# Chapter 25 Electric Currents and Resistance

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Chapter 25 Electric Currents and Resistance
Chapter 25 opener. The glow of the thin wire filament of a light bulb is caused by the electric current passing through it. Electric energy is transformed to thermal energy (via collisions between moving electrons and atoms of the wire), which causes the wire’s temperature to become so high that it glows. Electric current and electric power in electric circuits are of basic importance in everyday life. We examine both dc and ac in this Chapter, and include the microscopic analysis of electric current.

25-1 The Electric Battery Volta discovered that electricity could be created if dissimilar metals were connected by a conductive solution called an electrolyte. This is a simple electric cell. Figure Simple electric cell.

25-1 The Electric Battery A battery transforms chemical energy into electrical energy. Chemical reactions within the cell create a potential difference between the terminals by slowly dissolving them. This potential difference can be maintained even if a current is kept flowing, until one or the other terminal is completely dissolved.

25-1 The Electric Battery Several cells connected together make a battery, although now we refer to a single cell as a battery as well. Figure (a) Diagram of an ordinary dry cell (like a D-cell or AA). The cylindrical zinc cup is covered on the sides; its flat bottom is the negative terminal. (b) Two dry cells (AA type) connected in series. Note that the positive terminal of one cell pushes against the negative terminal of the other.

Unit of electric current: the ampere, A:
Electric current is the rate of flow of charge through a conductor: The instantaneous current is given by: Unit of electric current: the ampere, A: 1 A = 1 C/s.

25-2 Electric Current A complete circuit is one where current can flow all the way around. Note that the schematic drawing doesn’t look much like the physical circuit! Figure (a) A simple electric circuit. (b) Schematic drawing of the same circuit, consisting of a battery, connecting wires (thick gray lines), and a lightbulb or other device.

25-2 Electric Current Example 25-1: Current is flow of charge.
A steady current of 2.5 A exists in a wire for 4.0 min. (a) How much total charge passed by a given point in the circuit during those 4.0 min? (b) How many electrons would this be? Solution: a. Charge = current x time; convert minutes to seconds. Q = 600 C. b. Divide by electron charge: n = 3.8 x 1021 electrons.

25-2 Electric Current Conceptual Example 25-2: How to connect a battery. What is wrong with each of the schemes shown for lighting a flashlight bulb with a flashlight battery and a single wire? Solution: a. Not a complete circuit. b. Circuit does not include both terminals of battery (so no potential difference, and no current) c. This will work. Just make sure the wire at the top touches only the bulb and not the battery!

25-2 Electric Current By convention, current is defined as flowing from + to -. Electrons actually flow in the opposite direction, but not all currents consist of electrons. Figure Conventional current from + to - is equivalent to a negative electron flow from – to +.

25-3 Ohm’s Law: Resistance and Resistors
Experimentally, it is found that the current in a wire is proportional to the potential difference between its ends:

25-3 Ohm’s Law: Resistance and Resistors
The ratio of voltage to current is called the resistance:

25-3 Ohm’s Law: Resistance and Resistors
In many conductors, the resistance is independent of the voltage; this relationship is called Ohm’s law. Materials that do not follow Ohm’s law are called nonohmic. Figure Graphs of current vs. voltage for (a) a metal conductor which obeys Ohm’s law, and (b) for a nonohmic device, in this case a semiconductor diode. Unit of resistance: the ohm, Ω: 1 Ω = 1 V/A.

25-3 Ohm’s Law: Resistance and Resistors
Conceptual Example 25-3: Current and potential. Current I enters a resistor R as shown. (a) Is the potential higher at point A or at point B? (b) Is the current greater at point A or at point B? Solution: a. Point A is at higher potential (current flows “downhill”). b. The current is the same – all the charge that flows past A also flows past B.

25-3 Ohm’s Law: Resistance and Resistors
Example 25-4: Flashlight bulb resistance. A small flashlight bulb draws 300 mA from its 1.5-V battery. (a) What is the resistance of the bulb? (b) If the battery becomes weak and the voltage drops to 1.2 V, how would the current change? Figure Flashlight (Example 25–4). Note how the circuit is completed along the side strip. a. R = V/I = 5.0 Ω. b. Assuming the resistance stays the same, the current will drop to 240 mA.

25-3 Ohm’s Law: Resistance and Resistors
Standard resistors are manufactured for use in electric circuits; they are color-coded to indicate their value and precision. Figure The resistance value of a given resistor is written on the exterior, or may be given as a color code as shown above and in the Table: the first two colors represent the first two digits in the value of the resistance, the third color represents the power of ten that it must be multiplied by, and the fourth is the manufactured tolerance. For example, a resistor whose four colors are red, green, yellow, and silver has a resistance of 25 x 104 Ω = 250,000 Ω = 250 kΩ, plus or minus 10%. An alternate example of a simple code is a number such as 104, which means R = 1.0 x 104 Ω.

