Principles and Problems

Principles and Problems
PHYSICS Principles and Problems Chapter 23: Series and Parallel Circuits

Series and Parallel Circuits
CHAPTER23 Series and Parallel Circuits BIG IDEA Circuit components can be placed in series, in parallel or in a combination of series and parallel.

Section 23.1 Simple Circuits Section 23.2 Applications of Circuits
CHAPTER23 Table Of Contents Section Simple Circuits Section Applications of Circuits Click a hyperlink to view the corresponding slides. Exit

Simple Circuits MAIN IDEA Essential Questions
SECTION23.1 Simple Circuits MAIN IDEA In a series circuit, current follows a single path; in a parallel circuit, current follows more than one path. Essential Questions What are the characteristics of series and parallel circuits? How are currents, potential differences, and equivalent resistances in series circuits related? How are currents, potential differences and equivalent resistances in parallel circuits related?

Simple Circuits Review Vocabulary New Vocabulary
SECTION23.1 Simple Circuits Review Vocabulary resistance the measure of how much an object or material impedes the current produced by a potential difference; is equal to voltage divided by current New Vocabulary Series circuit Equivalent resistance Voltage divider Parallel circuit

Simple Circuits A River Model
SECTION23.1 Simple Circuits A River Model Although the connection may not immediately be clear to you, a mountain river can be used to model an electric circuit. From its source high in the mountains, the river flows downhill to the plains below. No matter which path the river takes, its change in elevation, from the mountaintop to the plain, is the same. Some rivers flow downhill in a single stream.

Simple Circuits A River Model (cont.)
SECTION23.1 Simple Circuits A River Model (cont.) Other rivers may split into two or more smaller streams as they flow over a waterfall or through a series of rapids. In this case, part of the river follows one path, while other parts of the river follow different paths. No matter how many paths the river takes, however, the total amount of water flowing down the mountain remains unchanged. In other words, the amount of water flowing downhill is not affected by the path the water takes.

SECTION23.1 Simple Circuits A River Model (cont.) How does the river shown in the figure model an electric circuit? The distance that the river drops is similar to the potential difference in a circuit. Photodisc Collection/Getty Images

SECTION23.1 Simple Circuits A River Model (cont.) The rate of water flow is similar to current in a circuit. Large rocks and other obstacles produce resistance to water flow and are similar to resistors in a circuit. Photodisc Collection/Getty Images

SECTION23.1 Simple Circuits A River Model (cont.) What part of a river is similar to a battery or a generator in an electric circuit? The energy source needed to raise water to the top of the mountain is the Sun. Solar energy evaporates water from lakes and seas leading to the formation of clouds that release rain or snow that falls on the mountaintops. Continue to think about the mountain river model as you read about the current in electric circuits.

Simple Circuits Series Circuits
SECTION23.1 Simple Circuits Series Circuits Three students are connecting two identical lamps to a battery, as illustrated in the figure. Before they make the final connection to the battery, their teacher asks them to predict the brightness of the two lamps.

Series Circuits (cont.)
SECTION23.1 Simple Circuits Series Circuits (cont.) Each student knows that the brightness of a lamp depends on the current through it. The first student predicts that only the lamp close to the positive (+) terminal of the battery will light because all the current will be used up as thermal and light energy. The second student predicts that only part of the current will be used up, and the second lamp will glow, but more brightly than the first.

SECTION23.1 Simple Circuits Series Circuits (cont.) The third student predicts that the lamps will be of equal brightness because current is a flow of charge and the charge leaving the first lamp has nowhere else to go in the circuit except through the second lamp. The third student reasons that because the current will be the same in each lamp, the brightness also will be the same. How do you predict the lights will behave?

SECTION23.1 Simple Circuits Series Circuits (cont.) If you consider the mountain river model for this circuit, you will realize that the third student is correct. Recall that charge cannot be created or destroyed. Because the charge in the circuit has only one path to follow and cannot be destroyed, the same amount of charge that enters the circuit must leave the circuit. This means that the current is the same everywhere in the circuit.

SECTION23.1 Simple Circuits Series Circuits (cont.) If you connect three ammeters in the circuit, as shown in the figure, they all will show the same current. A circuit in which there is only one path for the current is called a series circuit.

SECTION23.1 Simple Circuits Series Circuits (cont.) If the current is the same throughout the circuit, what is used by the lamp to produce the thermal and light energy? Recall that power, the rate at which electric energy is converted, is represented by P = IV. Thus, if there is a potential difference, or voltage drop, across the lamp, then electric energy is being converted into another form.

SECTION23.1 Simple Circuits Series Circuits (cont.) The resistance of the lamp is defined as R = V/I. Thus, the potential difference, also called the voltage drop, is V = IR.

SECTION23.1 Simple Circuits Series Circuits (cont.) From the river model, you know that the sum of the drops in height is equal to the total drop from the top of the mountain to sea level. In an electric circuit, the increase in voltage provided by the generator or other energy source, Vsource, is equal to the sum of voltage drops across lamps A and B, and is represented by the following equation: Vsource = VA + VB

SECTION23.1 Simple Circuits Series Circuits (cont.) To find the potential drop across a resistor, multiply the current in the circuit by the resistance of the individual resistor. Because the current through the lamps is the same, VA = IRA and VB = IRB. Therefore, VA = IRA + IRB, or Vsource = I(RA + RB).

