomb of electric charge that passes through the battery. Batteries The physical size of a battery is not related to the emf, but to its storage capacity the total amount of electric charge it can move before it must be replaced or recharged. Through a series of chemical reactions, they push electrons toward other electrons. This produces constant voltage. AAA, AA, C, D: All 1.5 Volts 9 V Battery The bigger the battery, the longer the time interval during which the battery works. 5

Unlike electric current, voltage does not exist at at certain point. Voltage is the difference in energy per charge between two points in a circuit. The voltage across the terminals of a battery is the work done by the battery per unit charge moved inside the battery from one terminal to the other. V = W q Units: Volts = J/C A 9 Volt battery is really six 1.5 Volt cells connected in series. As electrons pass through each cell, they gain a energy from each one, to a total of 9 J/C. f a battery has a voltage of 9 Volts, it means that 9 Joules of energy are added to the circuit for every Coulomb of electric charge that passes through the battery. Batteries and emf The language of physics: Force and work The emf equals the work done by a battery per Coulomb of electric charge q that is moved through the battery from one terminal to the other in order to maintain a potential difference at the battery terminals. The physical size of a battery is not related to the emf but to its storage capacity the total charge it can move before it must be replaced or recharged. 6

Units: Batteries and emf ε = W q ε = W [ J ] q [ C ] ε = [ Volts] The flow of charged particles between two locations is a physical phenomenon called electric current. A system of devices such as a battery, wires, and a bulb that allows for the continuous flow of charge is called an electric circuit. n most electric circuits, the moving charged particles are free electrons that are already in the wires and circuit elements. Crystal lattice model with free electrons Crystal lattice model with free electrons Drift When a wire is placed in an external electric field, the electrons accelerate in the direction opposite to the direction of the E field. They slow down when they "collide" with the ions. Electric Current The magnitude of the electric current flowing through a wire equals the magnitude of the electric charge q that passes through a cross section of the wire divided by the time interval needed for that amount of charge to pass. This drift motion occurs in the same direction for all of the electrons. = q t 7

WB Chapter 16 Lecture = q t The unit of current is the ampere A, equivalent to one coulomb per second C/s. A current of 1 A (one ampere, or amp) means that 1 C of charge passes through a cross-section of the wire every second. The direction of the current is in the direction positive charges would move. A typical lightning bolt has a current of up to 10 kiloamperes passing through it. f the flash takes place in a time of 0.2 seconds, how many electrons flow through the lightning bolt? e = 1.25x10 22 Each second, 1.0 x 10 17 electrons pass from right to left past a cross section of a wire connecting the two terminals of a battery. Determine the magnitude and direction of the electric current in the wire. = q t = en t = 1.6x1019 C 1.0x10 17 1 s = 0.016 A Direction: Left to Right WB You inspect an alkaline AA battery, and see that it is labeled 2.12 amp-hours. a) What does this quantity measure about the battery? b) How much electric charge can pass through the battery during its lifetime? c) f the voltage of a AA battery is 1.5 V, how much work does the battery do on the circuit during its lifetime? a) What does this quantity measure about the battery? The amount of electric charge that can flow through the battery during its lifetime b) How much electric charge can pass through the battery during its lifetime? = q q = 7632 C t c) f the voltage of a AA battery is 1.5 V, how much work does the battery do on the circuit during its lifetime? V = W q W = 11448 J Electric current Free electrons in a wire drift in the direction of the higher V field. Traditionally, the direction of electric current in a circuit is defined as opposite to the direction of the electrons' drifting motion. 8

Negatively charged particles moving in one direction are mathematically equivalent to positively charged particles moving in the opposite direction. Electric current is mathematically defined as the flow of positive charge. This is known as conventional current. Neon and incandescent bulbs A neon bulb consists of a glass bulb filled with lowpressure neon gas. An incandescent bulb has a metal filament inside it. 9

Electric current Electric current: the flow of charged particles moving through a wire between two locations that are at different electric potentials. Direct current: an electric current in which the charged particles move in the same direction. Electric circuit: a system of devices that allows for the continuous flow of charge. DC circuit: an electric circuit that has a direct current. Ammeters capacitor An ammeter acts like a water flow meter. f you wish to measure how much water flows through a cross section of a pipe, the water must pass through the flow meter. Voltmeters A voltmeter measures the electric potential difference between two points in a circuit. Using a voltmeter is analogous to using a pressure meter to measure the water pressure difference. Measuring current through and potential difference across a resistive wire 10

