chapter 20 electric current, resistance, and ohm's la...electric potential. thus, an equivalent...

Much of modern technology involves devices that function using electric currents. Electric current is a certain type of flow of electric charge, but not just any movement of electric charge.

As an example of movement of electric charge that does not qualify as an electric current, consider a piece of metal warming in sunlight. "Normally" the free electrons in a metal move about randomly, in a seemingly disorganized, chaotic motion. Free electrons bump into atoms and change directions suddenly. This chaotic motion is NOT an electric current. On the contrary, an

Key Concepts

what is an electric current; the reality, and our simplified model•definition of current; "conventional current"•the principle of conservation of current; Kirchhoff's junction law•the "charge escalator" model of a battery•resistivity and resistance of a wire or other circuit element•Ohm's law•the ideal wire model for analyzing simple circuits•energy and power in simple circuits•circuit diagrams; the symbols used to represent circuit elements on circuit diagrams

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Kirchhoff's junction law; Kirchhoff's loop law•series and parallel circuits•equivalent resistance for a series circuit•equivalent resistance for a parallel circuit•analyzing simple circuits•analyzing more complex circuits•

Besides these key concepts found in the textbook, my lecture notes below also present three analogies that some students have found helpful in understanding electric currents in circuits: the money analogy, the playground analogy, and the necklace analogy.

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Chapter 20 Electric Current, Resistance, and Ohm's LawJanuary-18-18 9:15 PM

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motion is NOT an electric current. On the contrary, an electric current is an organized movement of charged particles. In a metal, the charged particles that move are electrons, whereas in a liquid or gas, it could be either positive or negative charged particles (often ions) that constitute an electric current.

"Reality" (at least microscopically):

When an electric field acts on a metal wire, an electric current is caused. Although the free electrons in the metal wire travel in somewhat of a zig-zag path, they tend to move in the opposite direction of the applied electric field.

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Our idealized (macroscopic) picture of reality:

Definition of electric current (in an averaged, macroscopic sense) is the rate of flow of electric charge:

Alternative (more precise) definition of current (for calculus lovers only):

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SI unit for current: the ampere (A), where 1 A = 1 C/s._____________________________________________________Example: A net charge of 1.2 C passes by a certain point in an electric circuit every 0.46 s. Determine the electric current.

Solution:

Thus, the current in the electric circuit is A._____________________________________________________Q: A net charge of 3.8 C passes by a certain point in an electric circuit every 2.5 s. Determine the electric current.

Solution:

Q: A current of 4.2 A exists in an electric circuit. How much charge flows past a certain point in the circuit in a time interval of 2.7 s?

Solution:

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In metal conductors, the charge carriers are electrons. Some electrons in a metal are relatively free to move around the metal (they are called conduction electrons) although they are still strongly bound to the metal as a whole; most of the electrons are tightly bound to individual atoms.

How do we know that the charge-carriers in a metal are negatively charged? As with everything in science, observations and experiments are the final referees. But a logically prior question is: How do we know that the "electricity" that is produced when you rub a balloon on your head is the same "electricity" that comes from a battery, and is the same electricity that comes from a wall outlet, and is the same electricity produced by electric eels, and that flows in human nerves, and that is present in lightning, etc.

Faraday (1833) and others performed numerous experiments to test the hypothesis that these "electricities" are really all the same, and they concluded that they are all the same, so that there is only one type of electricity.

The electron was finally identified ("discovered") in 1897, independently by Wiechert, Kaufmann, and Thomson. In their experiments, the electrons were produced by either heating a metal cathode, or bombarding a metal plate with positive ions. The particles released in each case (now known to be electrons) had the same properties, so it was natural to assume that there were plenty of free electrons inside a metal. How could this assumption be proved?

