Physics A Unit 12 Circuits

Transcription

1 Physics A Unit 12 Circuits From unit 11 through unit 13 you will study another of the major branches of physics, electromagnetism. In the second unit on electromagnetism we study simple circuits. I. Ohm s Law & One Resistor A. Circuit Diagrams B. Calculations 1. Current and voltage 2. Power and Heat II. Multiple Resistor Circuits A. Pure Resistor Circuits B. Pure Parallel Circuits C. Combination Circuits III. The Capacitor A. Defined 1. Reduction Method 2. Kirchoff s Laws B. Calculations 1. Stored Charge 2. Stored Energy IV. RC Circuits initial and final values This unit should take approximately eight days to complete as follows. Day 1 is I. Day 2 is IIA&B. Day 3 is IIC.1. Day 4 is II.C.2. Day 5 is III. Day 6 is IV. Day 7 is practice test day and day 8 will be the test. A graphical lab on Ohm s Law is also necessary.

2 Lesson 3-09 A Simple Resistor Circuit Recall from the last unit we studied charges at rest. In this unit the charges will move through a circuit. A circuit consists of at least one source of voltage difference a.k.a. a battery. The voltage difference will rapidly establish an electric field through the circuit. The electric field will enable charged particles to move along a closed path or loop. Although the electric field is established at near the speed of light the charged particles (electrons in reality) will move along at mm/sec. In this first lesson we consider a single battery connected to a single load. We will see how to draw the circuit and calculate four measurable values of a simple circuit. Drawing the Circuit A simple resistor circuit consists of two parts connected in a closed loop. One part is the battery and the other part is the load. A load can be a light bulb, radio, TV or any other electrical device. All loads will have a measurable amount of resistance to electrical flow of charge. Because of this another name for a load is a resistor. The resistance of a resistor is measured in ohms, Ω. The Battery The Load or Resistor A Complete Circuit Assume that positive charges leave the long end of the battery and return through the short end of the battery. The symbol for a resistor is a wire bent into many elbows so that charge flow is slightly hindered but not completely shut off. 8 ohms 20V A 4 Ω V or 8 Ω The voltmeter with dashed lines is not part of the closed circuit. The ammeter is also not needed. The ammeter measures how much charge comes out of the battery and/or how much charge flows into the resistor each second. The voltmeter will read the difference in the before and after voltage of any device it is wired across. Do you notice that the wires are attached both before and after? Calculating the Values There are four measurements that can be calculated or determined using a watch and the two meters above. The first of these is electrical current, I. Electrical current tells us how many charges are flowing past a point each second. The amount of Coulombs/sec is known as Amperes. I = q / t Here I is in Amps, q is in Coulombs and t in seconds.

3 The other relation that can determine current is known as Ohm s Law in honor of a fellow with the last name of Ohm. This law recognizes that a bigger voltage will create more current while a bigger resistance will allow less current. V = IR V is a voltage difference across a resistor and I is the current through the resistor. From the example of the completed circuit you see that total current through the resistor is 5 Amperes or 5 A. At a rate of 5 Coulombs/sec flowing through the resistor we determine that 300 C of charge will exit the battery and enter the resistor each minute. We can also determine the rate at which electrical energy is converted to heat, light or mechanical energy in the load. The rate at which energy is converted is described as power. Power can be expressed in terms of current and voltage or current and resistance or voltage and resistance. Take your pick. No matter which of the three forms you choose to learn, recall that power is in Watts. 1 Watt = 1 Joule / second. Current & Voltage Current & Resistance Voltage & Resistance P = IV P = I 2 R P = V 2 / R No matter which of the three forms you use you can see that the battery in the example of the completed circuit is supplying 100 W of power to the load. In a quarter of a minute the load will convert 1500 Joules or 15 kj of electrical energy to some other form. I like to think of a simple circuit like a bicycle. The battery is like the front sprocket and rider. The chain links are like the individual charges that move along the circuit. The back tire and sprocket are much like the load. Realize that even after a bicycle chain has delivered its energy to the back sprocket it does not slow down. In a simple circuit the electrons do not slow down after going the resistor in the same manner. Homework Assignment #1 Work each of the following problems showing a completed circuit with meters. For the solutions, write the equation and show the substitutions before writing down your final answer. Compare your answers to another student in order to check for correct values. 1. An 8 ohms resistor is connected to a battery with an EMF of 24 Volts. Determine the power delivered to the battery and the current flowing through the battery. How much charge exits the battery in one minute? How much heat is generated in 15s? 2. A 12 ohms resistor is connected to a battery with an EMF of 48 Volts. Determine the power delivered to the battery and the current flowing through the battery. How much charge exits the battery in a half minute? How much heat is generated in 20s? 3. A 7 ohms resistor is connected to a battery with an EMF of 28 Volts. Determine the power delivered to the battery and the current flowing through the battery. How much charge exits the battery in one minute? How much heat is generated in 12s? 4. A 3 ohms resistor is connected to a battery with an EMF of 24 Volts. Determine power delivered to the battery and current flowing through the battery. How much charge enters the battery in a quarter minute? How much heat is generated in 30s?

