Lesson#23 Topic: Simple Circuits Objectives: (After this class I will be able to) 1. Explain the difference between wiring light bulbs in series and in parallel 2. Describe EMF and terminal voltage of batteries. 3. Describe how the difference between EMF and terminal voltage can be treated as internal resistance of the battery. 4. Find voltage, current, and resistance for a simple circuit 5. Explain the loop rule for a single loop. 11/13/06 Assignment:

Warm Up ► A simple series circuit consists of three identical lamps powered by a battery. When a wire is connected between points a and b. 1. What happens to the brightness of lamp 3? 2. Does current in the circuit increase, decrease, or remain the same? 3. What happens to the brightness of lamps 1 and 2? 4. Does the voltage drop across lamps 1 and 2 increase, decrease, or remain the same? 5. Is the power dissipated by the circuit increased, decreased, or does it remain the same? a b

Series vs. Parallel Circuits ► Light bulbs wired in series are lined up one right after the other. ► The current throughout the circuit remains the same. ► The circuit makes a single loop from the positive terminal of the battery to the negative terminal. ► Light bulbs wired in parallel are each individually connected to the battery terminals. ► The current is then split and divided amongst each light bulb.

Series vs. Parallel Circuits ► The voltage drop across identical light bulbs in series is divided evenly among each bulb. ► The voltage drop across identical light bulbs in parallel is equal to the original voltage of the battery. ► Batteries wired in series add voltage (or subtract if are reversed) ► Batteries wired in parallel do not add. ► Two batteries in parallel would light the bulb about the same as one battery by itself. ► Possibly only slightly brighter due to internal resistance.

EMF ► Batteries are rated by the voltage difference between terminals, but when current begins to flow between terminals, the voltage drops. ► This is similar to what happens in water pipes when you turn on a faucet. ► The water pressure is larger when the faucet is closed than when the water is running. ► The plumbing cannot maintain the same pressure as the water is removed. ► The more water that flows (current) the lower the pressure becomes. ► The voltage across the battery terminals goes down as current is drawn. ► The more current drawn, the more the drop in voltage.

EMF ► Technically, the chemical reactions in the battery responsible for producing the electric potential energy cannot proceed quickly enough to maintain the potential difference. ► The initial voltage that the battery is rated at for when there is no current is its EMF. ► The actual voltage between terminals when the circuit is closed is the terminal voltage. ► The more current drawn from the battery, the smaller the terminal voltage.

Internal Resistance ► What is typically done when quantifying the properties of circuits is to measure the internal resistance of the battery. ► The difference in voltage between the EMF and the terminal voltage can be treated as a tiny resistor located within the battery. ► We can then say that the EMF is the actual voltage of the battery. ► And that there is an internal resistance that causes a voltage drop as soon as the circuit is closed. ► This explains why two batteries in parallel will light a bulb brighter than a single battery in series. ► This is because when used in parallel, less current needs to be drawn from each battery. ► With less current, the internal resistance of each battery has less of an effect.

Internal Resistance ► A typical fresh AA dry cell has an EMF of 1.5V and an internal resistance of.31ohms 1. What is the terminal voltage of the battery if the battery is hooked up in series to a resistor such that a current of 58mA is measured through the resistor? 2. What is the resistance of the resistor? 3. Suppose the resistor was replaced by a wire of negligible resistance. What current will flow through the wire?

Loop Rule ► If a circuit contains a single battery and a single resistor it is simple to find the current through the circuit. ► For circuits with multiple elements a loop rule needs to be applied. ► This rule extends from the idea that the voltage between a point and itself is zero. ► As we go around a complete loop in a circuit (starting and ending at the same point), we find that each element is associated with a voltage across it. ► Loop Rule: The sum of all of the voltages across each element in the loop as you go around the loop should equal zero.

Loop Rule Example ► Find the current in the following circuit. 8Ω8Ω 2Ω2Ω 6V

Loop Rule Example ► Find the current in the following circuit. 8Ω8Ω 2Ω2Ω 6V 16V

Lesson #24 Topic: Lab: Simple Circuits Objectives: (After this class I will be able to) 1. Create a simple series circuit with light bulbs 2. Measure the voltage drop and current through a light bulb using a multimeter. 3. Create a simple parallel circuit with light bulbs 4. Compare the changes in voltage and current in a series circuit to a parallel circuit. 5. Mechanically create the power needed to generate the electricity needed to light four bulbs and make note of the difference in torque when bulbs are removed. 11/9/06 Lab Task: Complete the Hewitt lab #91. Make sure all tables are completed and questions are answered. Assignment: Lab due at the end of the period.