25-3 Ohm’s Law: Resistance and Resistors
This is the standard resistor color code. Note that the colors from red to violet are in the order they appear in a rainbow.

25-3 Ohm’s Law: Resistance and Resistors
Some clarifications: Batteries maintain a (nearly) constant potential difference; the current varies. Resistance is a property of a material or device. Current is not a vector but it does have a direction. Current and charge do not get used up. Whatever charge goes in one end of a circuit comes out the other end.

25-4 Resistivity The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area: The constant ρ, the resistivity, is characteristic of the material.

25-4 Resistivity This table gives the resistivity and temperature coefficients of typical conductors, semiconductors, and insulators.

25-4 Resistivity Example 25-5: Speaker wires.
Suppose you want to connect your stereo to remote speakers. (a) If each wire must be 20 m long, what diameter copper wire should you use to keep the resistance less than 0.10 Ω per wire? (b) If the current to each speaker is 4.0 A, what is the potential difference, or voltage drop, across each wire? Solution: a. R = ρl/A; you can solve this for A and then find the diameter. D = 2.1 mm. b. V = IR = 0.40 V.

25-4 Resistivity Conceptual Example 25-6: Stretching changes resistance. Suppose a wire of resistance R could be stretched uniformly until it was twice its original length. What would happen to its resistance? Solution: The total volume of the wire should stay the same; therefore if the length doubles, the cross-sectional area is halved. This increases the resistance by a factor of 4.

25-4 Resistivity For any given material, the resistivity increases with temperature: Semiconductors are complex materials, and may have resistivities that decrease with temperature.

25-4 Resistivity Example 25-7: Resistance thermometer.
The variation in electrical resistance with temperature can be used to make precise temperature measurements. Platinum is commonly used since it is relatively free from corrosive effects and has a high melting point. Suppose at 20.0°C the resistance of a platinum resistance thermometer is Ω. When placed in a particular solution, the resistance is Ω. What is the temperature of this solution? Solution: The resistance is proportional to the resistivity; use the temperature dependence of resistivity to find the temperature. T = 56.0 °C.

25-5 Electric Power Power, as in kinematics, is the energy transformed by a device per unit time: or

25-5 Electric Power The unit of power is the watt, W.
For ohmic devices, we can make the substitutions:

25-5 Electric Power Example 25-8: Headlights.
Calculate the resistance of a 40-W automobile headlight designed for 12 V. Solution: R = V2/P = 3.6 Ω.

25-5 Electric Power What you pay for on your electric bill is not power, but energy – the power consumption multiplied by the time. We have been measuring energy in joules, but the electric company measures it in kilowatt-hours, kWh: 1 kWh = (1000 W)(3600 s) = 3.60 x 106 J.

25-5 Electric Power Example 25-9: Electric heater.
An electric heater draws a steady 15.0 A on a 120-V line. How much power does it require and how much does it cost per month (30 days) if it operates 3.0 h per day and the electric company charges 9.2 cents per kWh? Solution: P = IV = 1800 W W x 3.0 h/day x 30 days = 162 kWh. At 9.2 cents per kWh, this would cost \$15.

25-5 Electric Power Example 25-10: Lightning bolt.
Lightning is a spectacular example of electric current in a natural phenomenon. There is much variability to lightning bolts, but a typical event can transfer 109 J of energy across a potential difference of perhaps 5 x 107 V during a time interval of about 0.2 s. Use this information to estimate (a) the total amount of charge transferred between cloud and ground, (b) the current in the lightning bolt, and (c) the average power delivered over the 0.2 s. Solution: a. The charge is the change in energy divided by the change in electric potential: Q = 20 C. b. I = Q/t = 100 A. c. P = energy/time = IV = 5 x 109 W = 5 GW.

25-6 Power in Household Circuits
The wires used in homes to carry electricity have very low resistance. However, if the current is high enough, the power will increase and the wires can become hot enough to start a fire. To avoid this, we use fuses or circuit breakers, which disconnect when the current goes above a predetermined value.

25-6 Power in Household Circuits
Fuses are one-use items – if they blow, the fuse is destroyed and must be replaced. Figure 25-19a. Fuses. When the current exceeds a certain value, the metallic ribbon melts and the circuit opens. Then the fuse must be replaced.