4 Lesson 3-10 Multiple Resistor Circuits What happens when you have more than one resistor in a circuit? The answer depends upon how the resistors are wired. In today s lesson we consider one of two possibilities, a series circuit. Recognizing Resistors in Series Resistors are wired in series if there are no junctions between the resistors. A junction would allow the current flowing through the circuit to split into parts. R 1 R 2 R 3 R 1 R 2 /\/\/\ /\/\/\ /\/\/\ /\/\/\ /\/\/\ R 3 /\/\/\ For the circuit on the left all of the current flowing out of one resistor must flow directly into the next resistor. This is a series of resistors. For the circuit on the right there is a junction of wires between R 1 and R 2. At this junction the current can split into parts. Since not all of the current flowing out of R 1 must go through R 2 these resistors are not in series. Rules for Resistors in Series There are only three rules for resistors in series that are important. Consider the circuit below to demonstrate each of the three rules. 6 Ω /\/\/\ A EMF= 36 V 2 Ω B /\/\/\ 4 Ω 1st. Total Resistance Rule: R TOTAL = R 1 + R 2 + R 3 + The total resistance of resistors in series is the sum of their resistance. In the above example the resistance of the entire circuit is 12 Ω. 2nd. Total Current Rule: I TOT = I 1 = I 2 = I 3 = The total current flowing out of the battery is the same as the current flowing through each resistor. This is because there is nowhere else for the current to flow. You can use Ohm s Law to find the total current. I TOTAL = V TOTAL / R TOTAL or I TOT = 36 Volts / 12 Ohms = 3 Amps. 3rd. Total Voltage Rule: V TOT = V 1 + V 2 + V 3 + The individual voltage differences across each resistor can be found using V = IR. The 6Ω resistor- V = 3*6 = 18V. The 2Ω resistor has V =3*2 = 6V. Of the original 36 V there are now only 12 V left in the loop to go across the 4Ω resistor. Alternately you could use 3*4=12V.

5 A voltmeter hooked from A to B in the circuit would read 18 Volts. This is because the wire at point A is at (+36 18V) while the wire at point B is at 0 Volts. Alternately the voltmeter is reading the combined voltage drop across the 2Ω and the 4Ω resistors. Use the previous steps to find a) the total resistance of a circuit, b) the total current flowing out of the battery and c) the potential difference across each resistor and d) potential difference between A and B for each of the following circuits. Circuit #1 3 Ω /\/\/\ A EMF= 36 V 4 Ω B /\/\/\ 5 Ω Circuit #2 6 Ω /\/\/\ A EMF= 64 V 3 Ω Circuit #3 B /\/\/\ 7 Ω 6 Ω 3 Ω /\/\/\ /\/\/\ A EMF= 75 V 4 Ω /\/\/\ /\/\/\ B 5 Ω 7 Ω If a wire is clipped into the circuit at points A and B what would be the resulting current through each resistor in circuit #3? Answers 1. a) 12 Ω b) 3A c) 9V ; 12V ; 15 V d) 27 Volts 2. a) 16 Ω b) 4A c) 24V ; 12V ; 28V d) 40 Volts 3. a) 25 Ω b) 3A c) 18V ; 9V ; 12V ; 21V ; 15V ; d) 48 Volts 4. The current through the first two resistors would be 8⅓ A. The current through the last three resistors in the circuit would be 0 A. Can you explain why?

