## 5 Steps to a 5: AP Physics 1: Algebra-Based 2017 (2016)

**STEP **__4__

__4__

**Review the Knowledge You Need to Score High**

**CHAPTER 16**

**Electricity: Coulomb’s Law and Circuits**

**IN THIS CHAPTER**

**Summary:** Coulomb’s law says that the force of one charged particle on another is . AP Physics 1 deals only with electrical forces between exactly two charges. But AP Physics 1 does require an understanding of direct-current circuits, including series resistors, parallel resistors, and combinations thereof. This chapter shows you how to deal with circuits, both using conceptual descriptions and using calculations.

**Definitions**

Electric **charge** (*Q* ) exists due to excess or deficient electrons on an object. Charge comes in two kinds: positive and negative. The unit of charge is the **coulomb** .

Electric **current** (*I* ) is the flow of (positive) charge per second. The units of current are **amperes.** One ampere means one coulomb of charge flowing per second.

**Resistance** (R), measured in ohms (Ω), tells how difficult it is for charge to flow through a circuit element.

**Resistivity** (*ρ* ) is a property of a material, which implies what the resistance would be of a meter-cube bit of that material.

**Voltage** is electrical potential energy per coulomb of charge.

Resistors are connected in **series** if they are connected in a single path.

Resistors are connected in **parallel** if the path for current divides, then comes immediately back together.

The AP Physics 1, Algebra-Based Exam requires you to learn about two aspects of electricity. First, you must understand how charged objects apply forces to each other in isolation, as when a balloon sticks to the wall. Next, you need to know about circuits, in which gazillions of submicroscopic flowing charges produce effects that can be measured and observed. The following picture shows such a circuit, in which flowing charge causes two bulbs to light up. The meter is positioned to read the voltage across one of the bulbs.

**Electric Charge**

All matter is made up of three types of particles: protons, neutrons, and electrons. Protons have an intrinsic property called “positive charge.” Neutrons don’t contain any charge, and electrons have a property called “negative charge.”

The unit of charge is the coulomb, abbreviated C. One proton has a charge of 1.6 × 10^{−19} coulombs.

Most objects that we encounter in our daily lives are electrically neutral—things like couches, for instance, or trees, or bison. These objects contain as many positive charges as negative charges. In other words, they contain as many protons as electrons.

When an object has more protons than electrons, though, it is described as “positively charged”; and when it has more electrons than protons, it is described as “negatively charged.” The reason that big objects like couches and trees and bison don’t behave like charged particles is because they contain so many bazillions of protons and electrons that an extra few here or there won’t really make much of a difference. So even though they might have a slight electric charge, that charge would be much too small, relatively speaking, to detect.

Tiny objects, like atoms, more commonly carry a measurable electric charge, because they have so few protons and electrons that an extra electron, for example, would make a big difference. Of course, you can have very large charged objects. When you walk across a carpeted floor in the winter, you pick up lots of extra charges and become a charged object yourself … until you touch a doorknob, at which point all the excess charge in your body travels through your finger and into the doorknob, causing you to feel a mild electric shock.

Electric charges follow a simple rule: *Like charges repel; opposite charges attract* . Two positively charged particles will try to get as far away from each other as possible, while a positively charged particle and a negatively charged particle will try to get as close as possible.

Only two types of charge exist. If a question on the exam purports to give evidence of a third type of charge, reject that evidence; if a question suggests a kind of charge that repels *both* positive and negative charges, reject that suggestion as silly.

**Coulomb’s Law**

Just *how much* do charged objects attract and repel? Coulomb’s law tells how much.

**FACT:** Coulomb’s law is an equation for the force exerted by one electrical charge on another:

The *Q* s represent the amount of charge on each object; the *d* represents the distance between the two objects. The variable *k* is the Coulomb’s law constant, 9.0 × 10^{9} N·m^{2} /C^{2} .

As with Newton’s law of universal gravitation for the force of one planet on another, you’re not going to be asked to calculate much with Coulomb’s law. Rather, the questions will be qualitative and semiquantitative. For example, if the amount of charge *A* is doubled, what happens to the force of charge *A* on charge *B* ? (It doubles.) Or, if you double the distance between two charges, what happens to the force of one on the other? (It is cut by one-fourth.)