SECTION23.1 Simple Circuits Series Circuits (cont.) The current through the circuit is represented by the following equation:

SECTION23.1 Simple Circuits Series Circuits (cont.) The same idea can be extended to any number of resistances in series, not just two. The same current would exist in the circuit with a singe resistor, R, that has a resistance equal to the sum of the resistances of the two lamps. Such a resistance is called the equivalent resistance of the circuit. The equivalent resistance of resistors in series equals the sum of the individual resistances of the resistors.

SECTION23.1 Simple Circuits Series Circuits (cont.) For resistors in series, the equivalent resistance is the sum of all the individual resistances, as expressed by the following equation. Equivalent Resistance for Resistors in Series R = RA + RB + …

SECTION23.1 Simple Circuits Series Circuits (cont.) Notice that the equivalent resistance is greater than that of any individual resistor. Therefore, if the battery voltage does not change, adding more devices in series always decreases the current.

SECTION23.1 Simple Circuits Series Circuits (cont.) To find the current through a series circuit, first calculate the equivalent resistance and then use the following equation. Current Current in a series circuit is equal to the potential difference of the source divided by the equivalent resistance.

SECTION23.1 Simple Circuits Series Circuits (cont.) As current moves through any circuit, the net change in potential must be zero. This is because the circuit’s electric energy source, the battery or generator, raises the potential an amount equal to the potential drop produced when the current passes through the resistors. Therefore, the net change is zero.

SECTION23.1 Simple Circuits Series Circuits (cont.) An important application of series resistors is a circuit called a voltage divider. A voltage divider is a series circuit used to produce a source of potential difference that is less than the potential difference across the battery.

SECTION23.1 Simple Circuits Series Circuits (cont.) For example, suppose you have a 9-V battery but need a 5-V potential source. Consider the circuit shown in the figure.

SECTION23.1 Simple Circuits Series Circuits (cont.) Two resistors, RA and RB, are connected in series across a battery of magnitude V. The equivalent resistance of the circuit is R = RA + RB.

SECTION23.1 Simple Circuits Series Circuits (cont.) The current is represented by the following equation:

SECTION23.1 Simple Circuits Series Circuits (cont.) The desired voltage, 5 V, is the voltage drop, VB, across resistor RB: VB = IRB.

SECTION23.1 Simple Circuits Series Circuits (cont.) Into this equation, the earlier equation for current is substituted.

SECTION23.1 Simple Circuits Series Circuits (cont.) Voltage dividers often are used with sensors, such as photoresistors. The resistance of a photoresistor depends upon the amount of light that strikes it.

SECTION23.1 Simple Circuits Series Circuits (cont.) Photoresistors are made of semiconductors, such as silicon, selenium, or cadmium sulfide. A typical photoresistor can have a resistance of 400 Ω when light is striking it compared with a resistance of 400,000 Ω when the photoresistor is in the dark.

SECTION23.1 Simple Circuits Series Circuits (cont.) The voltage output of a voltage divider that uses a photoresistor depends upon the amount of light striking the photoresistor sensor. This circuit can be used as a light meter, such as the one shown in the figure. Static graphic: Figure 23-5 (b) Compassionate Eye Foundation/Matt Hind/OJO Images Ltd/Photodisc/Getty Images

SECTION23.1 Simple Circuits Series Circuits (cont.) In this device, an electronic circuit detects the potential difference and converts it to a measurement of illuminance that can be read on the digital display. The amplified voltmeter reading will drop as illuminance increases.

Potential Difference in a Series Circuit
SECTION23.1 Simple Circuits Potential Difference in a Series Circuit Two resistors, 47.0 Ω and 82.0 Ω, are connected in series across a 45.0-V battery.

Potential Difference in a Series Circuit (cont.)
SECTION23.1 Simple Circuits Potential Difference in a Series Circuit (cont.) a. What is the current in the circuit? b. What is the potential difference across each resistor? c. If the 47.0-Ω resistor is replaced by a 39.0-Ω resistor, will the current increase, decrease, or remain the same? d. What is the new potential difference across the 82.0-Ω resistor?

SECTION23.1 Simple Circuits Potential Difference in a Series Circuit (cont.) Step 1: Analyze and Sketch the Problem Draw a schematic of the circuit.

SECTION23.1 Simple Circuits Potential Difference in a Series Circuit (cont.) Identify the known and unknown variables. Known: Vsource = 45.0 V RA = 47.0 Ω RB = 82.0 Ω Unknown: I = ? VA = ? VB = ?

SECTION23.1 Simple Circuits Potential Difference in a Series Circuit (cont.) Step 2: Solve for the Unknown

SECTION23.1 Simple Circuits Potential Difference in a Series Circuit (cont.) a. To determine the current, first find the equivalent resistance. and R = RA + RB

SECTION23.1 Simple Circuits Potential Difference in a Series Circuit (cont.) Substitute R = RA + RB Substitute Vsource = 45.0 V, RA = 47.0 Ω, RB = 82.0 Ω = A

SECTION23.1 Simple Circuits Potential Difference in a Series Circuit (cont.) b. Use V = IR for each resistor. VA = IRA Substitute I = A, RA = 47.0 Ω VA= (0.349 A)(47.0 Ω) = 16.4 V

SECTION23.1 Simple Circuits Potential Difference in a Series Circuit (cont.) VB = IRB Substitute I = A, RA = 82.0 Ω VB = (0.349 A)(82.0 Ω) = 28.6 V

SECTION23.1 Simple Circuits Potential Difference in a Series Circuit (cont.) c. Calculate current, this time using 39.0 Ω as RA.