Tolman and Stewart conducted a series of experiments in 1916 and 1917, initially using copper wire, confirming their results using silver wire and aluminum wire. In their experiments, they used a spool of wire, and spun it like you would a lettuce-dryer (centrifuge). The free electrons piled up at one end, creating an electric field (potential difference) that could be measured. Had the mobile particles had a positive charge instead of a negative charge, the electric field (and the potential difference) would

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charge, the electric field (and the potential difference) would have been in the opposite direction, and Tolman and Stewart would have detected this. Their results confirmed that the mobile particles in metals were negatively charged.

Besides determining that the free charges in metals are negative, Tolman and Stewart also determined that the charge-to-mass ratio of the free charges in metals is roughly the same as the electrons discovered (and measured) by Wiechert, Kaufmann, and Thomson in 1897. Their results, and other similar experimental results, convinced the scientific community that the free charges in metals are indeed electrons.

Insulators have no free electrons and therefore do not carry electric currents, unless they suffer a catastrophic event. In other words, if you make the applied electric field strong enough to rip bound electrons from the atoms of an insulator, you can force them to pass through the insulator, but the electrons will then have so much energy that they will smash their way through, breaking chemical bonds, and perhaps causing a fire as they ram through. An example of this is a lightning strike.

In conducting fluids, the charge carriers might be positive, negative, or both.

Semiconductors are intermediate between conductors and insulators, and are found in all kinds of electronic devices (such as computer chips), photovoltaic (solar) panels, and so on. One needs quantum mechanics to really understand how semi-conductors work, although you'll start studying them more carefully in second-year electronics courses if you are a physics major here at Brock. Semiconductors are essential to micro-miniaturization, and it's thanks to them that we have all the wonderful electronic gadgets that we enjoy nowadays.

How electric currents are created

Contrast the chaotic motion of free electrons in a metal when there is no current (see diagram earlier in these notes) with the chaotic motion of free electrons in a metal when there is an electric current. When there is an electric current in a metal, there is a "drift" of electrons in the direction opposite to the

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there is a "drift" of electrons in the direction opposite to the direction of the electric field, and this drift is superposed on the chaotic motion.

We typically oversimplify electric currents when we think about them, focussing our attention on the drift, and ignoring the chaotic motion. Watch out for this when you are reading textbooks; one hopes that the ubiquitous oversimplification won't lead your thinking astray, and it's OK to indulge in it, but just remember that reality is more complex.

The modern way to think about the cause of an electric current is to recognize that an external electric field causes an electric current in a metal wire. We have already discussed the fact that an electric field can also be described by a collection of values of electric potential. Thus, an equivalent viewpoint is to consider that an electric potential difference (due to an external electric field) causes an electric current in a metal wire. Whether you focus on the electric field itself, or on the electric potential difference between two points on the wire, ultimately it's an electric field that causes an electric current in the metal wire.

Back in a bygone era, it was popular to speak of "electromotive force" (emf for short) as the cause of an electric current. Now that we know about electric fields, we recognize that an electric field is what causes the force that moves electric charges. (Remember the relation F = qE for the force exerted on a particle of charge q by an electric field E.) Nevertheless, one still frequently sees the term "emf" in textbooks; just recognize this for what it is, which is a throwback to a term popular in olden times. When you see the term "emf" just replace it by "potential difference" (or, if you prefer, "voltage difference") and you'll be fine. The SI unit for emf is the volt (V), so an emf represents a voltage difference of some sort.

Conventional current

Alas, we humans evidently have some anxiety around negative signs, and so some of our mathematical conventions are intended to eliminate unwanted negative signs from certain formulas. "Conventional current" is one of these conventions.

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When it comes to electric current in a metal wire, the idea behind conventional current is to pretend that the current is actually a flow of positive charges. We know this is not right; we know that it is negatively-charged electrons that flow in a metal wire. Nevertheless, inherent in the definition of current is the convention that it is the direction of flow of hypothetical positive charges.

Thus, the actual direction of drift of electrons in a metal is opposite to the direction of flow of conventional current.