**Conservation of Charge**

**FACT:** The total amount of charge in a system (or in the universe itself) is always the same.

Equal amounts of positive and negative charge can cancel out to make an object neutral, but those charges still exist on the object in the form of protons and electrons. Charge can be transferred from one object to another, for example by touching two charged metal spheres together or scuffing your feet on the carpet, but the total amount of charge stays the same.

Conservation of charge is usually discussed in conjunction with circuits and Kirchoff’s junction rule, as discussed below.

**Circuits**

A circuit is any wire path that allows charge to flow. Technically, a current is defined as the flow of positive charge.^{ }^{1}^{ }Under what conditions would this charge flow through a wire? This would occur when a coulomb of charge has a potential energy that’s higher at one position in the wire than the other. We call this difference in potential energy per coulomb a “voltage,” and a battery’s job is to provide this voltage that allows current to flow. Current flows out of a battery from the positive side of the battery to the negative side.

**Resistance Versus Resistivity**

Resistance tells how difficult it is for charge to flow through something. Usually that “something” is a resistor, or a light bulb, or something that has a known or determinable resistance; usually the resistance of the wires connecting the “somethings” together is nothing, at least compared to the resistance of the things.

**FACT:** The resistance *R* of a wire of known dimensions is given by

Sometimes, though, it’s important to know how the properties of a wire affect its resistance. The longer the wire is, the more its resistance. The “wider” the wire is—that is, the bigger its cross-sectional area—the less its resistance. Two wires with the same shapes can have different resistances if they are made of different materials. Assuming the same shape, the wire with more resistance has a greater resistivity, represented by the variable *ρ* .

**Example 1:** The preceding circuit diagram contains a battery and three identical 100 Ω resistors.

Questions about circuits will occasionally ask for calculation: Find the voltage across each resistor and the current through each resistor. More often, though, you’ll be asked qualitative questions, like which bulb takes the greatest current, or rank resistors from largest to smallest voltage across.

**The Four Key Facts About Circuits**

**FACT: Series** resistors each carry the same current, which is equal to the total current through the series combination.

**FACT:** The voltage across **series** resistors is different for each but adds to the total voltage across the series combination.

**FACT:** The voltage across **parallel** resistors is the same for each and equal to the total voltage across the parallel combination.

**FACT: Parallel** resistors each carry different currents, which add to the total current through the parallel combination.

**Exam Tip from an AP Physics Veteran**

Many first-year physics students are more comfortable making calculations with circuit problems than with explaining effects in words. If you are confused by a qualitative circuit question, try answering with a calculation: “Well, with a 150-V battery here’s a calculation showing that I get 1 A of current in the circuit, but with a 75-V battery I only get 0.5 A. Thus, cutting the battery’s voltage in half also cuts the current in half.”

When you see a circuit, regardless of what questions about it are asked, it’s worth making a *V-I-R* chart listing the voltage, current, and resistance for each resistor. Then the four facts above and Ohm’s Law—the equation *V* = *IR*—can be used to find the missing value on any row of the chart. The easiest way to understand the *V-I-R* chart is to see it in action. Watch.

Start by sketching a chart and filling in known values. Right now, we know the resistance of each resistor. The voltage of the battery isn’t given, so make it up. Try 100 V.^{ }^{2}

Next, simplify the circuit, collapsing sets of parallel and series resistors into their equivalent resistors.

**FACT:** The equivalent resistance of series resistors is the sum of all of the individual resistors. The equivalent resistance of parallel resistors is less than any individual resistor.

*Strategy:* When two *identical* resistors are connected in parallel, their equivalent resistance is half of either resistor. Here, then, the parallel combination of resistors has equivalent resistance 50 Ω. Even with nonidentical parallel resistors, you can usually estimate their equivalent resistance enough to do qualitative problems. If you need to make the calculation, the equivalent resistance of parallel resistors is given by .

Since the other 100 Ω resistor is in series with the 50 Ω equivalent resistance, the equivalent resistance of the whole circuit is 150 Ω. We can put that equivalent resistance into the chart.

**Mistake:** The V-I-R chart is not a magic square—it’s merely a tool for organizing your calculations for a complicated circuit. You can *not* just add values up and down the columns.

Ah, progress: Two of the three entries in the “total” row are complete. Therefore, we can use Ohm’s Law to calculate the total current in this circuit.