OHM S LAW Georg Ohm (1789 1854) Georg Ohm was a German physicist. As a high school teacher, Ohm began his research with the recently invented electrochemical cell, invented by talian Count Allesandro Volta. Using equipment of his own creation, Ohm determined that for many objects and types of materials, there is a direct proportionality between the voltage applied across the object and the resulting electric current. This relationship is known as Ohm s Law. Ohm connected various objects to batteries of different voltages, and measured the resulting current through the circuit. object ΔV Metal block ΔV Ceramic block Rubber block Ohm discovered that for many objects, there is a linear relationship between voltage and current Metal block object ΔV = (slope) * ΔV Ceramic block ΔV Rubber block The slope of this graph tells how well the object conducts an electric current. t is called the conductance of the object. Metal block (high conductance) Ceramic block (medium conductance) Rubber block (low conductance) ΔV The inverse of the slope of this graph tells how much the object resists an electric current. That is called the resistance of the object. Metal Block (low resistance) Ceramic Block (medium resistance) Rubber Block (high resistance) ΔV slope = 1/R 11

Ohm s Law comes directly from this linear relationship Metal Block (low resistance) Ceramic Block (medium resistance) = 1 R DV MATH CLASS MATH MODEL y = m x + b Rubber Block (high resistance) ΔV = DV R R is the resistance of the object. PHYSCS = 1 R V Units: Ohms (Ω) 1 Ω = 1 V/A [A] 45 40 35 30 25 20 15 Slope = 1 R OHM S LAW The electric current passing through resistor is directly proportional to the electric potential difference [voltage] 10 5 0 0 5 10 15 20 25 30 V [V] The electric current is inversely proportional to the resistance. = V = ε R R OHM S LAW Resistors that follow Ohm s Law obey a LNEAR relationship. Not all objects obey Ohm s Law. Objects that have a constant resistance are called ohmic. Resistor or bulb Electric Current 0.50 0.45 0.40 = 0.005 DV 0.35 0.30 0.25 0.20 0.15 = 0.0025 DV 0.10 0.05 0.00 0 10 20 30 40 50 60 70 80 90 100 Voltage 12

Electric Current 2/6/2018 What is the magnitude of Resistance used in this single resistor circuit? 0.03 0.02 0.01-0.01-0.02-0.03 = 0.01 DV = 0.005 DV 0-3.00-2.00-1.00 0.00 1.00 2.00 3.00 R 1 = 100 R 2 = 200 Voltage R1 R2 Linear (R1) Linear (R2) Resistance Resistance characterizes the degree to which an object resists a current. We can write the relationship between current and potential difference using resistance R: R = V R = ε The unit of resistance is called the ohm; 1 ohm = 1 volt/ampere. When commercial resistor R 1 is connected to a 9-V battery, the current through the resistor is 0.1 A. When commercial resistor R 2 is connected to the same battery, the current through it is 0.3 A. What is the resistance of these resistors? R = ε R 1 = 90 Ω R 2 = 30 Ω Relationship between current and electric potential difference = ε R The larger the potential difference across the wire, the larger the E field inside. When a larger electric force is exerted on the electrons, it results in larger accelerations of the electrons between the collisions with the ions in the lattice and, therefore, a larger current through the wire. Diode A diode is a non-ohmic circuit element that consists of two kinds of materials that cause it to be asymmetrical with respect to the potential difference across it. RESSTORS Resistors are electric devices which change the electric potential energy of the electric charges and therefore creating a change in electric potential between two points. Using a diode in a circuit allows you to achieve one-directional current. 13

Microscopic model of resistivity Free electrons move chaotically inside a metal, and they collide with metal ions. These collisions increase the internal energy of the metal ions, causing them to vibrate more vigorously. The vibrating ions become even more of an obstacle for the electrons. This explains why a hotter metal has a higher resistivity than a cooler one and why the resistance of the lightbulb filament increases with temperature. What determines an object s resistance? L = length A = cross-sectional area ρ = resistivity (type of material) A longer wire has more nuclei for electrons to collide with (a longer path). Therefore a longer wire has a greater resistance. Rµ L f the wire has a circular cross section: r A = pr 2 r = radius of cross section R µ 1 R µ 1 A r 2 A wire with a greater cross-sectional area has more pathways for electrons to move in parallel. This is why a thicker wire has a lower resistance. R µ 1 A The type of material is defined by its resistivity The higher the resistivity, the more the atoms of a particular element will resist the flow of electrons. ρ = resistivity 14