If this is utterly confusing, I do apologize, but there is not much we can do about this, because virtually every physics textbook adopts the same convention. When we are analyzing circuits later in this chapter we will discuss current flow and current direction without giving much thought to the sign of the charge of the actual moving particles. Similarly, when we solve problems to determine the magnitudes of various currents, we can usually do so without thinking about the sign of the charge carriers.

In short, we will be able to solve just about all of the problems we're faced with without worrying too much about the direction of current. However, sometimes we'll have to pay attention to the direction of current. In such cases, just remember that when you see a current arrow in a diagram in the textbook that the direction of motion of the actual electrons is opposite. This is shown in the following diagram:

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The principle of conservation of current

As you know by now, conservation principles in physics provide us with powerful tools for furthering our understanding, and also for solving specific problems. Based on the principle of conservation of charge (the net charge in a system is constant), we can derive the principle of conservation of current. In any electric circuit, current does not appear or disappear.

Based on the principle of conservation of current, one can prove Kirchhoff's junction law: At a junction in an electric circuit, the total current entering the junction is equal to the total current leaving the junction.

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______________________________________________Q: Determine the unknown current.

Q: Determine the unknown current.

Q: Determine the unknown current.

Solution:

Solution:

Solution:

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How a battery works: The "charge escalator"

A battery is an energy storage device. In moving (inside the battery) from the top of the 1.5 V battery to the bottom, an electron's potential energy is increased by

U = q V = (-e)(-1.5V) = 1.5 eV

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Solution:

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U = q V = (-e)(-1.5V) = 1.5 eV

It's useful to contrast batteries with capacitors. Both are capable of storing charge and creating an electric current. A capacitor can typically create a much larger current, but for a much smaller time. A battery can typically create only a smaller current, but is able to sustain the smaller current for a much longer time.

For example, a capacitor is used for a camera flash, when you need a large current, but only for a short time. In this case a battery and a capacitor work together; the battery slowly charges the capacitor, and then the capacitor rapidly discharges by sending a large current through the flash light for a short time.

Resistance of a wire (R)

The electrical resistance of a wire is a measure of how hard it is to create a current in the wire for a given voltage difference. The resistance depends on the material that the wire is made of, and the shape and size of the wire. For a cylindrical wire, the resistance R can be calculated by:

where ρ is the resistivity of the material, L is the length of the wire, and A is the wire's cross-sectional area. The resistivity of the wire accounts for the fact that wires of the same dimensions but made of different materials may have different resistances.

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The SI unit for resistance is the ohm (Ω); the SI unit for

resistivity is the ohm-metre (Ωm).

Typical resistances:metal wires: less than 1 Ωgood insulators: more than 109 Ω_________________________________________________Example: Determine the resistance of a copper wire that has a radius of 1.3 mm and a length of 12.6 m. The resistivity of copper is 1.72 × 10--8 Ω.m.

Solution:

Ideal wire model

Because metal wires have very little resistance, it's often a good approximation when analyzing circuits to make the

________________________________________________Q: Determine the resistance of an aluminum wire that has a radius of 1.5 mm and a length of 11.8 m. The resistivity of aluminum is 2.65 × 10--8 Ω.m.

Solution

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good approximation when analyzing circuits to make the approximation that wires connecting other circuit elements have zero resistance. This is called the ideal wire model.

Question: Are the formulas we learned in earlier chapters for the electric field due to a point charge also valid for the electric field in a wire?

Answer: These formulas are not applicable to a wire in a simple way, because a wire is a collection of many, many point charges. We would have to apply the formula for the electric field for a point charge for each point charge in the wire, and then add up all of the fields to obtain the net electric field. This amounts to a calculus problem, and is beyond the level of this course.

A single point charge is a simple enough system that it's possible to write a simple formula for the electric field due to a single point charge. A hunk of metal is a very complicated system, and writing a simple formula for the electric field due to a hunk of metal is out of the question.

Ohmic and non-ohmic materials

For some materials, the resistance is nearly independent of temperature; such materials are called ohmic materials. Most metals are ohmic over "reasonable" temperature ranges; the wishy-washiness of this statement is a good sign that the concept of an ohmic material is just an approximation.

