unit ii - edl · web view2019-03-27 · physics for scientists and engineers, serway &...

Unit V – OHM'S LAW AND D.C. CIRCUITS

References:PHYSICS FOR SCIENTISTS AND ENGINEERS, Serway & Beichner, 5th ed., Ch. 27, 28

FUNDAMENTALS OF PHYSICS, Halliday, Resnick, & Walker, 6th ed., Ch. 27, 28

Unit ObjectivesWhen you have completed Unit V, you should be able to:

1. Define electric current in terms of the motion of charges.

2. Apply the microscopic model for charge conduction in a metal to problems where you are given several of the quantities: drift speed, current, charge, charge density, or cross-sectional area, and are asked to solve for another of these quantities.

3. Solve problems using the relationships among the following quantities: resistance, voltage, current, electric field, resistivity, and the physical dimensions of a conductor.

4. Apply the relation for electric power to problems dealing with circuit elements that obey Ohm's law.

5. Apply Ohm's law and Kirchhoff's rules to single or multi-loop direct current circuits in order to determine the potential difference between two points, the current in a branch of the circuit, the power dissipated in the circuit elements, and the potential difference across the terminals of the energy source.

6. Determine the equivalent resistance of two or more resistors connected in series or in parallel, or of a network of resisters which can be broken down into a series or parallel combination.

7. Describe the basic construction and properties of voltmeters and ammeters and given a galvanometer of specified internal resistance and full scale deflection, choose an appropriate series or shunt resister to be used to create a voltmeter or ammeter of a specified full scale deflection.

8. Discuss the charging and discharging of a capacitor through a resister and be able to:a. calculate the time constant of the circuit, b. sketch or identify graphs of stored charge or voltage for the capacitor, or of current or

voltage for the resister, and indicate on the graph the significance of the "RC" time constant

c. write down expressions to describe the time dependence of the stored charge or voltage for the capacitor, or the current or voltage for the resistor.

V-1


In the laboratory exercise entitled “K1 and K2 Introduction” (note: Warren edited this title) you were introduced to the following ideas as they apply to D.C. Circuits:

1) The concept and definition of electric current.2) Potential difference (or voltage) across a circuit element and what its meaning is in

terms of the energy of a charge moving in the circuit.3) The concept of power as it applies to electric circuits.4) The idea that the voltage increases equal the decreases around any closed loop in a

circuit.5) Ohm’s law and the concept of resistance.

This unit assumes that you have completed the entire write-up of “K1 and K2 Introduction”. If you have not completed it, you will do well to complete it before continuing.

CURRENTSuppose a long piece of wire is connected to a constant voltage source (like a battery, d.c. generator, or solar cell array, etc.) and suppose we have a lens that allows us to observe what is happening inside a small section of the wire. Since there is a ∆V between the ends of the wire there must

be an -field set up in the wire directed along its axis from the (+) terminal toward the (–) terminal. In conducting metals like copper, there are negatively charged electrons free to move around within the metal. Let the magnitude of the charge on an electron be denoted by the letter e. In our wire then each electron experience a force of

magnitude in a direction opposite to the direction of the -field. If n electrons, each having charge e move through plane C-C’ in time t, the current I is

where Qnet = ne. The current then, is the total or net charge passing a point per unit time. The expression above is true is the ratio Qnet /t (and thus the current I) is constant. If the amount of charge passing the point varies with time, that is, Qnet is a function of t, the current at any instant is given by

As you recall from your lab exercise the MKSA unit for electric current is the AMPERE (or amp for short, abbre.: A). When there is a current of 1 Ampere in our wire, this means that a net charge of 1 ___________ moves through the plane C-C’ during each second or that

unit______________ individual electrons pass through C-C’ in each second. how many? (V-2 #1)

There are situations (such as in a gas discharge tube or in a chemical solution containing (+) and (–) ions) where both (+) and (–) charges are moving past a given plane. Let’s illustrate this so we understand what the net means in the definition of current.