**FACT:** Ohm’s law says that voltage across a circuit element equals that element’s current times its resistance:

This equation can *only* be used across a single row in a *V-I-R* chart.

In the total row, (100 V) = *I* (150 Ω), giving a current in the circuit of *I* = 0.67 A.

Now what? We can’t use Ohm’s law because we don’t have any rows missing just one entry. We go back to the Four Key Facts.

Resistor *R* _{1} is in series with the battery; since current through series resistors is equal to the total current, *R* _{1} must take the entire current flowing from the battery, all 0.67 A. Aha! Put 0.67 A in the chart for the current through *R*_{1} , and we can use Ohm’s Law to calculate the voltage across *R* _{1} : that’s 67 V. The chart appears as follows.

Now to figure out *R* _{2} and *R* _{3} . Look at their 50 Ω equivalent resistor in the redrawn diagram. It’s in series with the 100 Ω resistor. Therefore, the 50 Ω resistor must add its voltage to 67 V to get the total voltage of 100 V and the voltage across the 50 Ω equivalent resistor is 33 V.

Then use the facts. Parallel resistors take the same voltage across each that is equal to the total voltage across the combination. Both *R* _{2} and *R* _{3} take 33 V across them. Fill in the chart.

Now use Ohm’s law across the rows for *R* _{2} and *R* _{3} to finish the chart.^{ }^{3}

This chart can now be used to answer any qualitative question. Sure, you should give more justification than just “look at my chart here.” The chart will ensure you get the *right* answers, and that you have a clue about how to approach the qualitative questions.

For example, the exam might ask the following: Rank the voltage across each resistor from largest to smallest. Easy: *R* _{1} > *R* _{2} = *R* _{3} . Justify that with a verbal description of why you decided to use the calculations that you did: “Look at the simplification of the circuit to two series resistors. The voltage across these two series resistors must add to the voltage of the battery, but the current through them must be the same. By *V* = *IR* , the bigger resistor must take the larger voltage; this is *R* _{1} . Then *R* _{2} and *R* _{3} take equal voltage because they are in parallel with one another.”

**Kirchoff’s Laws: Conservation of Charge and Energy**

**FACT:** Kirchoff’s junction rule says that the current entering a wire junction equals the current leaving the junction.

This fact is a statement of conservation of charge: Since charge can’t be created or destroyed, if 1 C of charge enters each second, the same amount each second must leave.

**FACT:** Kirchoff’s loop rule says that the sum of voltage changes around a circuit loop is zero.

This fact is a statement of conservation of energy because voltage is a change in the electrical potential energy of 1 C of charge. A battery can raise the electrical potential energy of some charge that flows; a resistor will lower the potential energy of that charge.^{ }^{4}^{ }But the sum of all these energy changes must be zero.

Look back at the Four Key Facts: These are just restatements of Kirchoff’s laws and thus of conservation of energy and charge.

The junction rule applies to the facts about current. Series resistors take the same current through each, because there’s no junction. The current through parallel resistors adds to the total because of the junction before and after the parallel combination.

The loop rule applies to the facts about voltage. The voltage across series resistors adds to the total voltage because the resistors can only drop the potential energy of 1 C of charge as much as the battery raised the charge’s potential energy. The voltage across parallel resistors must be the same because Kirchoff’s loop rule applies to all loops of the circuit. No matter which parallel path you look at, the sum of voltage changes must still be zero.

**Power in a Circuit**

Power is still defined as energy per second, just as it was in the chapter about energy. Resistors generally convert electrical energy to other forms of energy—the amount of power says how quickly that conversion occurs.

**FACT:** To determine the power dissipated by a resistor, use the equation

Of course, using Ohm’s law, you can show that *IV* is equivalent to *I* ^{2} *R* as well as .

**Exam Tip from an AP Physics Veteran**

If an AP question asks about power, or equivalently, “the energy dissipated by a resistor per unit time,” make a fourth column in your *V-I-R* chart. Use whichever of the power equations you can to calculate power.

Power doesn’t obey the Four Key Facts. The total power dissipated by a bunch of resistors is just the sum of the power dissipated by each, whether the resistors are in series, parallel, or whatever.

**Circuits from an Experimental Point of View**

When a real circuit is set up in the laboratory, it usually consists of more than just resistors—light bulbs and motors are common devices to hook to a battery, for example. For the purposes of computation, though, we can consider pretty much any electronic device to act like a resistor.