25-6 Power in Household Circuits
Circuit breakers, which are now much more common in homes than they once were, are switches that will open if the current is too high; they can then be reset. Figure (b) One type of circuit breaker. The electric current passes through a bimetallic strip. When the current exceeds a safe level, the heating of the bimetallic strip causes the strip to bend so far to the left that the notch in the spring-loaded metal strip drops down over the end of the bimetallic strip; (c) the circuit then opens at the contact points (one is attached to the metal strip) and the outside switch is also flipped. As soon as the bimetallic strip cools down, it can be reset using the outside switch. Magnetic-type circuit breakers are discussed in Chapters 27 and 29.

25-6 Power in Household Circuits
Example 25-11: Will a fuse blow? Determine the total current drawn by all the devices in the circuit shown. Solution: The current is given by I = P/V, where V = 120 V. Adding the currents gives 28.7 A, which exceeds the usual 20-A circuit breakers found in most household applications. The electric heater should probably be on its own circuit.

25-6 Power in Household Circuits
Conceptual Example 25-12: A dangerous extension cord. Your 1800-W portable electric heater is too far from your desk to warm your feet. Its cord is too short, so you plug it into an extension cord rated at 11 A. Why is this dangerous? Solution: An 1800-W heater operating at 120 V draws 15 A of current, which exceeds the rating of the extension cord. This creates a risk of overheating and fire.

25-7 Alternating Current Current from a battery flows steadily in one direction (direct current, DC). Current from a power plant varies sinusoidally (alternating current, AC). Figure (a) Direct current. (b) Alternating current.

25-7 Alternating Current The voltage varies sinusoidally with time:
, , as does the current:

25-7 Alternating Current Multiplying the current and the voltage gives the power: Figure Power transformed in a resistor in an ac circuit.

25-7 Alternating Current Usually we are interested in the average power: .

25-7 Alternating Current The current and voltage both have average values of zero, so we square them, take the average, then take the square root, yielding the root-mean-square (rms) value:

25-7 Alternating Current Example 25-13: Hair dryer.
(a) Calculate the resistance and the peak current in a 1000-W hair dryer connected to a 120-V line. (b) What happens if it is connected to a 240-V line in Britain? Figure A hair dryer. Most of the current goes through the heating coils, a pure resistance; a small part goes to the motor to turn the fan. Example 25–13. Solution: a. The rms current is the power divided by the rms voltage, or 8.33 A, so the peak current is 11.8 A. Then R = V/I = 14.4 Ω. b. Assuming the resistance would stay the same (probably wrong – the heat would increase it), doubling the voltage would mean doubling the current, and the power (P = IV) would increase by a factor of 4. This would not be good for the hair dryer.

25-8 Microscopic View of Electric Current: Current Density and Drift Velocity
Electrons in a conductor have large, random speeds just due to their temperature. When a potential difference is applied, the electrons also acquire an average drift velocity, which is generally considerably smaller than the thermal velocity. Figure Electric field E in a uniform wire of cross-sectional area A carrying a current I. The current density j = I/A.

25-8 Microscopic View of Electric Current: Current Density and Drift Velocity
We define the current density (current per unit area) – this is a convenient concept for relating the microscopic motions of electrons to the macroscopic current: If the current is not uniform: .

25-8 Microscopic View of Electric Current: Current Density and Drift Velocity
This drift speed is related to the current in the wire, and also to the number of electrons per unit volume: and

25-8 Microscopic View of Electric Current: Current Density and Drift Velocity
Example 25-14: Electron speeds in a wire. A copper wire 3.2 mm in diameter carries a 5.0-A current. Determine (a) the current density in the wire, and (b) the drift velocity of the free electrons. (c) Estimate the rms speed of electrons assuming they behave like an ideal gas at 20°C. Assume that one electron per Cu atom is free to move (the others remain bound to the atom). Solution: a. j = I/A = 6.2 x 105 A/m2 b. The number of free electrons per unit volume is equal to the number of atoms per unit volume. For pure copper, the number of atoms per unit volume can be written as the number of atoms in a mole divided by the volume of a mole of copper; the volume of a mole of copper is given by the atomic mass of copper in grams (63.5 g = 1 mole of copper) divided by the density of copper. This gives 8.4 x 1028 atoms/m3. Then the drift velocity is j/ne = 4.6 x 10-5 m/s. c. If we model the electrons as a three-dimensional single-atom gas, the rms speed is given by the square root of 3kT/m, which is 1.2 x 105 m/s. (This actually underestimates the speed of the electrons by about a factor of 10).