V-2


In our example above we mentioned that electrons were flowing left to right across our plane C-C’. Suppose we had 10 elections move across C-C’ in time t. If we isolated our little section of wire, this would mean that the left hand side of C-C’ would be 10 charges less negative (or 10 charges more positive and the right hand side would be 10 charges more (–) (or 10 chargesless positive. Do you see that the result would be exactly the same if we had 10 (+) charges moving through C-C’ to the left? Because there are many situations where we have (+) and (–) charges moving it is useful to use this idea to define CONVENTIONAL CURRENT. Conventional current is current where all the charges flowing are (+) and thus move from the (+) terminal of the voltage source to the (–) terminal.

This means that if there are (–) charges flowing to the right with a current of 2 A, we mentally replace it with an equivalent conventional current of 2 A to the left.

Sample Exercise V-1: When a sufficiently high potential difference is applied across a gas discharge tube, the gas in the tube is ionized and a current is created in the tube as the (+) ions move toward the (–) plate and the (–) electrons move toward the (+) plate. What is the magnitude and direction of the flow of conventional current in a hydrogen discharge tube in which 3.1 x 1018 electrons and 1.1 x 1018 protons move across the plane A-A’ in each second?

Solution: The magnitude of the charge on an electron is ____________ C

and on a proton is ____________ C. (V-3 #1)

The charge moving across A-A’ to the left in 1 second is

(____________)(____________) = ____________ C (V-3 #2)

Therefore, the current due to the electrons is

(V-3 #3)

The charge moving across A-A’ to the right in 1 second is

(____________)(____________) = ____________ C (V-3 #4)

Therefore, the current due to the protons is

(V-3 #5)

A current of 0.5 A of (–) charge moving to the left is equivalent of 0.5 A of (+) moving to the right. Thus the total conventional current in the tube is

ITOTAL = ____________ to the ____________ (V-3 #6)

If a hydrogen atom ionizes into 1 e- and 1 p+, why do more e- move across A-A’ per second then protons? (V-3 #7)

V-3


Back to our hunk of wire. If there is an -field exerting a force on

the electrons equal to qe why don’t the electrons accelerate along the wire causing more to cross C-C’ per unit time and hence cause the current to increase? The answer to this is the same as the reason that an object dropped in air doesn’t continually accelerate; because of collisions with the air molecules, the object reaches a

terminal constant speed. Similarly, the -field continually accelerates the charges, but due to collisions with the atoms of the metal wire, the charges end up with a constant drift speed, vd,

along the wire. Let’s write the current in terms of this drift speed. In our sketch above let

represent the distance an electron travels at speed vd in time t. Thus, = vdt. Now the number of electrons/volume is constant since just as many are entering through C-C’ as leave through D-D’. If n is the number of electrons/volume, then the number of electrons between C-

C’ and D-D’ is n A = nvdtA where A is the cross-sectional area of the wire. The total charge qT, then, in volume vdtA is equal to the number of charges in the volume times the charge on one electron, or

The current is the charge passing through D-D’ per unit time. In this case, charge q passes through D-D’ in time t, so the current is

Now let’s get an idea of roughly how fast the electrons travel in a length of wire. Is it at a snail’s pace, or close to the speed of light? Take a guess, then let’s do it.

Sample Exercise V-2: In a copper wire there is roughly 1 electron per copper atom that contributes to the current flow. The gram atomic mass of copper is 63.5 g/mole and its density is about 9 g/cm3. Find the average drift speed of electrons traveling through a copper wire having a cross-sectional area of 1 mm2 when carrying a current of 1A.

Solution: Solve the equation above for vd.vd = ____________ (V-4 #1)

where I = ___ A, qe = ____________ C and A = ___ mm2 = ______ m2 (V-4 #2)

If we can find n, then we can calculate vd.