6 Recognizing Resistors in Parallel Resistors are wired in parallel if both ends of the resistors are wired to common junctions without interruption. R 1 R 2 R 1 R 2 /\/\/\ /\/\/\ /\/\/\ /\/\/\ R 3 R 3 _/\/\/\ /\/\/\ For the circuit on the left the left ends of R 1 and R 3 are wired to a common point but not the right ends. The right ends of R 2 and R 3 are wired to a common point but not the left. There are no parallel resistors for the circuit on the left. For the circuit on the right the left ends of R 2 and R 3 are wired to common junctions. The right ends are also wired directly to a common junction. Since both ends go directly to common junctions then R 2 and R 3 are wired in parallel. Rules for Resistors in Parallel There are only three rules for resistors in parallel that are important. Consider the circuit below to demonstrate each of the three rules. EMF= 36 V 12 Ω 9 Ω 18 Ω 4th. 5th. 6th. Total Resistance Rule: (R TOTAL ) -1 = R R R The reciprocal of the total resistance of resistors in parallel is the sum of their reciprocal. In the above example the total resistance is 4 Ω. The keystrokes are then enter. This gives you 1/R TOTAL. Now hit the x -1 key to flip your number again. Total Voltage Rule: V TOT = V 1 = V 2 = V 3 = All objects in parallel have the same voltage difference. Since each of the above resistors is wired parallel to each other and to the battery they each have 36 V across them. Total Current Rule: I TOT = I 1 + I 2 + I 3 + The individual currents through each resistor will add to the total current flowing out of the battery. Use I = V/R. You find 3A flowing down the 12ohms branch. You notice 4A flowing down the 9ohms branch. There are 2A flowing down the 18ohms branch. The total current coming out of the battery is 3A+4A+2A = 9A. This is in agreement with I TOT =36Volts/4Ω. An ammeter wired into the circuit at in the above circuit would read 6A. This can be explained by recognizing that of the 9A leaving the battery 3A will go down the first branch leaving 6A to continue across the top of the circuit. Also, you could argue that the 2A and 4A both must flow through the ammeter to get to their respective resistors. Use the previous steps to find a) the total resistance of a circuit, b) the total current flowing out of the battery and c) the current through each resistor and d) current flowing through the indicated point for each of the circuits below.

7 Circuit #4 EMF= 48 V 16 Ω 8 Ω 12 Ω Circuit #5 EMF= 60 V 20 Ω 5 Ω 15 Ω Circuit #6 EMF= 72 V 12 Ω 9 Ω 18 Ω 8 Ω If a wire is clipped into the circuit between the 9 ohms branch and the 18 ohms branch what happens to all of the readings in circuit 3? Answers 4. a) 3.69 Ω b) 13A c) 3A ; 6A ; 4 A d) 10 Amps 5. a) 3.16 Ω b) 19A c) 3A ; 12A ; 4 A d) 16 Amps 6. a) 2.67 Ω b) 27A c) 6A ; 8A ; 4A ; 9A ; d) 21A by 1 st arrow and 13A by 2 nd arrow 7. The current through all resistors would drop to zero. The current coming out of the battery is approaching I = which destroys the battery. Can you explain why?

8 Lesson 3-11 Multiple Resistor Circuits II In the previous lesson pure parallel or pure series circuits were explored. In this lesson we will look at circuits which are part series and part parallel. These circuits are known as combination circuits. Our goal is the same as before. Find the currents through each resistor and the potential difference across each resistor. Before we start one must review some previously discussed items. Series Circuits Components are in series when there are no junctions to separate them. A lack of junctions causes the currents to be identical for all objects in series. R total = R 1 + R 2 + R 3 + Parallel Circuits Components are parallel if both ends are wired to common points or common wires. All objects in parallel have the same potential difference. 1/ R total = 1 / R 1 + 1/ R 2 + 1/ R 3 + Kirchoff s Circuit Rules The sum of the voltage differences around any closed loop must add to zero. Positive potential differences occur if you cross a battery from negative to positive or if you cross a resistor against the current. The sum of currents flowing into a junction equals the sum of currents flowing out of a junction. Combination Circuits There are two main steps to solving these type problems. For part I use the rules for parallel and series resistors to reduce the circuit to a battery with an equivalent resistance. For part II, start with Ohm s Law to get the total current and then attack the circuit using Kirchoff s Rules. Instead of learning both parts at the same time practice each part separately until you have that part down. Circuit I 5Ω Show that R equiv = 8 3/7 Ω /\/\/\ 59 V 6Ω 8Ω