But what if your purpose is *not* computation? Often on the AP Exam, as in the laboratory, you are asked about observational and measurable effects. The most common questions involve the brightness of light bulbs and the measurement (not just computation) of current and voltage.

**Brightness of a Bulb**

The brightness of a bulb depends solely on the power dissipated by the bulb. (Remember, power is given by any of the equations *I* ^{2} *R, IV* , or *V* ^{2} /*R* ). You can remember that from your own experience—when you go to the store to buy a light bulb, you don’t ask for a 400-ohm bulb, but for a 100-watt bulb. And a 100-watt bulb is brighter than a 25-watt bulb. But be careful—a bulb’s power can change depending on the current and voltage it’s hooked up to. Consider this problem.

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 the standard wall outlet voltage is 240 V, which of the following would be true?

(A) The bulb would be one-quarter as bright.

(B) The bulb would be one-half 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.

Your first instinct might be to say that because brightness depends on power, the bulb is exactly as bright. But that’s not right! The power of a bulb can change.

Under most operating conditions, the resistance of a lightbulb is a property of the bulb itself, and so it will not change much no matter to what the bulb is hooked.

That said, the resistance of a bulb can vary when the bulb’s temperature is very cold (i.e., room temperature) or very hot. You can assume a bulb has constant resistance unless a problem clearly asks you to consider temperature variation.

**Ammeters and Voltmeters**

Ammeters measure current, and voltmeters measure voltage. This is pretty obvious, because current is measured in amps, voltage in volts. It is *not* necessarily obvious, though, how to connect these meters into a circuit.

Remind yourself of the properties of series and parallel resistors—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*with the resistor. In the following figure, the meter labeled V_{2} measures the voltage across the 100 Ω resistor, while the meter labeled V_{1} measures the potential difference between points *A* and *B* (which is also the voltage across *R* _{1}).

**Measuring voltage with a voltmeter.**

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 that resistor. In the following figure, ammeter A_{1} measures the current through resistor *R* _{1} , while ammeter A_{2} measures the current through resistor *R* _{2} .

**Measuring current with an ammeter.**

As an exercise, ask yourself, is there a way to figure out the current in the other three resistors based only on the readings in these two ammeters? Answer is in the footnote.^{ }^{5}

** Practice Problems**

** Note: **Additional drills on circuit calculation can be found in

__Chapter 18__.

** 1 .** Three resistors are connected to a 9-V battery in the circuit shown in the preceding figure. Justify all answers thoroughly.

(a) Rank the resistors, from greatest to least, by the voltage across each.

(b) Rank the resistors, from greatest to least, by the current through each.

(c) Rank the resistors, from greatest to least, by the power dissipated by each.

(d) Is it possible to replace the 120 Ω resistor with a different resistor and change the voltage ranking? Answer thoroughly in a short paragraph.

(e) Is it possible to replace the 120 Ω resistor with a different resistor and change the current ranking? Answer thoroughly in a short paragraph.

** 2 .** Four resistors are connected to a 9-V battery in the circuit in the diagram.

(a) Calculate the equivalent resistance of the circuit.

(b) Calculate the voltage across each resistor.

(c) Calculate the current through each resistor.

(d) The 500 Ω resistor is now removed from the circuit. Describe in words, without using calculations, what effect this would have on the circuit. Be specific about each resistor.

** 3 .** Justify your answers to the following in short paragraphs.

(a) Should an ammeter be connected in series or parallel with the resistor it measures?

(b) Should a voltmeter be connected in series or parallel with the resistor it measures?

(c) Does an ideal ammeter have large or small resistance?

(d) Does an ideal voltmeter have large or small resistance?

** 4 .** Two positive charges +

*Q*and +2

*Q*are separated by a distance

*a*, as shown above.

(a) Which is greater, the force of the +*Q* charge on the +2*Q* charge, or the force of the +2*Q* charge on the +*Q* charge?

(b) In terms of given variables and fundamental constants, determine the magnitude and direction of the force of the +2*Q* charge on the +*Q* charge.

(c) By what factor would the force calculated in (b) change if the distance between the charges were increased to 3*a* ?

(d) Now the +*Q* charge is replaced by a negative charge of the same magnitude, and the distance between the charges is returned to *a* . Describe how the magnitude and direction of the force exerted by each charge on the other will change from the original situation.