For non-ohmic materials, the resistance depends on temperature. A good example of a non-ohmic resistor is the heating element in a toaster or in the rear window of your car, or the filament in an incandescent light bulb.

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current. As we shall see later, the greater the current passing through a material, the greater its temperature. Thus, when the current increases the temperature of the material also increases; for an ohmic material, this does not result in an increase of resistance.

Question: Does electric current occur along the surface of a wire, or throughout its "bulk"? Answer: We can get a hint from the formula R = ρL/A. If the current only flowed near the surface of the wire, we would expect that the circumference of the wire should be in the denominator of the formula, not the cross-sectional area. The presence of the cross-sectional area in the denominator of the formula suggests that the current flows throughout the bulk of the wire, not just near the surface. Careful experiments confirm this.

Full disclosure for advanced readers: The formula above (and conclusions above) are valid only for direct current (DC). For alternating current (AC), current tends to flow mainly near the surface of the wire; do a search on the term "skin effect" if you would like to learn more about this.

Ohm's law: Connection between voltage difference and current for an ohmic material

Example: The potential difference across the two ends of a resistor is 12.5 V, and the resistance of the resistor is 40 Ω. Determine the current through the resistor.

Solution:

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Superconductivity

When certain materials are cooled to extremely low temperatures, they exhibit bizarre properties. For example, when certain materials are cooled to below a certain critical temperature, their electrical resistance suddenly decreases to practically zero!

(We won't be able to explain why this is so in this course; you'll need to learn a fair bit of quantum mechanics first before you can really come to grips with why superconductivity occurs.)

Q: The potential difference across the two ends of a resistor is 30 V, and the resistance of the resistor is 18.5 Ω. Determine the current through the resistor.

Solution:

Q: The potential difference across the two ends of a resistor is 40 V, and the current through the resistor is 5 A. Determine the resistance of the resistor.

Solution:

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Zero electrical resistance means that if you set up a current in a super-conducting wire the current will remain (seemingly) indefinitely, and so will the resulting magnetic field. (As we'll learn later in the course, an electric current causes a magnetic field.) Superconductivity therefore gives us a way to create powerful (and seemingly permanent) electromagnets. The cost is keeping the material cold.

In the 1980s, so called "high-temperature" superconductors were discovered. Their critical temperatures are in the liquid-nitrogen range, and therefore much more easily (and inexpensively) achievable than the near-absolute-zero critical temperatures of "ordinary" superconductors.

Nowadays, superconductors are used in magnetically-levitated ("maglev") trains, MRI machines, and in many other applications.

If one of you could discover a material that became superconducting at a critical temperature that is near room temperature, it would be a major advance, and extremely useful technologically.

What are you waiting for? Get to work!

Energy and power in an electric circuit

Recall that power is interpreted in a variety of ways in physics:power is the rate of doing work•power is the rate at which energy is transferred from one system to another, or from one object to another

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power is the rate at which energy is transformed from one type to another type

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Note that the SI unit for potential, the volt, is defined so

that 1 V = 1 (N/C)m (which connects potential nicely to electric field), but it is also true that 1 V = 1 J/C (which connects potential nicely to work or energy).

Suppose a particle of charge q is moved through a potential

difference V. Then the magnitude of the change in

potential energy of the charge is U = qV. Thus, the work done on the charge (or maybe by the charge; it depends on

the sign of the work) has magnitude U = qV. The power transferred (or dissipated, or whatever) is therefore

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________________________________________________

Example: Determine the current through a 60-W light bulb connected to a 120-V power supply.

Solution:

________________________________________________Q: Continuing the previous problem, determine the resistance of the 60-W light bulb connected to a 120-V power supply.