9 Circuit 2 3Ω R equiv = 4 Ω /\/\/\ Volts Ω 9Ω --- Circuit 3 3Ω 5Ω R equiv = 9Ω /\/\/\ /\/\/\ 54 Volts 18Ω 4Ω Circuit 4 7 Ω 5 Ω R equiv = Ω /\/\/\ /\/\/\ 68 Volts Ω 2Ω 6Ω

10 Circuit 5 9 Ω R equiv = 3.5 Ω /\/\/\ 42 Volts Ω Ω 4Ω /\/\/\ 5Ω

11 Lesson 3-12 Capacitors Now that you have successfully mastered circuits containing resistors it is time to branch out into other circuit components. In this unit you will look at an electronic device known as a capacitor. In contrast to a resistor, the capacitor temporarily removes electrical energy from a circuit and returns that energy at a later time. The first lesson focuses on a single capacitor and a single battery. The last lesson studies the effects of a circuit with both resistors and capacitors. A capacitor is nothing more than two parallel plates that are separated by a nonconducting gap. Usually the gap is filled with something like glass. For this unit we will assume that the gap is filled with a vacuum unless otherwise stated. A capacitor can store charge and energy. The ability of the capacitor to store charge is known as the capacitance. Capacitance depends upon the stuff between plates, the surface area of plates (A), and the gap distance between the plates (d). d A figure 1 Aε o With a larger surface area the capacitor C = where ε o = 8.85E -12 f/m holds more charge. With a smaller gap d distance the capacitor holds more charge. If the gap between the plates is filled with something then the epsilon naught will change value. Capacitance is measured in farads which is short for Michael Faraday. A millifarad capacitor can hold a lot of charge and can be potentially lethal. When the ends of the capacitor are connected to a battery the plates will almost instantly become polarized. Equal amounts of positive and negative charge appear on opposite plates. An electric field shows up between the plates going from the positive to the negative plate. Often asked questions are How much charge, how much energy, and how strong is the electric field between the plates? figure 2

12 In the above diagram positive charges are on the top plate, negative charges on the bottom plate and field lines are red. The goals of this lesson are to answer the questions preceding figure 2. The amount of charge that collects on either plate depends on both capacitance and the size of the battery. If you double the size of the battery you will double the amount of charge as shown in the equation below. q = C V or q = CV The V in the second form is merely the voltage difference between the plates. It is customary in circuits to drop the deltas out of the equations. Example #1 Two, circular, parallel plates each have a radius of 10 centimeters. The plates are separated by a distance of 4 millimeters. If the plates are connected to a 9 Volts battery then how much charge will each plate hold? C = π(0.10m) E -12 farads/m 4 E -3 m =6.95 E -11 farads or 69.5 picofarads q = CV = (69.5pf) ( 9 Volts ) = E -10 Coulombs or 626 pc The electric field strength can be determined from an equation that you already know. E = - V / d The previous example has an electric field of strength E = 2250 Volts per meter. The energy stored in a capacitor is found by the work needed W = ½ q V or ½ C V 2 to charge the capacitor. Example #2 A 9 nanofarads capacitor has equivalent surface area on each plate of 78 m 2. The capacitor is hooked up to a 12 Volts power supply. Find a) the gap distance between the plates, b) charge on either plate, c) electric field strength between the plates and d) the energy stored in the capacitor. a) from C = Aε o /d you get d = Aε o /C d = 78 m E -12 farads/m 9 E 9 farads d = m or 7.67 cm b) from q = CV = 9nanofarads 12Volts q = ±108 nanocoulombs c) from E = - V / d = 12 Volts / m E = Volts/meter or N/C