SECTION23.1 Simple Circuits Potential Difference in a Series Circuit (cont.) Substitute Vsource = 45.0 V, RA = 39.0 Ω, RB = 82.0 Ω = A The current will increase.

SECTION23.1 Simple Circuits Potential Difference in a Series Circuit (cont.) d. Determine the new voltage drop in RB. VB = IRB Substitute I = A, RA = 82.0 Ω VB = (0.372 A)(82.0 Ω) = 30.5 V

SECTION23.1 Simple Circuits Potential Difference in a Series Circuit (cont.) Step 3: Evaluate the Answer

SECTION23.1 Simple Circuits Potential Difference in a Series Circuit (cont.) Are the units correct? Current is A = V/Ω; voltage is V = A·Ω. Is the magnitude realistic? For current, if R > V, I < 1. The voltage drop across any one resistor must be less than the voltage of the circuit. Both values of VB are less than Vsource, which is 45 V.

SECTION23.1 Simple Circuits Potential Difference in a Series Circuit (cont.) The steps covered were: Step 1: Analyze and Sketch the Problem Draw a schematic of the circuit. Step 2: Solve for the Unknown Step 3: Evaluate the Answer

Simple Circuits Parallel Circuits
SECTION23.1 Simple Circuits Parallel Circuits Look at the circuit shown in the figure below. How many current paths are there?

Parallel Circuits (cont.)
SECTION23.1 Simple Circuits Parallel Circuits (cont.) The current from the generator can go through any of the three resistors. A circuit in which there are several current paths is called a parallel circuit.

SECTION23.1 Simple Circuits Parallel Circuits (cont.) The three resistors are connected in parallel; both ends of the three paths are connected together. In the mountain river model, such a circuit is illustrated by three paths for the water over a waterfall.

SECTION23.1 Simple Circuits Parallel Circuits (cont.) Some paths might have a large flow of water, while others might have a small flow. The sum of the flows, however, is equal to the total flow of water over the falls.

SECTION23.1 Simple Circuits Parallel Circuits (cont.) In addition, regardless of which channel the water flows through, the drop in height is the same. Similarly, in a parallel electric circuit, the total current is the sum of the currents through each path, and the potential difference across each path is the same.

SECTION23.1 Simple Circuits Parallel Circuits (cont.) What is the current through each resistor in a parallel electric circuit? It depends upon the individual resistances. For example, in the figure, the potential difference across each resistor is 120 V.

SECTION23.1 Simple Circuits Parallel Circuits (cont.) The current through a resistor is given by I = V/R, so you can calculate the current through the 24-Ω resistor as I = (120 V)/(24 Ω) = 5.0 A and then calculate the currents through the other two resistors. The total current through the generator is the sum of the currents through the three paths, in this case, 38 A.

SECTION23.1 Simple Circuits Parallel Circuits (cont.) What would happen if the 6-Ω resistor were removed from the circuit? Would the current through the 24-Ω resistor change? The current depends only upon the potential difference across it and its resistance; because neither has changed, the current also is unchanged.

SECTION23.1 Simple Circuits Parallel Circuits (cont.) The same is true for the current through the 9-Ω resistor. The branches of a parallel circuit are independent of each other. The total current through the generator, however, would change. The sum of the currents in the branches would be 18 A if the 6-Ω resistor were removed.

SECTION23.1 Simple Circuits Parallel Circuits (cont.) How can you find the equivalent resistance of a parallel circuit? In the figure shown, the total current through the generator is 38 A.

SECTION23.1 Simple Circuits Parallel Circuits (cont.) Thus, the value of a single resistor that results in a 38-A current when a 120-V potential difference is placed across it can easily be calculated by using the following equation: = 3.2 Ω

SECTION23.1 Simple Circuits Parallel Circuits (cont.) Notice that this resistance is smaller than that of any of the three resistors in parallel. Placing two or more resistors in parallel always decreases the equivalent resistance of a circuit. The resistance decreases because each new resistor provides an additional path for current, thereby increasing the total current while the potential difference remains unchanged.

SECTION23.1 Simple Circuits Parallel Circuits (cont.) To calculate the equivalent resistance of a parallel circuit, first note that the total current is the sum of the currents through the branches. If IA, IB, and IC are the currents through the branches and I is the total current, then I = IA + IB + IC. The potential difference across each resistor is the same, so the current through each resistor, for example, RA, can be found from:

SECTION23.1 Simple Circuits Parallel Circuits (cont.) Therefore, the equation for the sum of the currents is as follows:

Equivalent Resistance for Resistors in Parallel
SECTION23.1 Simple Circuits Parallel Circuits (cont.) Dividing both sides of the equation by V provides an equation for the equivalent resistance of the three parallel resistors. Equivalent Resistance for Resistors in Parallel The reciprocal of the equivalent resistance is equal to the sum of the reciprocals of the individual resistances.