Solution:

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"people carrying money" model (reminds us that current is not consumed; the people don't disappear, they just "pay their money" and keep going; that is, the electrons don't disappear, they pay their electrical potential energy, which is converted to thermal energy and/or light in a resistor) and they keep going

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"necklace model (reminds us that current begins to flow everywhere in a circuit simultaneously, so there is no time delay between one light bulb lighting up and another one

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Simplified models (analogies) for electric current that many people find useful

Q: Determine the power output of a light bulb that has a resistance of 120 Ω and is connected to a 120-V power supply.

Solution:

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Chapter 21: Electric Circuits________________________________________________

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delay between one light bulb lighting up and another one "farther along" in the circuit lighting up)"playground" model (helps us to see how the electrical potential energy of flowing electrons changes as the electrons go through various circuit elements; the kinetic energy of the electrons is assumed not to change as the electrons travel around the circuit, only their electric potential energy changes)

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"fluid flow" model; this model was commonly used to teach electric circuits a generation ago, and it's not a bad model, but I won't emphasize it much because I find that not many people have a good feel for water flow in a pipe, so it's perhaps not as useful as the other models; however, if you like it, use it

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Here's an example of the playground model:

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Gradually reduce the given circuit, step-by-step, to a simpler equivalent one that is more easily solved. Once you determine the current coming from the battery, you can then go back to the original circuit to determine other quantities of interest.

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Use Kirchhoff's laws; this relies on your abilities to solve systems of equations.

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A kind of hybrid method, that starts with labelling the potentials at key nodes of the circuit.

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Now we'll begin analyzing electric circuits, starting with relatively simple ones and moving on to more complex ones. There are three main methods:

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For a series circuit, there is a short cut for finding the equivalent resistance of the circuit without going through this detailed analysis. What we do is go through the detailed analysis once, for a general circuit, and then draw the appropriate conclusion. The derivation follows:

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Notice that with each addition of a new parallel branch to the Ch20-21L

Notice that with each addition of a new parallel branch to the circuit, each branch having a resistance of 10 ohms, the current from the battery increases. This seems strange to many students at first; after all you are "adding" resistance to a circuit, and yet the total (equivalent, effective) resistance of the circuit decreases.

One way to understand this is to use the water analogy; if we assume that a battery is a constant-voltage device (has the same potential difference across its terminals no matter which circuit you attach), this is like having a constant-pressure water source for your water pipes. Thus, if you use such a water source to irrigate your farm, if you only turn on one hose, you get a certain water flow (analogous to electric current). If you turn on two hoses, you get more flow, and if you turn on more than two hoses, you get even more flow.

If some of the hoses have different diameters, then they will naturally carry more water than the smaller hoses if the pressure is sufficient. It's the same with parallel electric circuits; branches with less resistance will carry more current than branches with more resistance.

Be careful about the slogan "electricity seeks the path of least resistance." What it means is exactly what is described in the previous paragraph. It does not mean that current will only flow through the one path that has the least resistance; current will flow on every parallel path, but more current will flow along paths that have less resistance.

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Now that we've done a few examples, we'll derive a general formula for the equivalent resistance for resistors in parallel.

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______________________________________________Q: Determine the equivalent resistance if resistors with resistance 5 Ω, 10 Ω, and 20 Ω are connected in series.

Solution:

Q: Determine the equivalent resistance if resistors with resistance 5 Ω, 10 Ω, and 20 Ω are connected in parallel.

Solution:

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More complex circuits___________________

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Using Kirchhoff's laws to solve a complex circuit problem

Determine the currents in the following circuit. Assume that V1 = 20 V, V2 = 40 V, R1 = 10 Ω, R2 = 20 Ω, and R3 = 30 Ω.

Use equation (1) to replace I2 in equations (2) and (3):

Solution: Apply Kirchhoff's junction law at point A. Then apply Kirchhoff's loop law to any two of the three possible loops; i.e., C to D to A to C, or C to A to B to C, or C to D to A to B to C. (Only two lead to independent equations; if you use all three loops you end up with one of the three resulting equations being dependent, so there is no advantage to doing the extra work of using all three loops. And, of course, you can start at other points and go in other senses, but it all works similarly.)