The connecting wires that we use in electric circuit experiments are usually made of copper. What is the resistance of a 10-cm-long piece of copper (1.7x10-8 m) connecting wire that has a diameter of 2.0 mm. R = rl A R = ρ L A R = 5.41x10 4 Ω P = U Thermal t U Thermal = U q U q = q V ELECTRC POWER Joule's law P = P = q V t q t V P = V ELECTRC POWER Joule's law The rate P at which electric potential Energy is converted into thermal energy DU THERMAL in a resistive device equals the magnitude of the potential difference across the device multiplied by the electric current through the device. P = U Thermal t P = V P = ε f P = ε ELECTRC POWER Joule's law determine the units of power. P = watts ELECTRC CURRENT Ohm's law = ε R Use Joule s Law and Ohm s law and write an expression for P: P = ε = ε R n terms of, R P = ε R 2 ELECTRC POWER Joule's law P = ε P = ε R 2 n terms of, R P = 2 R P = 2 R 15

RESSTORS N SERES Circuits in series Two resistors are connected in series, when they are connected ONLY by one point. Adding more bulbs must increase the resistance of the circuit, reducing the electric current everywhere in the circuit. RESSTORS N SERES RESSTORS N PARALLEL stays the same T = 1 + 2 + R T = R 1 + R 2 +. P T = P 1 + P 2 +. Two resistors are connected in PARALLEL, when they are connected by TWO points. Circuits in parallel For each identical bulb to be equally bright and have the same brightness as a single bulb, the current must be the same through all of the bulbs. RESSTORS N PARALLEL T = 1 + 2 +. stays the same R T = 1 R 1 + 1 R 2... 1 P T = P 1 + P 2 +. 16

ELECTRC POTENTAL DFFERENCE 2/6/2018 (in terms of R & ) Assume that we have a battery and two identical lightbulbs, each of resistance R. We connect the bulbs to the battery in three different configurations: (a) one bulb, (b) two bulbs in series, and (c) two bulbs in parallel 1. Determine the equivalent resistance of a single resistor that will produce the same current through the battery as produced in arrangements (b) and (c). 2. Compare the total power output of the bulbs in arrangements (b) and (c) with the output in (a). (a) one bulb R T = R (in terms of R & ) (b) two bulbs in series R T = 2R (c) two bulbs in parallel R T = R 2 P T = ε2 R P T = ε2 2R P T = 2ε2 R Qualitative investigations of electric circuits Adding more circuit elements in series decreases the total current through all the elements and increases the effective total resistance of the elements. Adding more elements in parallel increases the total current through the parallel elements and reduces the effective total resistance of the elements. NTERNAL RESSTANCE OF A BATTERY nternal resistance of a battery 10 The potential difference across a battery in an open circuit will be greater than that across the same battery in a closed circuit. This is due to internal resistance of the battery. The idea of internal resistance is consistent with observations that batteries feel warm to the touch after they have been operating for a while. 9 8 7 6 5 4 3 2 1 0 y = -3.21x + 8.8396 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 ELECTRC CURRENT 17

ELECTRC POTENTAL DFFERENCE 2/6/2018 MATH CLASS PHYSCS MATH MODEL y = m x + b DV = r + ε The emf ( ) is the electric potential difference (DV) of a source when no current is flowing y = -3.21x + 8.8396 = -3.21 + 8.8396 = 0 A = 0 v = 8.8396 v y-intercept mf of the battery = 2.753805 A x-intercept Maximum drawn from battery Slope nternal resistance of the battery ABOUT BATTERES The battery is not a source of constant current: it drives a smaller current through circuit elements arranged in series and a larger total current through circuit elements arranged in parallel. Changing the number of circuit elements changes the whole circuit. The total current may decrease or increase depending on the arrangement of circuit elements. You buy a 9.0-V battery. To check whether it really produces a 9.0-V potential difference across its terminals, you use a voltmeter and find that it reads 9.0 V. Satisfied, you build a simple series circuit with a 5.0-ohm resistor. An ammeter indicates a 1.5-A current through the resistor. A voltmeter reads 7.5 V across the battery and 7.5 V across the resistor. 1. Graph DV vs and write a math model 2. What s the internal resistance of the battery? 3. What s the emf of the battery? 4. What is the max. electric current that can be drawn? 10 9 8 7 6 5 4 3 2 1 0 y = -x + 9 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 ELECTRC CURRENT DV = r + ε = 1 DV + 9 JOULE S LAW and Electric Circuits 18