13 d) from W = ½ qv = ½ 108 nc 12Volts W = 648 nanojoules Drawing Capacitor Circuits The symbol for a capacitor is two parallel lines or one straight and one curved line if the capacitor has a preferred negative plate. The circuit for example #2 is drawn below. 12 V C = 9 nf Homework 1. Two, parallel plates are separated by a distance of 2 mm. Each plate has a surface area of 120 m 2. A 400 volts power supply is connected across the plates. Find a) capacitance of parallel plates, b) charge on the positive plate, c) electric field between the plates, d) energy stored in the capacitor. 2. A capacitor has an equivalent surface area of 300 m 2 per plate with a capacitance of 332 µfarads. The capacitor is connected to a 30 Volts power supply. Determine a) the plate separation distance, b) charge on either plate, c) electric field strength between the plates, d) work done by power supply to charge the capacitor. 3. A parallel plate capacitor has a capacitance of 48 nanofarads with a gap distance of 360 micrometers. The capacitor is connected to a 60 Volts battery. Calculate a) the surface area of one of the parallel plates, b) charge on the positive plate, c) energy stored in the charged capacitor, d) electric field between the plates. 4. The maximum electric field that can be sustained in air is 3 million volts / meter. Any value higher than three million volts per meter will cause the air to become a conducting path, ZOT! Suppose two, parallel plates are separated by a distance of 2 millimeters and have a surface area of 280 m 2 each. Find a) the maximum electric potential difference that the capacitor can sustain before sparking occurs, b) maximum charge and c) maximum stored energy before breakdown occurs. 5. The bottom of a cloud has a surface area of 4E6 m 2 and is located 3000 meters above the surface of the earth. Treat the bottom of the cloud and the earth as parallel plates just before electric breakdown occurs. In other words assume that the field between earth and cloud is at 3 millions volts per meter. Calculate a) the potential difference in volts between earth and cloud, b) maximum charge that can be transferred between cloud and ground, c) capacitance between cloud and earth, and d) maximum energy that could be transferred in a single bolt of lightning. 1. a) 531 nanofarads b) 212 µc, c) 2 E +5 Volts/meter, d) 42.4 mj 2. a) 8 µm, b) ±9.96 mc, c) 3.76E+6 V/m, d) millijoules 3. a) 1.95 m 2, b) 2.88 µc c) 86.4µJ d) 1.67E5 V / m 4. a) 6 kvolts, b) 7.43 mc, c) 22.3 J

14 5. a) 9E9 volts, b) C c) 11.8 nanofarads, d) 478GJ Lesson 3-13 Suppose that a circuit contains both resistors and capacitors together. How do you analyze a circuit of that nature? Consider the example below. 3 Ω A /\/\/\ 12 V S B 8mf We assume that the capacitor is always empty at the beginning of a problem unless stated otherwise. Initial Values As soon as switch S, is thrown to position A the circuit is complete. The 8 millifarad capacitor begins to collect charge. An empty capacitor on any branch of a circuit can be replaced by pure wire during initial conditions. For this example the current just after the switch is thrown to position A is found using I = V/R = 12 V / 3 ohms or 4 amps. As soon as the capacitor begins to collect charge the voltage of the capacitor starts to rise. At the same time the voltage on the resistor is falling which means less current. Voltage vs. time graphs for both capacitor (C ) and resistor (R) are shown below: RC Circuits As you can tell by the graph the capacitor will slowly cutoff voltage to the resistor. In the beginning there is no charge and no voltage across the capacitor according to q=cv. The entire 12 Volts shows up across the resistor. The following conditions are true. q = 0 C, V C = 0 Volts; I = 4 A, V R = 12 V As q the voltage across the capacitor, V C. Simultaneously the voltage across the resistor, V R, causing the current to also drop. Eventually the entire 12 Volts will show up across the capacitor leaving no voltage for the resistor according to the loop rule. The following conditions will eventually take place. q = 96 mc, V C = 12 V; I = 0A, V R = 0 V where the charge is found using q = CV. The final conditions are known as steadystate conditions. Steady-State Conditions During steady state conditions the capacitor is assumed to be full charged. A fully charged capacitor acts like an open switch cutting off current flow to the entire branch where it is located. With no current flowing down the branch of a fully charged capacitor the V R s will be zero. Any closed loop formed with the capacitor branch will have all of the potential of the branch showing up as V C. Discharging Circuits After the steady-state condition is reached in our example circuit the switch can be thrown from position A to position B. A discharging capacitor acts like a battery with a decaying voltage.

15 For the above circuit the initial current after the switch reaches position B will be 4 Amps. Your goal in this section is to be able to list the initial values, steady state values and initial discharging values for charge, current, voltage across the resistor and voltage across the capacitor. We will look at four examples. This is the final goal for this unit.