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Use equation (1) to replace I2 in equations (2) and (3):

Thus,

You can check the results by substituting them into the original three equations.

Alternative solution:

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Alternative solution:

We can try to solve the previous problem more in the spirit of the method we have been using throughout this chapter; that is, attempting to label all key points in the circuit with a potential value, and then solving from there. If we try this method with this problem, we can solve it, provided we're willing to live with the small complication that we won't immediately know the potential at one of the points of the circuit. Here are the details:

Begin by identifying the potential at key points of the circuit, as shown in the diagram below. You'll note that it's not immediately apparent what the potential is at the upper right corner; this is a pain. If we knew what the potential were at that point, we would be able to solve the problem in a straightforward way.

After thinking for a bit, and failing to figure out what the potential is at the upper right corner, we accept the fact that we probably can't determine this potential immediately, but we don't totally give up. Rather, we label the potential by x. This is a good problem-solving strategy: If you wish to know something, but can't figure out how to determine it, call it something, so that you can have a hope of determining it through writing equations involving the variable you just introduced.

Now that we've called the unknown potential x, we can proceed to apply Ohm's law to each resistor; this is done to the right of the diagram.

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The solutions we obtain by this method are the same as the ones obtained using the other method, so this is comforting.

Try each method for a number of complex problems and decide

Now we have three equations for four unknowns (x and the three currents), so we need another independent equation to be able to solve the problem. We've used Ohm's law for each resistor (which is equivalent to using Kirchhoff's loop laws for various loops), but we haven't used Kirchhoff's junction law, which relates the three currents. This will provide us the needed fourth equation.

So, write Kirchhoff's junction law for the junction labelled x, then insert the expressions derived above for each current into this fourth equation, and we'll end up with one equation for x. After solving this equation for x, we can back-substitute into the other three equations to determine the three currents. Here are the details:

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Try each method for a number of complex problems and decide for yourself which method is easier for you. You're free to use any valid method to solve problems, so play with the various methods and choose a method that is best for you._____________________________________________________

Capacitors in Parallel

In the same way that resistors in a series circuit can be replaced by a single equivalent resistor, capacitors in a parallel circuit can be replaced by a single equivalent capacitor.

In the circuit on the left, the voltage across each capacitor is the same, and is equal to the potential difference across the battery. However, the magnitude of the charge on one plate of each capacitor is not necessarily the same, but is related to the capacitance as follows:

If the circuit on the right is to be equivalent to the circuit on the left, then the magnitude of the charge on one plate of the single capacitor must equal the sum of the charges on the capacitors in the left circuit:

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The same kind of argument shows that for any number of capacitors placed in parallel, the equivalent capacitance is

Capacitors in Series

In the same way that resistors in a parallel circuit can be replaced by a single equivalent resistor, capacitors in a series circuit can be replaced by a single equivalent capacitor.

For capacitors in series, each capacitor has the same magnitude of charge on each plate. The battery pushes out charge of magnitude Q onto the nearest plate of the nearest capacitor, while pulling out charge of magnitude Q from the nearest plate of the nearest capacitor on the opposite side of the battery. Each capacitor subsequently pulls or pushes the same amount of charge onto or from each subsequent capacitor. Thus,

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Selected problems and solutions

Problem:

By Kirchhoff's loop law,

If the circuit on the right is to be equivalent to the circuit on the left, then the same amount of charge must leave the battery, and so

Substituting the expression from equation (2) into equation (1), and simplifying, we obtain

The same kind of argument shows that for any number of capacitors placed in series, the equivalent capacitance is

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Electrical safety

Grounding, ground-fault interruptors: we'll discuss these concepts in a later chapter.

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Problem:

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Problem:

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________________________________________________

Problem: In a typical lightning strike, 2.5 C flows from cloud to ground in 0.20 ms. What is the current during the strike?