Joule's law t is usually most convenient to use: P = R 2 The expression when elements are connected in parallel (DV = constant) Joule's law and bulb brightness The brightness of a bulb is directly related to its power. The greater power, the brighter the bulb. Bulbs can have different brightness depending on the type of connection. P = 2 R The expression when comparing power in elements connected in series ( = constant). P = 2 R P = ε R 2 Short circuit How is your house wired? magine that you accidentally connect the terminals of a battery with a connecting wire of approximately zero resistance. Turning any one device on or off will not affect the others. Series or parallel? The wire and the battery become very hot because the current becomes very large. 1 2 3 The language of physics: Electricity charge, current, potential difference, and power 1 2 3 T DV = 3 V R DV P 0.5 1 0.5 0.25 0.5 0.5 2 3 1 1.5 0.5 0.75 0.5 6 3 1.5 19

R 2R 3R 1 2 6 DV = 3 V DV = 1 2 3 T R DV P /6R R /6 2 /36R /6R /6R 2R 3R /3 /2 2 /18R 2 /12R /6R 6R 2 /6R 1 2 3 T R DV P 3 1 3 9 1.5 0.5 2 6 3 3 4.5 1.5 5 0.6 3 15 R 2 = 4 R 2R 6R DV = R 1 = 6 R 3 = 12 1 2 3 T R DV P /R R 2 /R /2R /6R 2R 6R 2 /2R 2 /6R 5 /3R 3R/5 5 2 /3R 1 2 3 T 9 V R DV P 1 6 6 6 3/4 1/4 4 12 3 3 9/4 3/4 1 9 9 9 R 1 = 5R R 2 = 6R R 3 = 12R a) b) Rank the Brightnesses 1 2 3 T R DV P /9R 2 /27R /27R /9R 5R 5 /9 6R 4 /9 12R 4 /9 9R 5 2 /81R 8 2 /243R 4 2 /243R 2 /9R n each of the circuits shown, rank the light bulbs from least bright to brightest. Show your reasoning on your whiteboards. 20

P = DV Both branches will receive the same current, resistor #3 will have the full voltage of the battery across it. Voltage will be divided among resistors 1 and 2, and resistor #2 will receive twice the voltage as resistor #1. P 1 < P 2 < P 3 P = DV The branch of #2 and #3 is equivalent to one resistor of 2R. This means that resistors #2 and #3 will each receive twice the voltage as resistor #1. However, the current splits, so resistors #2 and #3 will each receive half the current as resistor #1. P 1 = P 2 = P 3 Double the bulbage! f you were to connect another light bulb in parallel to the first, as shown by the dotted lines, what would happen to the brightness of the first light bulb? Discuss with your partner to come up with a prediction. The first bulb is unaffected!! 2 The battery is now providing a voltage ΔV to two light bulbs. Each light bulb will receive a current of ΔV/R, and the total electric current through the battery will double. 2 2 FND RT: A lamp, a voltmeter V, an ammeter A, and a battery with zero internal resistance are connected as shown above. Connecting another lamp in parallel with the first lamp as shown by the dashed lines would (A) increase the ammeter reading (B) decrease the ammeter reading (C) increase the voltmeter reading (D) decrease the voltmeter reading (E) produce no change in either meter reading We need techniques to determine the electric current through each circuit element in a complicated circuit: Kirchhoff's rules 21

Kirchhoff's rules Kirchhoff's junction rule Junction Rule Loop Rule Kirchhoff s rules apply to any electric circuit. Charged particles making up a current cannot vanish or be created out of nothing (electric charge is a conserved quantity). The sum of the currents into a junction must equal the sum of the currents out of the junction. Kirchhoff s rules help us solve circuits that cannot be reduced to simple series-parallel combinations of resistors Kirchhoff's junction rule Kirchhoff's junction rule The total rate at which electric charge enters a junction equals the total rate at which electric charge leaves the junction. junction 1-2 - 3 = 0 Conservation of electric charge Conservation of electric charge You connect a 9.0-V battery to a small motor. Describe the changes in electric potential in the circuit with a graph. 22