SECTION23.1 Simple Circuits Parallel Circuits (cont.) Series and parallel connections differ in how they affect a lighting circuit. Imagine a 60-W and a 100-W bulb are used in a lighting circuit. Recall that the brightness of a lightbulb is proportional to the power it dissipates, and that P = I2R. When the bulbs are connected in parallel, each is connected across 120 V and the 100-W bulb glows more brightly.

SECTION23.1 Simple Circuits Parallel Circuits (cont.) When connected in series, the current through each bulb is the same. Because the resistance of the 60-W bulb is greater than that of the 100-W bulb, the higher-resistance 60-W bulb dissipates more power and glows more brightly.

Simple Circuits Kirchhoff’s Rules
SECTION23.1 Simple Circuits Kirchhoff’s Rules Gustav Robert Kirchhoff was a German physicist who, in 1845, when he was only 21 years old, formulated two rules that govern electric circuits: The loop rule The junction rule You can use these two rules to analyze complex electric circuits.

Kirchhoff’s Rules (cont.)
SECTION23.1 Simple Circuits Kirchhoff’s Rules (cont.) The loop rule describes electric potential differences and is based on the law of conservation of energy. The sum of increases in electric potential around a loop in an electric circuit equals the sum of decreases in electric potential around that loop.

SECTION23.1 Simple Circuits Kirchhoff’s Rules (cont.) For an application of the loop rule, look at the figure. Picture an electric current traveling clockwise around the red loop. Electric potential increases by 9V as this charge travels through the battery, and electric potential drops by 5V as this charge travels through resistor 1.

SECTION23.1 Simple Circuits Kirchhoff’s Rules (cont.) What will be the change in potential as the charge travels through resistor 2? Since the increases in electric potential around a loop must equal the decreases in electric potential around that loop, the drop in electric potential across resistor 2 must be 9V – 5V = 4V. Note that resistor 3 does not affect our answer because resistor 3 is not a part of the loop that includes the battery, resistor 1 and resistor 2.

SECTION23.1 Simple Circuits Kirchhoff’s Rules (cont.) The junction rule describes currents and is based on the law of conservation of charge. Recall the law of conservation of charge states that charge can neither be created or destroyed. This means that, in an electric circuit, the total current into a section of that circuit must equal the total current out of that same section.

SECTION23.1 Simple Circuits Kirchhoff’s Rules (cont.) A junction is a location where three or more wires are connected together. The figure shows a circuit with two such junctions, at point A and B. How does the total current into a junction relate to the total current out of that junction?

SECTION23.1 Simple Circuits Kirchhoff’s Rules (cont.) According to Kirchhoff’s rule, the sum of currents entering a junction is equal to the sum of currents leaving that junction. Otherwise, charge would build up at the junction. Therefore, I1 = I2 + I3 at junction A, and I2 + I3 = I1 at junction B.

SECTION23.1 Section Check What is a series circuit? How is it different from a parallel circuit?

SECTION23.1 Section Check Answer A circuit in which all the current travels through each device is called a series circuit. A diagram of a series circuit is shown below.

SECTION23.1 Section Check Answer On the other hand, a circuit in which there are several current paths, as shown in the following figure, is called a parallel circuit. In a parallel circuit, the total current is the sum of the currents through each path. In the case of the above circuit, I = I1 + I2 + I3

SECTION23.1 Section Check Three resistors of resistance 2 Ω each are connected in series with a battery. What is the equivalent resistance of the circuit? C. A. D. B.

SECTION23.1 Section Check Answer Reason: Equivalent resistance for resistors in series is given by R = RA + RB The equivalent resistance of resistors in series is equal to the sum of the individual resistances of the resistors. In this case, R = 2 Ω + 2 Ω + 2 Ω = 6 Ω

SECTION23.1 Section Check What will be the ammeter reading in the following circuit? C. A. B. D.

SECTION23.1 Section Check Answer Reason: The given circuit is a series circuit and the current through a series circuit is constant. Current in a series circuit is equal to the potential difference of the sources divided by the equivalent resistance. That is,

SECTION23.1 Section Check Answer Reason: In this case, Vsource = 12 V and equivalent resistance R = 10 Ω + 10 Ω.

Section Check What is the voltmeter reading in the following circuit?
B. C. D.

SECTION23.1 Section Check Answer Reason: The given circuit is a voltage divider. A voltage divider is a series circuit used to produce a voltage source of desired magnitude from a higher-voltage battery. To calculate the voltage across RB (10-Ω resistor), we first calculate the current flowing through the circuit. That is,

SECTION23.1 Section Check Answer Reason: We need to calculate the voltage VB (voltage across RB). Using Ohm’s law, we can write, VB = IRB. Substituting the value of I from the above equation, we get

Applications of Circuits
SECTION23.2 Applications of Circuits MAIN IDEA Most circuits are combination series-parallel circuits. Essential Questions How do fuses, circuit breakers, and ground-fault interrupters protect household wiring? How can you find currents and potential differences in combined series-parallel circuits? How do you use voltmeters and ammeters to measure potential differences and currents in circuits?