Conceptual Question: A light bulb is connected to a battery by two copper wires of equal lengths but different thicknesses. A thick wire connects one side of the light bulb to the positive terminal of the battery and a thin wire connects the other side of the bulb to the negative terminal. (a) Which wire carries a greater current? (b) If the two wires are switched, will the bulb get brighter, dimmer, or stay the same? Explain.

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to ground in 0.20 ms. What is the current during the strike?

Solution:

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Problem: In an ionic solution, 5.0 × 1015 positive ions with charge +2e pass to the right each second while 6.0 × 1015

negative ions with charge e pass to the left. What are the magnitude and direction of current in the solution?

Solution:

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Problem: A car battery is rated at 90 Ahr, meaning that it can (for example) supply a 90 A current for 1 hr before being completely discharged. If you leave your headlights on until the battery is completely dead, how much charge leaves the positive terminal of the battery?

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Solution:

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Conceptual Question: A wire carries 4 A of current. What is the current in a second wire that carries twice as much charge in half the time?

Solution:

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Conceptual Question: The wires in the figure are all made of the same material; the length and radius of each wire is noted. Rank in order, from largest to smallest, the resistances of these wires.

Solution:

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Problem: Rank in order, from largest to smallest, the currents through the four resistors in the figure.

Solution:

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Conceptual Question: When lightning strikes the ground, it generates a large electric field along the surface of the ground directed toward the point of the strike. People near a lightning

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directed toward the point of the strike. People near a lightning strike are often injured not by the lightning itself but by a large current that flows up one leg, through the body, and down the other due to this electric field. To minimize this possibility, you are advised to stand with your feet close together if you are trapped outside during a lightning storm. Explain why this is beneficial.

Solution:

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Problem: A motorcyclist is making an electric vest that, when connected to her motorcycle's 12 V battery, will warm her on cold rides. She is using 0.25 mm diameter copper wire, and she wants a current of 4 A in the wire. What length of wire must she use?

Solution:

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Problem: A 70 W electric blanket runs at 18 V. (a) What is the resistance of the wire in the blanket? (b) How much current does the wire carry?

Solution:

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Problem: The total charge a household battery can supply is

given in units of mAhr. For example, a 9.0 V alkaline battery is

rated at 450 mAhr, meaning that such a battery could supply a 1 mA current for 450 hr, a 2 mA current for 225 hr, etc. How much energy, in joules, is this battery capable of supplying?

Solution:

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Problem: A sculptor has asked you to help electroplate gold onto a brass statue. You know that the charge carriers in the ionic solution are monovalent (charge e) gold ions, and you've calculated that you must deposit 0.50 g of gold to reach the necessary thickness. How much current do you need, in mA, to plate the statue in 3.0 hr? (atomic mass of gold = 197.0; Avogadro's number = 6.02 × 1023)

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Problem: The hot dog cooker described in the chapter heats hot dogs by connecting them to 120 V household electricity. A typical hot dog has a mass of 60 g and a resistance of 150 Ω. How long will it take for the cooker to raise the temperature of the hot dog from 20°C to 80°C? The specific heat of a hot

dog is approximately 2500 J/kgK.

Solution:

gold = 197.0; Avogadro's number = 6.02 × 1023)

Solution:

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Additional problems and solutions

Problem: The ammeter in the figure reads 3.0 A. Determine

I1, I2, and .

Problem: Two 75 W light bulbs are wired in series, then the

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Problem: Two 75 W light bulbs are wired in series, then the combination is connected to a 120 V power supply. How much power is dissipated by each bulb? (Additional question: Repeat if the two bulbs are connected in parallel.)

Solution:

Problem: When two resistors are connected in parallel across a battery of unknown voltage, one resistor carries a current of 3.2 A and the second carries a current of 1.8 A. How much current will be supplied by the same battery if these two resistors are connected to it in series?

Solution:

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Problem: The 10 Ω resistor in the figure dissipates 40 W of power. How much power do the other two resistors dissipate?

Ch20-21L

Interesting (and more advanced) examples:

the resistor cube1.the infinite array of resistors2.For the resistor cube, assume that each edge of the cube has resistance 1 ohm, and determine the resistance between the points A and B. For the infinite array of resistors in the other diagram, assume that each resistor has resistance 1 ohm, and determine the resistance between points A and B.

