Physics 25 Chapters Dr. Alward

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1 Physics 25 Chapters Dr. Alward Electric Circuits Batteries store chemical energy. When the battery is used to operate an electrical device, such as a lightbulb, the chemical energy stored in the battery is converted to electric energy, which then is converted to heat and light energy in the lightbulb filament. In a 9-volt battery, the new battery starts off with about 20,000 joules of chemical energy. A car battery, depending on its rating, might store two million joules. Electric Current In the figure below, which constitutes what we call an electrical circuit, electrons produced from chemical reactions in the battery leave the negative terminal of the battery and circulate around the circuit. The current from the battery is the absolute value of the amount of charge exiting one terminal, and entering the other terminal each second: Current Symbol: I Units: coulombs/second (C/s) I = Q/t I = 3.2 C/10 s = 0.32 C/s Change units name: Let 1.0 ampere (A) = 1.0 C/s The absolute value of this charge is Q = 2 x x -1.6 x C = 3.2 C I = 0.32 A One-tenth of an ampere flowing for two seconds through a human can cause death. Ten milliamperes can paralyze the muscles. Typical lightbulb currents are about one ampere. 1

2 Battery Electromotive Force and Battery Energy In our earlier study of air pressure in Physics 23, it was noted that the greater the air pressure difference between two points, the greater will be the speed of the wind that blows from the higher toward the lower pressure area. Batteries have a kind of built-in electrical pressure difference built in across its terminal. We call this pressure electromotive force, often abbreviated EMF or emf. Symbol: ε Units: volts (V) Another name for electromotive force: battery voltage The greater the battery voltage, the greater will be the current out of the battery. The number of joules of chemical energy expended each second during the operation of the battery is called the power output of the battery. Power output is symbolized as P, and its units are J/s, or watts (W). I = current out of the battery ε = battery voltage P = Iε Example A: The energy stored in a new 9-volt battery is about 20,000 J. The current out of such a battery is A. After how many hours will this battery be drained of all of its energy? Solution: P = Iε = (9) = watts = J/s Example B: What is the power output in milli-watts of a battery whose voltage is 2.0 volts, and out of which flow 7.5 milli-amperes? P = 2.0 (7.5 x 10-3 ) = 15.0 x 10-3 watts = 15.0 milli-watts Each second of operation, joules of chemical energy are consumed. time = 20,000 J /0.135 J/s = 148,148 s = 41 hours 2

3 Resistance Objects through which current pass have a property called resistance, so-called because these resistors resist, restrict, or limit the flow of electrons (current) from the battery. The greater the resistance, the smaller the current from the battery. Symbol: R Units: ohms (Ω). Examples of Resistances Object R (Ω) 100 W light bulb W light bulb 240 Dry human 100,000 Wet human 1,000 1 meter copper wire mm diameter AA battery 0.10 In the simplest circuit, there is one battery and one resistor, as shown at the right. Ohm s Law: I = /R or ε = IR The symbol for a resistor reminds one of the saw-tooth shape of a lightbulb filament: Two symbols for a battery: The positive and negative signs on the symbol are usually omitted. The circuit above is represented below: Sometimes the symbol V is used to represent the emf (voltage) of a battery: Example: A 6 Ω resistor (light bulb) is connected to the terminals of a 24 V battery. What is the power output of the battery? Solution: I = /R = 24/6 = 4 A P = I = 4 (24) = 96 watts Alternative Power Output Equation P = I = ( /R) P = 2 /R = 24 2 /6 = 96 watts This power output of a battery equation only works if there is only one resistor connected to the battery. 3