16 40V 5 Ω A /\/\/\ 24 V S 8 Ω /\/\/\ B 3Ω 6µf 4Ω 5nf For the instant just after the switch is thrown to position A determine the current through each resistor, the voltage across each resistor and the voltage across the capacitor. Assume that the capacitor is initially uncharged. V C = 0 since q = VC = V(6µf) = 0 I 3 & V 3 both are zero since they do not form a complete loop with the battery. V 5 = 40 Volts since the 5 ohm resistor along with the capacitor form a closed loop with the battery & V C =0 already. Using V = IR on the 5 ohms resistor gives an initial current of I 5 = 8 Amps. Find the values again after steady-state conditions have been reached and the capacitor is fully charged. V 3 & I 3 are same as before for same reason as before. V 5 and I 5 are zero since now all of the 40 Volts appears across the capacitor rather than the resistor. See graph from last page for V R. V C = 40 Volts due to the loop rule on the outer loop. Using q = VC on the fully charge capacitor gives q = 40V 6µf = 240µC. The switch is thrown from the steady-state conditions at position A to position B to allow the capacitor to discharge. Capacitor acts like 40V battery pushing current through a total of 8Ω. As a result both resistors experience a 5 A current initially. V 5 = I 5 R 5 = 25 Volts initially V 3 = I 3 R 3 = 15 Volts initially All of these values will decay e -kt Assume capacitor is initially uncharged before switch S is thrown to horizontally closed position. Find the initial value just after S is closed. This time the 4Ω resistor forms a closed loop with the battery, V 4 =24V. Using V=IR gives I 4 = 6Amps. The outer loop starts with q=0 & V C =0 so all of the 24 Volts in the loop starts out at the 8 ohm resistor, V 8 = 24 V. Using V=IR gives I 8 = 3 Amps. The initial current flowing out of the battery is 6A + 3A = 9 Amps. Find values after a steady-state condition with a fully charge capacitor is reached. The conditions for the 4ohm resistor remain as before since it still forms an independent loop with the battery. The charged capacitor has now shut down current flow into its branch so that I 8 = 0Amps and V 8 = 0Volts. The outer branch still forms a loop with the battery however. All 24 Volts shows up across capacitor, V C =24V. Using q = CV = (5nf)(24V) = 120 nc. If switch S is reopened after steady-state conditions the capacitor can discharge back through the other loop. Initial charge on the capacitor gives it 24V across the plates. The 24V capacitor now acts like a battery discharging through a 12Ω loop. The current through both resistors is I = V/R = 24V /12Ω = 2A. Using V=IR gives 2A 8Ω = 16V = V 8 Similarly, V 4 = 8V All of these values will rapidly decay and only exist at instant S is opened.

17 Each of the following circuits will show ammeters and voltmeters to read the current flowing through a branch or the potential difference across a component on a branch. Assume that all capacitors are initially empty. Find meter readings for each circuit a) just after a switch is closed, b) steady-state values with maximum charge on capacitor and c) discharging values immediately after the switch is reopened.. 36 V V 2 4 Ω /\/\/\ A 1 8 Ω V 1 3µf V 3 30V V 1 6 Ω /\/\/\ A 1 10Ω V 2 4nf A 3 A 2 Initial values when switch is raised. Initial values when switch is lowered. A 1 = Amps V 1 = Volts A 1 = A A 2 = A A 3 = A V 2 = Volts V 3 = Volts V 1 = Volts V 2 = Volts Steady-State Conditions after switch has been raised for a long time? A 1 = Amps V 1 = Volts Steady-State Conditions? A 1 = A A 2 = A A 3 = A V 2 = Volts V 3 = Volts V 1 = Volts V 2 = Volts q max = microcoulombs Values immediately after switch is lowered to contact central branch? A 1 = Amps V 1 = Volts q max = nanocoulombs Values just after switch is reopened? A 1 = A 2 = A 3 = V 2 = Volts V 3 = Volts V 1 = Volts V 2 =

18 1. a) A 1 =9A, V 1 = 0V, V 2 =36V V 3 = 0V b) A 1 =0A, V 1 = 36V, V 2 = V 3 = 0V Maximum charge is 108 µc. c) A 1 = 3A, V 1 = +36V V 2 = -12V V 3 = -24V 2. a) A 1 = 5 A, A 2 = 3 A, A 3 = 8 A V 1 = 30 V V 2 = 30 V b) A 1 = 0 A, A 2 = 3 A, A 3 = 3 A V 1 = 0V, V 2 = 30 V Max charge = 120 nc c) A 1 = A 2 = A, A 3 = 0A V 1 = V, V 2 = V This concludes unit 302 on circuits. Expect a practice test tomorrow and a unit test very soon after the practice test.