** Solutions to Practice Problems**

** 1 .** (a) The circuit can be redrawn, simplifying the two parallel resistors to their equivalent resistance of 67 Ω. Then what’s left is two resistors—100 Ω and 67 Ω—in series with the 9-V battery. Since series resistors take the same current through each, by

*V*=

*IR*with

*I*constant, the larger resistance takes the larger voltage. That’s the 100 Ω resistor. Parallel resistors take the same voltage across each, equal to the voltage across the equivalent resistor. Ranking by voltage gives

*V*

_{100 Ω}>

*V*

_{120 Ω}=

*V*

_{150 Ω}.

(b) Current only runs through a wire. All the current in the circuit must run through the 100 Ω resistor and then split at the junction with the parallel resistors. The parallel resistors each take the same voltage; by *V* = *IR* , with constant *V* , the smaller resistor takes the larger current. That’s the 120 Ω resistor. The ranking by current is *I* _{100 Ω} > *I* _{120 Ω} > *I* _{150 Ω} .

(c) Power is current times voltage. The 100 Ω resistor has the largest current *and* the largest voltage, so it has the greatest power. The other two resistors take the same voltage, but the 120 Ω resistor takes more current; thus, it has more power than the 150 Ω resistor. Ranking by power is *P* _{100 Ω} > *P* _{120 Ω} > *P* _{150 Ω} .

(d) On one hand, it’s not possible for the two parallel resistors to take different voltages, by definition (or by Kirchoff’s loop rule, if you will). But it’s totally possible for the parallel combination to take more voltage than the 100 Ω resistor. We need the equivalent resistance of the parallel combination to be greater than 100 Ω; then by *V* = *IR* with *I* constant, the equivalent resistors would take higher voltage. Replace the 120 Ω resistor with, say, a 1,000 Ω resistor. Then the equivalent resistance of the parallel combination would be 130 Ω and would take more voltage than the 100 Ω resistor.

(e) There’s no way to avoid having the largest current go through the 100 Ω resistor; by Kirchoff’s junction rule, it must take all the current in the circuit, while current splits between the other two resistors. Now, nothing says that the parallel resistor that comes first in the diagram has to take the larger current. Replace the 120 Ω resistor with, say, a 200 Ω resistor—the voltage will still be the same for both parallel resistors, but now the 150 Ω resistor is the smaller one, taking the larger current.

** 2 .** (a) First simplify the two parallel combinations using . This gives 120 Ω for the top branch, and 220 Ω for the bottom branch. These two equivalent resistors are in series with one another, so their resistances add to the equivalent to give 340 Ω.

(b) Each pair of parallel resistors takes the same voltage which is equal to that across the equivalent resistance of the pair. Treat this as series resistors of 120 Ω and 220 Ω connected to 9 V. The total current in the circuit is found with Ohm’s law with the battery and the equivalent resistance of the whole circuit: . (That’s easier to understand as 26 mA.) The same 26 mA goes through each of the 120 Ω and 220 Ω resistors. Use Ohm’s law with 0.026 A and each of these resistances to get voltages of 3.2 V and 5.8 V. The final answer:

200 Ω resistor: 3.2 V.

300 Ω resistor: 3.2 V

400 Ω resistor: 5.8 V

500 Ω resistor: 5.8 V

(c) We know the voltage and resistance for each individual resistor. Ohm’s law can be used to get the current through each by just dividing *V* /*R* . The answers are as follows (don’t worry if you rounded slightly differently than I did):

200 Ω resistor: 16 mA.

300 Ω resistor: 11 mA

400 Ω resistor: 15 mA

500 Ω resistor: 12 mA

(d) Start with the entire equivalent circuit. The 9-V battery is unchanged. The total resistance increases—the bottom branch previously had a resistance of 220 Ω but now is just the remaining 400 Ω. By *V* = *IR* with constant *V* , a larger total resistance causes a smaller total current. And the 400 Ω resistor is now in series in the battery, and so it would take this entire current. The 200 Ω and 300 Ω resistors would still split the current, and in the same proportion; but the total current is smaller than before, so both will take smaller current now. As for voltage, the equivalent series circuit has more resistance in the second branch, so the 400 Ω resistor takes more of the 9 V than before, and the first parallel combination takes less voltage than before.