SECTION23.2 Applications of Circuits Review Vocabulary Electric current a flow of charged particles New Vocabulary Short circuit Fuse Circuit breaker Ground-fault interrupter Combination series-parallel circuit

SECTION23.2 Applications of Circuits Safety Devices In an electric circuit, fuses and circuit breakers act as safety devices. They prevent circuit overloads that can occur when too many appliances are turned on at the same time or when a short circuit occurs in one appliance. A short circuit occurs when a circuit with very low resistance is formed. The low resistance causes the current to be very large.

SECTION23.2 Applications of Circuits Safety Devices (cont.) When appliances are connected in parallel, each additional appliance placed in operation reduces the equivalent resistance in the circuit and increases the current through the wires. This additional current might produce enough thermal energy to melt the wiring’s insulation, cause a short circuit, or even begin a fire.

SECTION23.2 Applications of Circuits Safety Devices (cont.) A fuse is a short piece of metal that melts when too large a current passes through it. The thickness of the metal used in a fuse is determined by the amount of current that the circuit is designed to handle safely. If a large, unsafe current passes through the circuit, the fuse melts and breaks the circuit.

SECTION23.2 Applications of Circuits Safety Devices (cont.) A circuit breaker is an automatic switch that opens when the current reaches a threshold value.

SECTION23.2 Applications of Circuits Safety Devices (cont.) If there is a current greater than the rated (threshold) value in the circuit, the circuit becomes overloaded. The circuit breaker opens and stops the current.

SECTION23.2 Applications of Circuits Safety Devices (cont.) Current follows a single path from the power source through an electrical appliance and back to the source. Sometimes, a faulty appliance or an accidental drop of the appliance into water might create another current pathway. If this pathway flows through the user, serious injury could result. A current as small as 5 mA flowing through a person could result in electrocution.

SECTION23.2 Applications of Circuits Safety Devices (cont.) A ground-fault interrupter in an electric outlet prevents such injuries because it contains an electronic circuit that detects small differences in current caused by an extra current path and opens the circuit. Electric codes for buildings often require ground-fault interrupters to be used in bathrooms, kitchens, and exterior outlets.

SECTION23.2 Applications of Circuits Safety Devices (cont.) The figure diagrams a parallel circuit used in the wiring of homes and also shows some common appliances that would be connected in parallel. The current in any one circuit does not depend upon the current in the other circuits. Richard Hutchings/Digital Light Source

SECTION23.2 Applications of Circuits Safety Devices (cont.) Suppose that a 240-W television is plugged into a 120-V outlet. The current is represented by I = P/V. For the television, I = (240 W)/(120 V) = 2.0 A. When a 720-W curling iron is plugged into the outlet, its current draw is I = (720 W)/(120 V) = 6.0 A. Finally, a 1440-W hair dryer is plugged into the same outlet. The current through the hair dryer is I = (1440 W)/(120 V) = 12 A.

SECTION23.2 Applications of Circuits Safety Devices (cont.) The resistance of each appliance can be calculated using the equation R = V/I. The equivalent resistance of the three appliances is as follows.

SECTION23.2 Applications of Circuits Safety Devices (cont.) A fuse is connected in series with the power source so that the entire current passes through it. The current through the fuse is calculated using the equivalent resistance.

SECTION23.2 Applications of Circuits Safety Devices (cont.) If the fuse in the circuit is rated as 15 A, the 20-A current would exceed the rating and cause the fuse to melt, or “blow,” and cut off current. Fuses and circuit breakers also protect against the large currents created by a short circuit. Without a fuse or a circuit breaker, the current caused by a short circuit easily could start a fire. For example, a short circuit could occur if the insulation on a lamp cord becomes old and brittle.

SECTION23.2 Applications of Circuits Safety Devices (cont.) The two wires in the cord might accidentally touch, resulting in a resistance in the wire of only Ω. This resistance results in a huge current. Such a current would cause a fuse or a circuit breaker to open the circuit, thereby preventing the wires from becoming hot enough to start a fire.

SECTION23.2 Applications of Circuits Combined Series-Parallel Circuits Have you ever noticed the light in your bathroom or bedroom dim when you turned on a hair dryer? The light and the hair dryer are connected in parallel across 120 V. Because of the parallel connection, the current through the light should not have changed when you turned on the hair dryer. Yet the light did dim, so the current must have changed. The dimming occurred because the house wiring had a small resistance.

SECTION23.2 Applications of Circuits Combined Series-Parallel Circuits (cont.) As shown in the figure, this resistance was in series with the parallel circuit. A circuit that includes series and parallel branches is called a combination series-parallel circuit.

SECTION23.2 Applications of Circuits Series-Parallel Circuit A hair dryer with a resistance of 12.0 Ω and a lamp with a resistance of 125 Ω are connected in parallel to a 125-V source through a 1.50-Ω resistor in series. Find the current through the lamp when the hair dryer is on.

SECTION23.2 Applications of Circuits Series-Parallel Circuit (cont.) Step 1: Analyze and Sketch the Problem

SECTION23.2 Applications of Circuits Series-Parallel Circuit (cont.) Draw the series-parallel circuit including the hair dryer and the lamp. Replace RA and RB with a single equivalent resistance, Rp.