Lesson #25 Topic: Multi-loop Circuits Objectives: (After this class I will be able to) 1. Describe and draw a schematic of a multi-loop circuit 2. Pick out single loops from a multiple loop schematic 3. Write voltage equations for each individual loop 4. Solve for multiple unknown currents in different sections of a circuit 11/16/06 Warm Up: Draw a circuit with a 12V battery and 3 resistors wired in parallel with one another. If the resistance of each resistor is 2ohms, 4 ohms, and 6ohms, can we say that the current through each section is the same? How can we predict the current in each section? Assignment: Holt Physics p757 #43 b, c, d, e

Multi-loop Circuits ► Most real circuits have multiple loops and are much more complicated than the light bulb circuits we have been dealing with. ► The complication arises when there are different size resistors located at within different loops. ► This means that the current will be different for different sections of the circuit. ► To solve for the current in each section, we will have to create multiple loop equations to solve for the multiple unknown variables. ► To create these equations we will follow the loop rule, or specifically: ► Kirchoff’s loop rules

Multiple Loops ► Consider the following circuit. ► Pick out two loops seen within this circuit. ► Then use the loop rule to write an equation for each individual loop. ► For this loop there are three unknowns, so you should have at least three equations. ► Notice that there are three independent loops in this circuit. The top loop, the bottom loop, and the outer loop. 2Ω2Ω 8Ω8Ω 4Ω4Ω 6V 16V I top = ? I mid = ? I bot = ?

Junction Rule ► The loop rule can only be used to get two of the equations. ► In order to get the third independent equation we need to examine a point in the circuit where the current is split or comes back together. ► Because current is conserved, we can add up the current from two divided sections to obtain the current from the original source. 2Ω2Ω 8Ω8Ω 4Ω4Ω 6V 16V I top = ? I mid = ? I bot = ? I top + I mid = I bot

Solving for multiple unknowns ► Use the following schematic and direction of current to create two more equations besides the junction rule, and solve for each of the unknowns using algebraic substitution. 2Ω2Ω 8Ω8Ω 4Ω4Ω 6V 16V I top = ? I mid = ? I bot = ? Equations:

Solution

Practice A simple circuit is set up as shown. If the ammeter reads 1A, what is the resistance R? 4Ω4Ω 1Ω1Ω 2Ω2Ω R A 7V 6V

Lesson #26 Topic: Lab: Predicting Resistance Objectives: (After this class I will be able to) 1. Build a circuit from a schematic 2. Correctly wire a multimeter to measure the current in a section of wire. 3. Write equations and solve for multiple unknown variables 4. Compare the calculated prediction of resistance to the measured resistance of a light bulb. 11/29/06 Lab Task: In addition to the usual lab report, complete the calculations and questions found on the lab sheet. Assignment: Lab Report due tomorrow. DC Exam Review due Friday.

Lesson #27 Topic: Simplifying Circuits Objectives: (After this class I will be able to) 1. Add resistance of resistors that are wired in series 2. Add resistance of resistors that are wired in parallel. 3. Simplify a complex circuit to a simple circuit by calculating total resistance. 11/30/06 Warm Up: If you add resistors to a circuit wired in series, will the overall resistance go up or down? What if the circuit and the added resistors were wired in parallel with each other? Assignment: Holt Physics p755,756 #23, 24, 25, 26, 33

Adding Resistance in Series ► When you add resistors to a circuit in series, the overall resistance of the circuit will increase ► This is because the same current is forced through each resistor ► The voltage drop across each resistor is the total voltage divided by each resistor. ► If the resistance of each resistor is constant, and the voltage drop per resistor goes down with each resistor added, then the current must drop as well. ► Hence, as you add bulbs to a series circuit, bulbs get dimmer. ► The total resistance of a series circuit is the simple addition of the resistance of each individual resistor.

Practice ► Find the total resistance in the circuit. ► Find the total current drawn from the battery. 1Ω1Ω 5Ω5Ω 2Ω2Ω3Ω3Ω4Ω4Ω 7Ω7Ω6Ω6Ω 7V

Adding Resistance in Parallel ► When you add resistors to a circuit in parallel, the overall resistance of the circuit will decrease. ► This is because the current is split through each resistor. ► The voltage drop across each resistor is the same as the total voltage of the circuit. ► If the resistance of each resistor is constant, and the voltage drop per resistor remains the same with each resistor added, then the current drawn from the battery must increase. ► If the voltage is held constant and the overall current increases, then the resistance must drop as bulbs are added. ► The total resistance of a parallel circuit is the inverse of the addition of the inverses of each individual resistor.

Practice ► Find the total resistance in the circuit. ► Find the total current drawn from the battery. 1Ω1Ω 2Ω2Ω 3Ω3Ω 6Ω6Ω 7V

Simplifying Circuits ► With these addition rules, we can simplify circuits to a single battery and resistor. ► This is useful for finding the total voltage, resistance, or current drawn from a single battery in a circuit. ► This cannot be used to solve for the current within sections of a parallel circuit.

Practice Find the equivalent resistance of circuit and the current drawn from the battery. 4Ω4Ω 1Ω1Ω 2Ω2Ω 8 Ω 8Ω8Ω 6V 4Ω4Ω