4 As shown above, there are two ways to calculate the power output of a battery connected to just one resistor: P = Iε and P = ε 2 /R. Use the former if you know I and ε; use the latter if you know ε and R. Another Kind of Voltage Until now, the only voltage we ve discussed is the EMF of a battery, also called the voltage of the battery. The EMF ε appeared in the following equations: ε = IR P = Iε P = ε 2 /R (if there is only one resistor) There is a second kind of voltage, and it s the not the voltage of a battery, but instead the voltage that exists when current travels through a resistor. Because this type of voltage is measurable using the two leads of a voltage placed across a resistor, we speak of the voltage across a resistor. Everywhere an ε appears the equations at the left, which relate to batteries, place a V instead to obtain equations that related to a resistor. For example, while the voltage ε of a battery is a number printed on the battery, and it s a fixed value that never changes, the voltage across a resistor depends on how much current passes through it: ε = IR (voltage of a battery) V = IR (voltage across a resistor) P = Iε (power output by battery) P = IV (power consumed by resistor) P = ε 2 /R (power output of battery) P = V 2 /R (power consumed by a resistor) 4

5 Example A: What is the current in the resistor below, the voltage across the resistor, and the power consumed by the resistor? I = ε/r V = IR P = IV = 30/15 = 2 (15) = (2)30 = 2 A = 30 volts = 60 watts Example B: Calculate the voltages across each resistor below and the powers consumed in each. Also calculate the power produced. The power consumed by the resistor can also be calculated using P = V 2 /R, which gives P = 30 2 /15, which is 60 watts. The power produced is P = Iε = 60 watts, the same as the power consumed, as expected. 3 Ω: V = 12 volts P = 48 watts 6 Ω: V = 12 volts P = 24 watts Battery: P = Iε: P = 72 watts Note: the sum of powers consumed equals the power produced, as expected. Resistors Connected in Series Two resistors are connected in series if all of the current that goes through the first resistor has no other place to go except through the second one: what goes through one, has to go through the other. The three resistors below are in series: Resistors connected in series have the same effect on the current out of the battery as does a single resistor--called the equivalent resistor --whose resistance is the sum of the resistances. The series circuit at the left is equivalent to the one below: 5

6 Example A: What is the power output of the battery in the circuit below, and the powers consumed in each resistor? Solution: Knowing now that the current out of the battery is 2 A, we have The equivalent circuit is shown below. 5 Ω: V = 10 volts P = 20 watts 4 Ω: V = 8 volts P = 16 watts 6 Ω: V = 12 volts P = 24 watts Note that the sum of the powers consumed equals the power produced, 60 watts, as expected. I = 30/15 = 2 A P = 2 (30) = 60 watts Example B: The power output of a battery is 100 W. One of the two resistors in the circuit consumes 65 W of power. What is consumed by the other resistor? The sum of powers consumed equals the power produced: Answer: = 35 W 6

7 Another Way to Calculate the Power Consumed by a Resistor As we have already seen, the power consumed by a resistor is Example A: In the example below, the output power is 60 watts. Calculate the individual powers consumed, and then add them. P = IV If the voltage across a resistor is not given, one can calculate the power consumed this way, using Ohm s Law: P = IV = I (IR) = I 2 R R (Ω) I (A) P = I 2 R (W) = 60 W The sum of the powers consumed equals the power produced. Resistors Connected in Parallel Example B: What is the power output of the battery below? R = (3)(6)/(3+6) = 2 Ω Equivalent Circuit: I = 12/2 = 6 A P = Iε = (6)(12) = 72 watts How are the currents in the two resistors determined? See description below. 7

8 Calculating Currents at a Branch Point Branching Rule: Current = (Other/Sum) I The current through one resistor in a parallel combination with another resistor equals the ratio of the other resistance to the sum of the two resistances, times the current entering the branch point. I1 = [R2/(R1+R2)] I I2 = [R1/(R1+R2)] I I1 = (6/9)6 I2 = (3/9)6 = 4 A = 2 A 8