Kirchhoff's loop rule Kirchhoff's loop rule Trace the change in electric potential as we move with these charges around the circuit. The sum of the electric potential differences DV across the circuit element that make up closed path (called a loop) in a circuit is zero. The changes in potential for each step along the trip must add to zero. Conservation of energy Kirchhoff's loop rule Kirchhoff's loop rule Loop Through a battery: Low to high (negative to positive) The voltage is positive lift. High to low (positive to negative) the voltage is negative drop. loop loop - DV = 0 Conservation of energy Loop Through a resistor: f loop matches the direction of the labeled electric current, then DV is negative drop f loop opposes the direction of the labeled electric current, then DV is positive lift. Typically there will be one more loop than number of junctions. 80 V 20 Junction Loop A Loop B 1 + 2 = 3 80 50 3 20 2 = 0 120 + 50 3 = 0 2 1 3 120 V LOOP A LOOP B 50 3 2 1 80 50 12 5 20 2 = 0 80 120 20 2 = 0 3 = 12 5 A Junction 1 + 2 = 3 Loop A 80 50 3 20 2 = 0 Loop B 120 + 50 3 = 0 2 = 2 A 1 = 22 5 A f direction is negative, it means electric current flows in the opposite direction 23

- + 2 3 LOOP A 2 3 1 1 LOOP B + - Junction 1 + 2 = 3 Loop A 60 20 3 20 2 = 0 Loop B +90 + 20 3 + 20 1 = 0 Junction 1 + 2 = 3 Loop A 60 20 3 20 2 = 0 Loop B +90 + 20 3 + 20 1 = 0 Solve for the one that does not repeat 2 = 3 3 1 = 9 5 2 3 5 9 5 3 + 3 3 = 3 3 2 = 3 3 2 = 7 2 A 1 = 4A 3 = 1 2 A - 1-2 = 9 5 A = 1 11 A = 3 5 A = 5 11 A = 12 5 A = 4 11 A 24

3 Find the magnitude of the missing = 16 V 20 9 A 8 V 4 6 = 20 9 A Use the loop rule to predict the potential difference between point A and point B in the electric circuit shown in the figure. 12 = 4 9 A Use Kirchhoff s rules to determine the electric current through Resistors R 1, R 2, & R 3, in the following circuit. Problem 23 & 24 Page 615 Write two loop rules equations and one junction rule equation for the circuit: Problem 23 & 24 Page 615 B. Use these equations to determine the current in each branch of the circuit for the case in which R 1 = 0, R 2 = 18, R 3 = 9 and = 60 V. Problem 27 & 28 Page 615 Write Kirchhoff s loop rule for the circuit: C. Use these equations to determine the current in each branch of the circuit for the case in which R 1 = 50, R 2 = 30, R 3 = 15 and = 120 V. 25

Problem 27 & 28 Page 615 Problem 30 Page 615 B. Use this equation to determine the current of the circuit for the case in which 1 = 20 V, 2 = 8 V, R 1 = 30, R 2 = 20, and R 3 = 10. Write the loop rule for two different loops in the circuit, and the junction rule for point A: C. Use this equation to determine the current of the circuit for the case in which 1 = 12 V, 2 = 3 V, R 1 = 1, R 2 = 1, and R 3 = 16. Problem 30 & 31 Page 615-616 B. Use these equation to determine the current in each loop for the case in which 1 = 3 V, 2 = 6 V, R 1 = 10, R 2 = 20, and R 3 = 30. Problem 39 Page 616 Write Kirchhoff s loop rule for two loops and one junction in the circuit: C. Determine the value of R2, so that the current through R3 equals twice that through R2. The values of other circuit elements are: 1 = 12 V, 2 = 15 V, R 1 = 15, and R 3 = 30. Problem 39 Page 616 Applying Kirchhoff's rules in the problemsolving strategy: Sketch and translate B. Solve for the current in each branch of the circuit when 1 = 10 V, 2 = 2 V, R 1 = 50, R 2 = 200, and R 3 = 20. C. Determine the value of R2, so that the current through R3 equals twice that through R2. The values of other circuit elements are: 1 = 12 V, 2 = 15 V, R 1 = 15, and R 3 = 30. Draw the electric circuit described in the problem statement and label all the known quantities. Decide which resistors are in series with each other and which are in parallel. 26

Applying Kirchhoff's rules in the problemsolving strategy: Simplify and diagram Applying Kirchhoff's rules in the problemsolving strategy: Represent mathematically Decide whether you can neglect the internal resistance of the battery and the resistance of the connecting wires. Draw an arrow representing the direction of the electric current in each branch of the circuit. f possible, replace combinations of resistors with equivalent resistors. Apply the loop rule for the potential changes as you move around one or more different loops of the circuit. Each additional loop you choose must include at least one branch of the circuit that you have not yet included. Applying Kirchhoff's rules in the problemsolving strategy: Represent mathematically Applying Kirchhoff's rules in the problemsolving strategy: Solve and evaluate Once you have included branches, apply the junction rule for one or more junctions. n total, you will need the same number of independent equations as the number of unknown currents. Solve the equations for the unknown quantities. Check whether the directions of the current and the magnitude of the quantities make sense. f necessary, use expressions for electric power. Capacitors in Parallel Have the same voltage as one another. CAPACTORS A C 1 B C 2 27