SECTION23.2 Applications of Circuits Series-Parallel Circuit (cont.) Identify the known and unknown variables. Known: RA = 125 Ω RB = 12.0 Ω RC = 1.50 Ω Vsource = 125 V Unknown: I = ? R = ? IA = ? RP = ?

SECTION23.2 Applications of Circuits Series-Parallel Circuit (cont.) Step 2: Solve for the Unknown

SECTION23.2 Applications of Circuits Series-Parallel Circuit (cont.) Find the equivalent resistance for the parallel circuit, then find the equivalent resistance for the entire circuit, and then calculate the current. Substitute RA = 125 Ω, RB = 12.0 Ω RP = 10.9 Ω

SECTION23.2 Applications of Circuits Series-Parallel Circuit (cont.) R = RC + RP Substitute RC = 1.50 Ω, RP = 10.9 Ω R = 1.50 Ω Ω = 12.4 Ω

SECTION23.2 Applications of Circuits Series-Parallel Circuit (cont.) Substitute Vsource = 125 V, R = 12.4 Ω = 10.1 A

SECTION23.2 Applications of Circuits Series-Parallel Circuit (cont.) VC = IRC Substitute I = 10.1 A, RC = 1.50 Ω VC = (10.1 A)(1.50 Ω) = 15.2 V

SECTION23.2 Applications of Circuits Series-Parallel Circuit (cont.) VA = Vsource − VC Substitute Vsource = 125 V, VC = 15.2 V VA= 125 V − 15.2 V = 1.10×102 V

SECTION23.2 Applications of Circuits Series-Parallel Circuit (cont.) Substitute VA = 1.10×102 V, RA = 125 Ω = A

SECTION23.2 Applications of Circuits Series-Parallel Circuit (cont.) Step 3: Evaluate the Answer

SECTION23.2 Applications of Circuits Series-Parallel Circuit (cont.) Are the units correct? Current is measured in amps, and potential drops are measured in volts. Is the magnitude realistic? The resistance is greater than the voltage, so the current should be less than 1 A.

SECTION23.2 Applications of Circuits Series-Parallel Circuit (cont.) The steps covered were: Step 1: Analyze and Sketch the Problem Draw the series-parallel circuit including the hair dryer and lamp. Replace RA and RB with a single equivalent resistance, Rp.

SECTION23.2 Applications of Circuits Series-Parallel Circuit (cont.) The steps covered were: Step 2: Solve for the Unknown Step 3: Evaluate the Answer

SECTION23.2 Applications of Circuits Ammeters and Voltmeters An ammeter is a device that is used to measure the current in any branch or part of a circuit. If, for example, you wanted to measure the current through a resistor, you would place an ammeter in series with the resistor. This would require opening the current path and inserting an ammeter.

SECTION23.2 Applications of Circuits Ammeters and Voltmeters (cont.) Ideally, the use of an ammeter should not change the current in the circuit. Because the current would decrease if the ammeter increased the resistance in the circuit, the resistance of an ammeter is designed to be as low as possible.

SECTION23.2 Applications of Circuits Ammeters and Voltmeters (cont.) The figure shows an ammeter as a meter placed in parallel with a 0.01-Ω resistor. Because the resistance of the ammeter is much less than that of the resistors, the current decrease is negligible.

SECTION23.2 Applications of Circuits Ammeters and Voltmeters (cont.) Another instrument, called a voltmeter, is used to measure the voltage drop across a portion of a circuit.

SECTION23.2 Applications of Circuits Ammeters and Voltmeters (cont.) To measure the potential drop across a resistor, a voltmeter is connected in parallel with the resistor. Voltmeters are designed to have a very high resistance so as to cause the smallest possible change in currents and voltages in the circuit.

Section Check What is a fuse?
A. A fuse is a short piece of metal that melts when a circuit with a large resistance is formed. B. A fuse is a switch that opens if there is a current greater than the rated value in the circuit. C. A fuse is an electronic circuit that detects small differences in current. D. A fuse is a short piece of metal that melts when too large a current passes through it.

SECTION23.2 Section Check Answer Reason: A fuse is a short piece of metal that melts when too large a current passes through it. The thickness of the metal used in a fuse is determined by the amount of current that the circuit is designed to handle safely. If a large, unsafe current passes through the circuit, the fuse melts and breaks the circuit.

SECTION23.2 Section Check When does a short circuit occur?

SECTION23.2 Section Check Answer A short circuit occurs when a circuit with a very low resistance is formed. The low resistance causes the current to be very large. When appliances are connected in parallel, each additional appliance placed in operation reduces the equivalent resistance in the circuit and increases the current through the wires. This additional current might produce enough thermal energy to melt the wiring’s insulation, cause a short circuit, or even begin a fire.

Section Check Why does an ammeter always have a small resistance?
A. to protect the ammeter from damage B. so that the current decrease in the ammeter is negligible C. to cause the smallest possible change in the voltage of the circuit D. to reduce the potential drop across the ammeter

SECTION23.2 Section Check Answer Reason: An ammeter is a device used to measure the current in any branch or part of a circuit. If, for example, you want to measure the current through a resistor, you would place an ammeter in series with the resistor. This would require opening the current path and inserting an ammeter. Ideally, the use of an ammeter should not change the current in the circuit. Because the current would decrease if the ammeter’s resistance increased, the resistance of an ammeter is designed to be as low as possible.