** 3 .** (a) Series resistors take the same current as each other, which is equal to the total—that’s Kirchoff’s junction rule. An ammeter measures current; it should be connected in series so that all current that passes through the resistor also passes through the ammeter.

(b) Parallel resistors take the same voltage as each other, which is equal to the total—that’s Kirchoff’s loop rule. A voltmeter measures voltage; it should be connected in parallel so that it also takes the same voltage as the resistor it’s measuring.

(c) An ammeter is in series with a resistor. If the ammeter has a large resistance, then it also takes a large voltage, leaving less voltage to go across the resistor, and affecting the circuit. If the ammeter has a very *small* resistance, then it takes hardly any of the total voltage, leaving the resistor to have the same voltage and current as in a circuit without the ammeter.

(d) A voltmeter is in parallel with a resistor. If the voltmeter has a small resistance, then more current will choose the parallel path with the voltmeter than with the resistor, affecting the circuit. But if the voltmeter has a very *large* resistance, then almost none of the current would take the parallel path with the voltmeter in it, leaving the resistor to have the same voltage and current as in a circuit without the voltmeter.

** 4 .** (a) These forces are the same—Newton’s third law applies to all forces, even electrical forces.

(b) Coulomb’s law is the relevant equation. Using the variables as given, the force of one charge on the other is . You can rearrange this to if you want, but it’s not mandatory for AP-style credit, but notice that you should *not* be including a plus sign anywhere, or a minus sign if the charges were negative. The magnitude of a force should not have any signs on it. The direction is repulsive, so the +2*Q* charge pushes the +*Q* charge to the left.

(c) The distance between charges is in the denominator, so increasing the distance decreases the force. The distance term is squared, so that multiplying *a* by 3 multiplies the whole denominator by 3^{2} = 9. Therefore, the magnitude of the force is reduced by a factor of 9.

(d) Now the force between charges is attractive, meaning that the –*Q* charge will be pulled to the right. The magnitude of the force will not change from Part (b), since the amount of charge of each item and the distance between charges will not change.

** Rapid Review**

- Coulomb’s law is an equation for the force exerted by one electrical charge on another:
- The total amount of charge in a system (or in the universe itself) is always the same.
- The resistance
*R*of a wire of known dimensions is given by . **Series**resistors each carry the same current, which is equal to the total current through the series combination.- The voltage across
**series**resistors is different for each but adds to the total voltage across the series combination. - The voltage across
**parallel**resistors is the same for each and equal to the total voltage across the parallel combination. **Parallel**resistors each carry different currents, which add to the total current through the parallel combination.- The equivalent resistance of series resistors is the sum of all of the individual resistors. The equivalent resistance of parallel resistors is less than any individual resistor (and, if you have to do the calculation, is given by ).
- Ohm’s law says that voltage across a circuit element equals that element’s current times its resistance:
*V*=*IR*. This equation can*only*be used across a single row in a*V-I-R*chart. - Kirchoff’s junction rule says that the current entering a wire junction equals the current leaving the junction.
- Kirchoff’s loop rule says that the sum of voltage changes around a circuit loop is zero.
- To determine the power dissipated by a resistor, use the equation
*P*=*IV*.

^{1}^{ }No, not the flow of protons … okay, look, you really want to know? It’s the flow of “holes in the electron sea.” Understand? No? Neither do I. For the AP Exam, who cares. Current is the flow of positive charge, *capish* ?

^{2}^{ }Why is it okay to make up a voltage that wasn’t given? Say you’re asked to rank the currents in the resistors. It doesn’t matter how much current you calculate in each resistor, all that matters is whether *R* _{1} or *R* _{2} takes a bigger current. Any reasonable values will do for answering qualitative questions.

^{3}^{ }Um, is that really right? The current through parallel resistors should add to the total current. But 0.33 A plus 0.33 A gives 0.66 A, and we said the total current was 0.67 A. This is *fine* . Expect that rounding in a *V-I-R* chart will not allow for ten-figure accuracy. Who cares—the whole point is generally to help answer conceptual questions, anyway.

^{4}^{ }The potential energy can be converted by the resistor to internal energy of the resistor or the surrounding air, raising the temperature, as in a toaster oven; or could be converted to light, as in a lamp; or could even be converted to mechanical energy, as in an electric motor.

^{5}^{ }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 Kirchoff’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.

****