9 Combination of Series and Parallel-Connected Resistors Example A: The 40.0 and 20.0 Ω resistors are in series, so their equivalent resistance is 60.0 Ω. Likewise, the 70.0 and 20.0 ohms resistors are equivalent to a 90 Ω resistor. Thus, we have 60.0 Ω in parallel with 90.0 Ω. The equivalent resistance between Points A and B, therefore, is What is the current out of the battery? (Solution is provided at the right.) R = (60.0) (90.0)/ (60+90) = 36 Ω We now have the equivalent of three resistors in series: = 106 Ω Out of the battery: I = 212/106 = 2 A Example B: What are the currents in each resistor in the circuit above? 2.0 A leave the battery, so 2.0 A enter the battery, which means that the flow through the 20.0-ohm and 50.0-ohm resistors is also 2.0 A. At branch point B, 2.0 A enters. Using the other divided by the sum branching rule, we get Upper branch: (90/150) 2 = 1.2 A Lower branch: (60/150) 2 = 0.8 A 9

10 Kirchoff s Laws Kirchoff s First Law: In traveling around a closed path (loop), the sum of the signed voltages of the batteries and the resistors encountered equals zero. Example: Apply Kirchoff s Law to the loop ABEFA, then to loop BCDEB, then ABCDEFA. In traveling through a resistor with the guessed current direction arrow, the sign of the voltage IR is negative. In traveling against the resistor s current direction arrow, the sign of the voltage IR is positive. If the travel through a battery is from the negative to the positive terminal, the sign is positive; otherwise, from positive to negative the choice is negative Kirchoff s Second Law The sum of the currents at a branch point is zero. Currents entering are positive, while currents leaving are negative Arbitrarily assign unknown current names and (guess) directions for the various parts of the circuit. Note that the sum of the currents at branch point B is zero: x - y - (x-y) = Begin by picking any loop and traveling around it to create the equation for that loop. Do this for as many loops as there are unknown currents symbols. In this case, there are two: x, and y. ABEFA: -4x - 3y +12 = 0 BCDEB: -8-2(x-y) + 3y = 0 ABCDEFA: -4x - 8-2(x-y) +12 = 0 Pick any two of these equations to solve for the two unknowns x, and y. x = 1.38 A y = 2.15 A We guessed that the current was from B to C, but x-y = A, which means the current is 0.77 A, from C to B. The correct current directions are shown below. After the equations are solved, if any of the currents are negative, the wrong direction was guessed for the current. The correct value is found by removing the negative sign and reversing the current arrow. Recalculation is not necessary. 10

11 Use Kirchoff s Theorem to prove that the voltages across parallel resistors are the same. ABCD: -I2 R2 + I1R1 = 0 -V2 + V1 = 0 V2 = V1 Example: What is the current in the 4-ohm resistor? I (15) = 6 (5) I = 2 A 11

12 Internal Resistance of a Battery In electric circuit jargon, the greater the amount of current a battery is forced to provide, the greater is the burden, or load it has to carry. The smaller the resistance seen by the battery, the greater is the current, and therefore the greater is the battery s load. In the circuit above, the voltage experienced by the 1-Ω resistor is volts less than it would have been if the internal resistance of the battery were zero. We say the battery has been loaded down by volts. The battery is equivalent to an ideal battery whose emf is volts. Note that the amount by which the battery is loaded down equals the product of the current and the internal resistance: Ir = (0.20) = volts Example: Use Kirchoff s Law to show that the voltage across R is 1.667: Starting at the battery s negative terminal and moving clockwise: ε -Ir - IR = 0 IR = ε - Ir (This is actuallywhat the load gets. ) Thre promised voltage is ε = 10 volts Amount loaded down = Expected - Actual = ε - IR 12

13 Example: A nine-volt battery connected across a 0.40-ohm resistor is loaded down by 1.2 volts. What is the internal resistance of the battery? Equivalent resistance: As the work above t shows, the amount by which a battery is loaded down equals Ir, where I is the current out of the battery. Apply Ohm s Law to Get I: I = 9.0 / (0.4 + r) Ir = r [9.0 / (0.4 + r)] 1.20 = rs [9.0 / (0.4 + r)] Solve for r: r = Ω 13