CHAPTER23 Series and Parallel Circuits Resources Physics Online Study Guide Chapter Assessment Questions Standardized Test Practice

SECTION23.1 Simple Circuits Study Guide The current is the same everywhere in a simple series circuit. If any branch of a parallel circuit is opened, there is no current in that branch. The current in the other branches is unchanged. The current in a series circuit is equal to the potential difference divided by the equivalent resistance. The sum of the voltage that drops across resistors that are in series is equal to the potential difference applied across the combination. The equivalent resistance of a series circuit is the sum of the resistance of its parts.

SECTION23.1 Simple Circuits Study Guide In a parallel circuit, the total current is equal to the sum of the currents in the branches. The voltage drops across all branches of a parallel circuit are the same. The reciprocal of the equivalent resistance of parallel resistors is equal to the sum of the reciprocals of the individual resistances.

SECTION23.2 Applications of Circuits Study Guide A fuse, a circuit breaker, and a ground-fault interrupter create an open circuit and stop current when dangerously high currents flow. To find currents and potential differences in complex circuits, a combination of series and parallel branches, any parallel branch first is reduced to a single equivalent resistance. Then, any resistors in series are replaced by a single resistance.

SECTION23.2 Applications of Circuits Study Guide A voltmeter measures the potential difference (voltage) across any part or combination of parts of a circuit. A voltmeter always has a high resistance and is connected in parallel with the part of the circuit being measured. An ammeter is used to measure the current in a branch or part of a circuit. An ammeter always has a low resistance and is connected in series.

CHAPTER23 Series and Parallel Circuits Chapter Assessment Three bulbs, A, B, and C, are connected in series with a battery as shown below.

CHAPTER23 Series and Parallel Circuits Chapter Assessment Which statement is true? A. Only bulb A will glow. B. Only bulb C will glow. C. All the bulbs will glow but the intensity of bulb A will be more than the intensities of bulbs B and C. D. All the bulbs will glow with equal intensity.

CHAPTER23 Series and Parallel Circuits Chapter Assessment Reason: We know that charge can neither be created nor destroyed. Therefore, the current into each lightbulb must be the same as the current out of each light bulb. This means that the current is the same everywhere in the circuit. Hence, all the bulbs will glow with equal intensity.

CHAPTER23 Series and Parallel Circuits Chapter Assessment In the following figure three resistances, RA , RB, and RC, are connected parallel to each other across a generator. What will be the reading on the third ammeter, I3? A. 20 A B. 5 A C. 10 A D. 15 A

CHAPTER23 Series and Parallel Circuits Chapter Assessment Reason: In a parallel circuit, the total current is the sum of the currents through each path. In this case, I = I1 + I2 + I3  I3 = I – (I1 + I2) = 20 A – (5 A + 10 A) = 5 A

CHAPTER23 Series and Parallel Circuits Chapter Assessment A 60-W bulb and a 100-W bulb were first connected in series with a 120-V battery and then connected parallel across the same 120-V battery. Which of the following statements is true?

CHAPTER23 Series and Parallel Circuits Chapter Assessment A. In both the series and parallel connections, the intensity of the 60-W bulb will be more. B. In both the series and parallel connections, the intensity of the 100-W bulb will be more. C. In a series connection, the intensity of the 100-W bulb will be more, and in parallel connection, the intensity of the 60-W bulb will be more. D. In a series connection, the intensity of the 60-W bulb will be more, and in parallel connection, the intensity of the 100-W bulb will be more.

CHAPTER23 Series and Parallel Circuits Chapter Assessment Reason: Recall that the brightness of a lightbulb is proportional to the power it dissipates, and that P = I2R. When the bulbs are connected in parallel, each is connected across 120 V, and the 100-W bulb glows more brightly. When connected in series, the current through each bulb is the same. Because the resistance of the 60-W bulb is greater than that of the 100-W bulb, the higher resistance 60-W bulb dissipates more power and glows more brightly.

CHAPTER23 Series and Parallel Circuits Chapter Assessment Explain how a ground-fault interrupter protects household wiring.

CHAPTER23 Series and Parallel Circuits Chapter Assessment Answer: Current follows a single path from the power source through an electrical appliance and back to the source. Sometimes, a faulty appliance or an accidental drop of the appliance into water might create another current pathway. If this pathway flows through the user, serious injury could result. A current as small as 5 mA flowing through a person could result in electrocution.

CHAPTER23 Series and Parallel Circuits Chapter Assessment Answer: A ground-fault interrupter in an electric outlet prevents such injuries because it contains an electronic circuit that detects small differences in current caused by an extra current path and opens the circuit. Electric codes for buildings often require ground-fault interrupters to be used in bathroom, kitchen, and exterior outlets.

CHAPTER23 Series and Parallel Circuits Chapter Assessment What is a voltmeter? How is it connected in a circuit?