25-8 Microscopic View of Electric Current: Current Density and Drift Velocity
The electric field inside a current-carrying wire can be found from the relationship between the current, voltage, and resistance. Writing R = ρ l/lA, I = jA, and V = E l , and substituting in Ohm’s law gives:

25-8 Microscopic View of Electric Current: Current Density and Drift Velocity
Example 25-15: Electric field inside a wire. What is the electric field inside the wire of Example 25–14? (The current density was found to be 6.2 x 105 A/m2.) Solution: E = ρj = 0.01 V/m.

25-9 Superconductivity In general, resistivity decreases as temperature decreases. Some materials, however, have resistivity that falls abruptly to zero at a very low temperature, called the critical temperature, TC. Figure A superconducting material has zero resistivity when its temperature is below TC, its “critical temperature.” At TC, the resistivity jumps to a “normal” nonzero value and increases with temperature as most materials do (Eq. 25–5).

25-9 Superconductivity Experiments have shown that currents, once started, can flow through these materials for years without decreasing even without a potential difference. Critical temperatures are low; for many years no material was found to be superconducting above 23 K. Since 1987, new materials have been found that are superconducting below 90 K, and work on higher temperature superconductors is continuing.

25-10 Electrical Conduction in the Nervous System
The human nervous system depends on the flow of electric charge. The basic elements of the nervous system are cells called neurons. Neurons have a main cell body, small attachments called dendrites, and a long tail called the axon.

25-10 Electrical Conduction in the Nervous System
Signals are received by the dendrites, propagated along the axon, and transmitted through a connection called a synapse. Figure A simplified sketch of a typical neuron.

25-10 Electrical Conduction in the Nervous System
This process depends on there being a dipole layer of charge on the cell membrane, and different concentrations of ions inside and outside the cell. Figure How a dipole layer of charge forms on a cell membrane.

25-10 Electrical Conduction in the Nervous System
This applies to most cells in the body. Neurons can respond to a stimulus and conduct an electrical signal. This signal is in the form of an action potential. Figure Action potential.

25-10 Electrical Conduction in the Nervous System
The action potential propagates along the axon membrane. Figure Propagation of an action potential along an axon membrane.

Review Questions

ConcepTest 25.2 Ohm’s Law You double the voltage across a certain conductor and you observe the current increases three times. What can you conclude? A) Ohm’s law is obeyed since the current still increases when V increases B) Ohm’s law is not obeyed C) this has nothing to do with Ohm’s law Answer: 2

Follow-up: Where could this situation occur?
ConcepTest 25.2 Ohm’s Law You double the voltage across a certain conductor and you observe the current increases three times. What can you conclude? A) Ohm’s law is obeyed since the current still increases when V increases B) Ohm’s law is not obeyed C) this has nothing to do with Ohm’s law Ohm’s law, V = IR, states that the relationship between voltage and current is linear. Thus, for a conductor that obeys Ohm’s law, the current must double when you double the voltage. Follow-up: Where could this situation occur?

ConcepTest 25.3a Wires I A) dA = 4dB
B) dA = 2dB C) dA = dB D) dA = 1/2dB E) dA = 1/4dB Two wires, A and B, are made of the same metal and have equal length, but the resistance of wire A is four times the resistance of wire B. How do their diameters compare? Answer: 4

ConcepTest 25.3a Wires I A) dA = 4dB
B) dA = 2dB C) dA = dB D) dA = 1/2dB E) dA = 1/4dB Two wires, A and B, are made of the same metal and have equal length, but the resistance of wire A is four times the resistance of wire B. How do their diameters compare? The resistance of wire A is greater because its area is less than wire B. Since area is related to radius (or diameter) squared, the diameter of A must be two times less than the diameter of B.

ConcepTest 25.3b Wires II A) it decreases by a factor of 4
B) it decreases by a factor of 2 C) it stays the same D) it increases by a factor of 2 E) it increases by a factor of 4 A wire of resistance R is stretched uniformly (keeping its volume constant) until it is twice its original length. What happens to the resistance? Answer: 4

ConcepTest 25.3b Wires II A) it decreases by a factor of 4 B) it decreases by a factor of 2 C) it stays the same D) it increases by a factor of 2 E) it increases by a factor of 4 A wire of resistance R is stretched uniformly (keeping its volume constant) until it is twice its original length. What happens to the resistance? Keeping the volume (= area x length) constant means that if the length is doubled, the area is halved. This increases the resistance by a factor of 4.

Ch. 25 Homework Finish reading Ch. 25. Begin looking over Ch. 26
Group problems #’s 6, 14, 18, 40 HW Problems #’s 5, 9, 13, 15, 19, 25, 35, 39, 51