These final two examples are definitely for advanced students; don't worry if you find them challenging.

Ch20-21L

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Further practice problems

For each circuit, determine the current through each resistor.

Problem:

Problem: This circuit can be solved in a straight-forward way by first determining the equivalent resistance of the circuit by simplifying the circuit step-by-step, identifying parallel and series pieces of the circuit at each stage.

Solution: By this time you should have done so many practice problems that this one is quite easy for you; if not, you are definitely behind and need to step up your practice!

The equivalent resistance of the circuit is 10 + 20 + 10 = 40 ohms, and therefore the current from the battery is 100/40 = 2.5 A. Because it's a simple series circuit, the current is the same at each point in the circuit._________________________________________________

Ch20-21L

Problem: The same method as in the previous problem works out fine here too.

Once you know the current flowing from the battery you can proceed to determine the current through each resistor. I'll let you determine these on your own. My results are

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Ch20-21L

Ch20-21L

Problem:

Solution: We benefit from the symmetry of the circuit. You might try solving this using Kirchhoff's laws, but I'll try our "hybrid" method by labelling potentials and using Ohm's law.

Thus, the current flowing from the battery is

Now continue to determine the individual currents through each resistor. My results are

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Ch20-21L

Problem: The "hybrid method" worked well for the last problems, so I'll try it again this time. As usual, you can solve all of these problems by using just Kirchhoff's laws. Try this yourself for an alternative.

Solution 1:

Once each node is labelled with its potential, the problem is seen to be straightforward, even though it may have been intimidating at first. The current through the top resistor is zero, because the voltage drop across it is 10 - 10 = 0, and you can use Ohm's law to show that the current through each of the other two resistors is 1 A. Thus, the current flowing from each battery is 1 A.

Ch20-21L

First eliminate I1 and I3 from the junction equations; then substitute the remaining currents from the voltage drop equations and solve for x.

Ch20-21L

Solution 2: Redraw the circuit as follows, then simplify the circuit by identifying parallel and series parts.

Therefore,

Ch20-21L

Challenge Problem:

Solution: I guessed the directions of all the currents, as indicated in the diagram below. If some currents come out negative, it means that the corresponding guesses were wrong, and the corresponding current actually goes in the opposite direction.

Now proceed to determine the current from the battery, and then continue to determine all of the rest of the currents. You can check your results against Solution 1.

Ch20-21L

Notice that there are 11 unknowns (8 currents and 3 potentials) and we have written 12 equations, so presumably we have enough equations to solve the problem, but not all of the equations are independent. There are various ways to solve the system of equations, and after you've got the hang of things by doing some hand-calculations, using computer software is advisable. But there is definitely much to learn by working things out by hand, especially when you are just beginning to learn a subject; you really develop good intuition for situations by playing with them.

After some reflection, I decided that a good strategy is to eliminate the currents first, because then the potential equations are very easy to solve. So I chose to first eliminate currents 1 and 5, because they don't appear in the potential equations, then substitute the currents from the potential equations into the remaining junction equations, with the aim of ultimately determining the potentials. From this we can then substitute back to determine all the currents.

Allons enfants!

Eliminating currents 1 and 5 from the junction equations, we end up with

Ch20-21L

Combining the first two equations above results in the third; thus the third equation can be discarded.

Next, substitute the expressions for the currents in the voltage drop equations into the first two equations above, to get:

Clearing fractions and doing other minor simplification results in

One voltage drop equation has not been used yet:

So now we have three equations for the three unknown potentials. Solving the three equations, you will obtain

Back substituting, you will obtain the following currents:

Ch20-21L

Notice that I got the direction of current 4 wrong, but all the other currents indicated in the diagram are in the correct direction.

Ch20-21L

chapter 20 electric current, resistance, and ohm's la...electric potential. thus, an equivalent...

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