AP Physics Chapter 18 Basic Electric Circuits

Chapter 18: Basic Electric Circuits 18. 1 Resistances in Series, Parallel, and Series-Parallel 18. 2 Multiloop Circuits and Kirchoff’s Rules 18. 4 Ammeters and Voltmeters

Homework for Chapter 18 • Read Chapter 18 • HW 18: p. 592 -596: 7, 10, 11, 25, 27, 32, 44, 45, 72, 73.

Answer: Each appliance needs its own path so that it does not affect the operation of other appliances. As you add more appliances to a parallel circuit, the total current increases.

18. 1 Resistances in Series, Parallel, and Series-Parallel

equivalent series resistance (Rs) connected in series in a circuit: R s = R 1 + R 2 + R 3 + … = i. R i - when resistors are on Gold Sheet • The equivalent series resistance is the algebraic sum of the individual resistances. • The equivalent series resistance is larger than that of the largest resistor in the series combination. • In a series circuit, the current is the same through all resistors: I = I 1 = I 2 = I 3 = … • The total voltage is the sum of the voltages of the individual resistors: V = V 1 + V 2 + V 3 +… = i Vi = i I Ri

Resistors (Light Bulbs) in Series a) When resistors are connected in series, the current through each of them is the same. The sum of the voltage drops across each of them is equal to the voltage of the battery. b) The equivalent resistance Rs of the resistors in series.

equivalent parallel resistance (Rp) connected in parallel in a circuit: 1 1 1 R p = R 1 + R 2 + R 3 + … = i. R i - when resistors are on Gold Sheet • The reciprocal of the equivalent parallel resistance is equal to the sum of the reciprocals of the individual resistances. • The equivalent parallel resistance is less than that of the smallest resistor in the parallel combination. • In a parallel circuit, the voltage is the same across all resistors: V = V 1 = V 2 = V 3 = … • The total current is the sum of the currents of the individual resistors: I = I 1 + I 2 + I 3 +… = i Ii

Resistors (Light Bulbs) in Parallel a) When resistors are connected in parallel, the voltage drop across each of the resistors is the same. The current from the battery divides among the resistors. b) The equivalent resistance Rp of the resistors in parallel.

• Series connections provide a way to increase total resistance. • Parallel connections provide a way to decrease total resistance. Example 18. 1: Solution:

Example 18. 2: Solution:

Series – Parallel Resistor Combinations (5) Use a V-I-R chart. V R 1 R 2 Total I R

Example 18. 3: Solution:

Example 18. 4: For the circuit shown, complete a V-I-R table. R 1 = 5 R 2 = 7 12 V R 3 = 2

Check for Understanding 1. Which of the following is always the same for resistors in series? a) voltage b) current c) power d) energy Answer: b 2. Which of the following is always the same for resistors in parallel? a) voltage b) current c) power d) energy Answer: a

Check for Understanding 3. Are the voltages across resistors in series generally the same? Answer: No 4. Could they be so in certain cases? Answer: Yes, if all the resistors are equal 5. Are the currents in resistors in parallel generally the same? Answer: No 6. Could they be so in certain cases? Answer: Yes, if all the resistors are equal

Check for Understanding 7. Three resistors that have values of 5 , 2 , and 1 are connected in series to a battery. Which gets the most power, and why? Answer: The 5 resistor gets the most power because: P=IV all the resistors get the same current the voltage drop is the highest for the largest resistor (V=IR) 8. Three resistors that have values of 5 , 2 , and 1 are connected in parallel to a battery. Which gets the most power, and why? Answer: The 1 resistor gets the most power because: P=IV all the resistors have the same voltage it gets the most current

Answer: Five can safely be used. Each cooker draws 7. 5 amps.

18. 2 Multiloop Circuits and Kirchoff’s Rules

Kirchhoff’s rules - a general method for analyzing multiloop circuits • developed by the German physicist Gustav Kirchhoff (1824 -1887) • rules are based on the conservation of charge and energy junction or node - a point in a multiloop circuit at which three or more connecting wires are joined together. • current either divides or merges at a junction branch - a circuit path between two junctions, and may contain one or more elements.

A Multiloop Circuit a) Always reduce a circuit as far as possible before analyzing it. This circuit can be reduced to the one shown in b. b) At junction A the current divides, and at B the current merges. The paths between junctions A and B are called branches. How many branches does this circuit have? Answer: 3

current in = current out (Junction Theorem) current in is positive and current out is negative

Sign Convention for Kirchoff’s Rules a) The voltage across a battery is taken to be positive (+) if a battery is traversed from the negative to the positive terminal. The voltage across a battery is taken to be negative (-) if a battery is traversed from the positive to the negative terminal. b) The voltage across a resistor is taken to be negative (-) if the resistance is traversed in the direction of the assigned branch currrent. The voltage across a resistor is taken to be positive (+) if the resistance is traversed in the direction opposite that of the assigned branch currrent.