CHAPTER23 Series and Parallel Circuits Chapter Assessment Answer: A voltmeter is an instrument used to measure the voltage drop across a portion of a circuit. To measure the potential drop across a device such as a resistor, a voltmeter is connected in parallel with the resistor. Voltmeters are designed to have very high resistance so as to cause the smallest possible change in currents and voltages in the circuit.

CHAPTER23 Series and Parallel Circuits Chapter Assessment Answer: Consider the circuit shown in following figure.

CHAPTER23 Series and Parallel Circuits Chapter Assessment Answer: A voltmeter is shown as a meter in series with a 10 k-Ω resistor. When the voltmeter is connected in parallel with RB, the equivalent resistance of the voltmeter-RB combination is smaller than RB alone. Thus, the total resistance of the circuit decreases slightly, and the current increases slightly.

CHAPTER23 Series and Parallel Circuits Chapter Assessment Answer: The value of RA does not change, but the current through it increases, thereby increasing the potential drop across it. The battery, however, holds the potential drop across RA and RB constant. Thus, the potential drop across RB must decrease. The result of connecting a voltmeter across a resistor is to reduce the potential drop across it. The higher the resistance of the voltmeter, the smaller the voltage change. Practical meters have resistances of 10 kΩ or more.

CHAPTER23 Series and Parallel Circuits Standardized Test Practice What is the equivalent resistance of the circuit? A. C. 1.5 Ω B. 1.0 Ω D. 19 Ω

CHAPTER23 Series and Parallel Circuits Standardized Test Practice How much current flows through R3? A A C. 2.0 A B. 1.5 A D. 4.0 A

CHAPTER23 Series and Parallel Circuits Standardized Test Practice What is the equivalent resistance of the circuit? A Ω B Ω C Ω D Ω Edit this image. Use the correct multiply sign.

CHAPTER23 Series and Parallel Circuits Standardized Test Practice What is the current in the circuit? A A B A C A D A Edit this image. Use the correct multiply sign.

CHAPTER23 Series and Parallel Circuits Standardized Test Practice A series circuit has an 8.0-V battery and four resistors, R1 = 4.0 Ω, R2 = 8.0 Ω, R3 = 13.0 Ω, and R4 = 15.0 Ω. Calculate the current and power in the circuit. Answer: I = 0.20 A; P = 1.6 W

CHAPTER23 Series and Parallel Circuits Standardized Test Practice Test-Taking Tip Take a Break If you have the opportunity to take a break or get up from your desk during a test, take it. Getting up and moving around will give you extra energy and help you clear your mind. During the break, think about something other than the test so you’ll be able to begin again with a fresh start.

CHAPTER23 Series and Parallel Circuits Chapter Resources Two Lamps Connected to a Battery

CHAPTER23 Series and Parallel Circuits Chapter Resources Series Circuit

CHAPTER23 Series and Parallel Circuits Chapter Resources Voltage Divider Circuit

CHAPTER23 Series and Parallel Circuits Chapter Resources Voltage Divider

CHAPTER23 Series and Parallel Circuits Chapter Resources Voltage Drops in a Series Circuit

CHAPTER23 Series and Parallel Circuits Chapter Resources Voltage Divider

CHAPTER23 Series and Parallel Circuits Chapter Resources Parallel Circuit

CHAPTER23 Series and Parallel Circuits Chapter Resources Resistance in a Parallel Circuit

CHAPTER23 Series and Parallel Circuits Chapter Resources Equivalent Resistance and Current in a Parallel Circuit

CHAPTER23 Series and Parallel Circuits Chapter Resources Circuit with Four Identical Resistors

CHAPTER23 Series and Parallel Circuits Chapter Resources Circuit Breaker

CHAPTER23 Series and Parallel Circuits Chapter Resources Parallel Wiring Arrangement used for Home Appliances

CHAPTER23 Series and Parallel Circuits Chapter Resources Balanced Circuit

CHAPTER23 Series and Parallel Circuits Chapter Resources Combined Series-Parallel Circuits

CHAPTER23 Series and Parallel Circuits Chapter Resources Reducing Circuit Diagrams

CHAPTER23 Series and Parallel Circuits Chapter Resources Series-Parallel Circuit

CHAPTER23 Series and Parallel Circuits Chapter Resources Ammeter and Voltmeter

CHAPTER23 Series and Parallel Circuits Chapter Resources Circuit Containing Three Bulbs and Ammeters

CHAPTER23 Series and Parallel Circuits Chapter Resources Ground Fault Circuit Interrupter

CHAPTER23 Series and Parallel Circuits Chapter Resources Voltage Drops in a Series Circuit Two resistors, 47.0 Ω and 82.0 Ω, are connected in series across a 45.0-V battery. a. What is the current in the circuit? b. What is the voltage drop across each resistor? c. If the 47.0-Ω resistor is replaced by a 39.0-Ω resistor, will the current increase, decrease, or remain the same? d. What is the voltage drop across the 82.0-Ω resistor? Question of the Example problem.

CHAPTER23 Series and Parallel Circuits Chapter Resources Series-Parallel Circuit A hair dryer with a resistance of 12.0 Ω and a lamp with a resistance of 125 Ω are connected in parallel to a 125-V source through a 1.50-Ω resistor in series. Find the current through the lamp when the hair dryer is on. Question of the Example problem.

Question of the Example problem.

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