Combining Capacitors in Parallel Combining Capacitors in Parallel Add like resistors in series. C 1 C 1 C 2 C 1 + C 2 Connecting capacitors in parallel is like creating one big capacitor! C 2 C eq = C 1 + C 2 + C 1 + C 2 Charge will split proportionally to capacitance CAPACTORS N PARALLEL Q 1 C 1 C 2 Q 2 Q = CDV Since voltage is the same for both, Q a C = stays the same q T = q 1 + q 2 + C T = C 1 + C 2 + Capacitors in Series Adding capacitors in series shares the voltage of the battery between the two capacitors. Capacitors in Series Have the same charge as one another. This will decrease their capacitance. -Q +Q -Q +Q C 1 C 2 æ 1 + 1 ö ç èc 1 C 2 ø -1 Think about it! This piece of metal is net neutral and totally isolated. f a charge +Q is attracted to one side, the opposite charge Q must be left on the other side. 28

Capacitors in Series Add like resistors in parallel. Voltage will split in inverse proportion to capacitance ΔV 1 ΔV 2 DV = Q C C 1 C 2 æ 1 + 1 ö ç èc 1 C 2 ø -1 C 1 C 2 Since charge is the same for both, 1 C eq = 1 C 1 + 1 C 2 +... DV a 1 C CAPACTORS N SERES Equivalent Capacitance is Opposite to Equivalent Resistance! T = 1 + 2 + q = stays the same C T = 1 C 1 + 1 C 2... 1 n series R eq = R 1 + R 2 +... 1 C eq = 1 C 1 + 1 C 2 +... n parallel 1 R eq = 1 R 1 + 1 R 2 +... C eq = C 1 +C 2 +... The charge stored in the 5-microfarad capacitor is most nearly C = Q DV For the 5 μf capacitor, ΔV = 100 V (it is directly connected to the battery) Q = 500 μc 1) The equivalent capacitance for this network is most nearly (A) 10/7 F (B) 3/2 F (C) 7/3 F (D) 7 F (E) 14 F 2) The charge stored in the 5-microfarad capacitor is most nearly (A) 360 C (B) 500 C (C) 710 C (D) 1100 C (E) 1800 C 29

Simplify, simplify, simplify. How capacitors behave in circuits An uncharged capacitor behaves like a wire of zero resistance. 100 V 5 μf 4 μf 2 μf 100 V 5 μf 6 μf (This is only for the first instant that it is being charged) 3 μf 3 μf As it becomes charged, it increasingly resists current flow. 100 V 5 μf 2 μf 100 V 7 μf A charged capacitor behaves like an open circuit. (infinite resistance) This happens as time approaches infinity. What happens when you close the switch? At the very first instant that the switch is closed, the capacitor behaves like a wire of zero resistance. R C S R C ΔV Explain why the circuit behaved as it did. ΔV However, as the capacitor becomes more charged, the current through the circuit drops to zero. Calculate the amount of charge stored on the capacitor when the circuit has been allowed to run for a very long time. For the resistor, ΔV = R. When the current is zero, the voltage across the resistor is also zero! R C R C ΔV You will need to use the relationship C = Q DV ΔV As the current drops to zero, the full voltage of the battery will be the voltage across the capacitor. 30

At the instant that the switch is closed, the uncharged capacitor behaves like a wire. R R C S Predict what will happen when the switch is closed! C S ΔV ΔV t initially diverts the full current through itself! However, as the capacitor charges, it behaves like a branch of greater and greater resistance. R 200 Ω RC Circuits! The circuit shown is connected for a very long time. C S 400 Ω 2 μf What is the amount of charge stored in the capacitor? ΔV Eventually, all of the current travels steadily through the bulb. 12 V C = Q DV Use the same concepts as always! 200 Ω 1) n this circuit, the capacitor will have the same voltage as the 200 Ω resistor. 200 Ω Final Result ΔV 400 = 8 V ΔV cap = ΔV 200 = 4 V 400 Ω 2 μf 2) Calculate the voltage across the 200 Ω resistor by just treating the charged capacitor as an open circuit (R = ). 400 Ω 2 μf Q = 8 μc 12 V 3) Finally use Q = CΔV 12 V 31