Application of Kirchhoff’s Rules

Example 18. 5: Solution:

I 2 I 1

Example 18. 6: Solution:

Check for Understanding 1. A multiloop circuit has more than one a) b) c) d) junction branch current all of these Answer: d 2. By our sign convention, if a resistor is traversed in the direction of the current in it a) the current is negative b) the current is positive c) the voltage is negative d) the voltage is positive Answer: c

Check for Understanding 3. If you traversed a battery from the negative to the positive terminal, is the voltage positive or negative? Answer: positive

Hint: the 60 W bulb has a higher resistance than the 75 W bulb. Answer: The greater the power, the brighter the bulb. In the series circuit, the current is the same for both bulbs. Since P = I 2 R, the greater the power. Therefore, the 60 W bulb burns brighter. In a parallel circuit, the voltage is the same for both bulbs. Since P = V 2/R, the greater the R, the lower the power. Therefore, the 75 W bulb burns brighter.

Brightness of a Bulb • The brightness of a bulb depends solely on the power dissipated by the bulb. You can remember that from your own experience – when you go to the store to buy a light bulb, you as for a 60 watt or 75 watt bulb. Watt is the unit of power. • A 75 watt bulb is brighter than a 60 watt bulb, but be careful: a bulb’s power can change depending on the current and voltage it’s hooked up to. • Power is given by the equations P = IV, P = I 2 R, P = V 2/R. • The resistance of a bulb can be calculated by hooking it up to a 120 V source: For a 60 W bulb: I = P = 60 W = 0. 5 A so, R = V = 120 V = 240 V 120 V I 0. 5 A For a 75 W bulb: I = 75 W = 0. 625 A so, R = 120 V = 192 120 V The resistance of a bulb is a property of the bulb itself, and never changes.

Question: A light bulb is rated at 100 W in the United States, where the standard wall outlet voltage is 120 V. If this bulb were plugged in in Europe, where standard wall outlet voltage is 240 V, which of the following would be true? a) The bulb would be ¼ as bright. b) The bulb would be ½ as bright. c) The bulb’s brightness would be the same. d) The bulb would be twice as bright. e) The bulb would be four times as bright. Answer: e The resistance does not change because it is a property of the bulb itself. It will not vary no matter what the bulb is hooked up to. Since P = V 2/R, if the voltage is doubled, the power is quadrupled.

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18. 4 Ammeters and Voltmeters

ammeter - a low resistance device to measure current The circuit symbol for an ammeter is a circle with an A inside it. • Current is the same for any resistors in series with one another. So, if you’re going to measure the current through a resistor, the ammeter must be in series with the resistor whose current you want to measure. • The ammeter does not appreciably affect the current measurement due to its low resistance.

• In this circuit, ammeter A 1 measures the current through resistor R 1, while ammeter A 2 measures the current through resistor R 2. Question: Is there a way to figure out the current in the other three resistors based only on the readings in these two ammeters? Answer: The current through R 5 must be the same as through R 1, because both resistors carry whatever current came directly from the battery. The current through R 3 and R 4 can be determined from Kirchhoff’s junction rule: subtract the current in R 2 from the current in R 1 and that’s what’s left over for the right-hand branch of the circuit.

voltmeter - a high resistance device used to measure voltage. The symbol for a voltmeter is a circle with a V inside it. • Voltage is the same for any resistors in parallel with each other. So, if you’re going to measure the voltage across a resistor, you must put the voltmeter in parallel resistor. with the • Since the voltmeter has high resistance, it does not appreciably affect the voltage across and element.

In the circuit, the meter labeled V 2 measures voltage across the 100 resistor, while the meter labeled V 1 measures the potential difference between points A and B (which is also the voltage across R 1). • Both ammeter and voltmeter functions can be combined in a single multimeter.

Check for Understanding 1. A voltmeter has a a) large capacitance b) large resistance c) small resistance d) small capacitance Answer: b 2. To measure the current through a circuit element, an ammeter is connected a) in series with the element b) in parallel with the element c) between the high potential side of the element and ground d) none of these Answer: a

Check for Understanding 3. An ammeter has a a) large resistance b) large capacitance c) small resistance d) small capacitance Answer: c 4. To measure the voltage across a circuit element, a voltmeter is connected a) in series with the element b) in parallel with the element c) between the high potential side of the element and ground d) none of these Answer: b HW 18. A: p. 592 -596: 7, 10, 11, 25, 27, 32, 44, 45, 72, 73.