From givens and previous courses: = _____ , = ____________ ,

for copper = ____________ , = ____________ ,

for copper = ____________ (V-4 #3)

V-4


Now multiply these together and make sure your final answer is in MKSA units.

n = _________________ e-/cm3 = _________________ e-/m3 (V-4 #4)

V-5


Calculate vd:

vd = _________________ m/s = ______________ mm/s (V-5 #1)

The snail would probably pass the electron. Your answer should be that the electrons, on the average, are moving less than the thickness of this line “|” per second! Or another way of putting it: in a wire of this size it will take an electron about 50 minutes to travel 1 foot! The drift speed of the electrons must not be confused with the speed that signals travel along a wire (which is due to changes in the -field in the wire) a speed which approaches the speed of light (light takes roughly a nanosecond to travel 1 foot). By analogy, when a water filled garden house is turned on a pressure wave travels rapidly along the house causing water to squirt out the other end almost instantaneously. However, the speed with which the individual water molecules travel the length of the hose is much lower.

At this point try problems 1 through 3 at the back of this unit.

RESISTIVITY & RESISTANCE

Consider a section of wire of length l and uniform cross-sectional area A. Since the -field is constant within the wire, it follows that

Now I = nqevdA and since (the number of e-/vol) and qe are constants for our hunk of wire, then

vd ∝ I/A. (Remember ∝ means “is proportional to”)Also, the drift speed depends on the strength of the -field; in fact

vd ∝ .Combining these relationships we can write:

where (the Greek letter rho, like in “row, row, row your boat”) is a constant of proportionality which depends on the intrinsic properties of the material our hunk of wire is made of. is

called the RESISTIVITY of the material. Its MKSA units will be those of or

. (Note: so we don’t get confused, in this section I’ll let A mean area and I’ll write Amp as the unit of current instead of the usual abbreviation A.)

You should recognize the equation ∆V = (l/A)I as OHM’S LAW from your lab exercise where the RESISTANCE, R, is

V-6


From this relation it is easy to see where the more common of units of resistivity come from:

Since an ohm [symbol: Ω, the Greek letter capital omega, aka “horseshoe”, or the logo on the back pockets of True Religion brand blue jeans (circa 2014)] was defined in your lab exercise as a volt/amp.

The -field within a material can be written in terms of and other quantities in the following way. Since

,

then with a little algebraic finageling,

Just as a little side comment, the RESISTIVITY () and RESISTANCE (R) of a given material can be thought of in the following way. For a given ∆V across a hunk of material, the greater the and R, the greater the samples resistance to flow of current through the sample. That is, the greater or R the smaller the current flowing through the sample. The following table lists the resistivity at room temperature of a sample of materials. Those with resistivities in the

range of 10-8 obviously would be classified as “conductors,” those in the range of >105 “non-conductors” and those in the broad mid-range are “semi-conductors.

RESISTIVITY AT ROOM TEMP.

MATERIAL Silver 1.47 x 10 -8

Copper 1.72 x 10-8

Aluminum 2.83 x 10-8

Tungsten 5.51 x 10-8

Iron 10 x 10 -8

Manganin (Cu 84%,Mn 12%,Ni 4%) 44 x 10-8

Constantan (Cu 60%, Ni 40 % 44.1 x 10-8

Nichrome 100 x 10-8

Graphite 8 x 10 -6

Carbon 3.5 x 10 -5

Germanium 0.43Silicon 2.6 x 103

Rock (Granite) 105 - 107

Wood (Maple) 4 x 1011

Mica 9 x 1013

Quartz (Fused) 5 x 1016

V-7


Sample Exercise V-3: A rectangular carbon block has dimensions 1 x 1 x 50 cm.

a. Find its R between its two square ends.

Solution: Write R in terms of , l and A and substitute the known quantities (watch the units!)

(V-7 #1)

b. Find its R between two opposing rectangular faces.

Solution: Again R is given by

(V-7 #2)

Sample Exercise V-4: One end of an iron wire and one end of a tungsten wire are butted together and a ∆V of 60 V is applied across their free ends. The iron wire is 2 m in length and has a diameter of 1 mm. The tungsten is 3 m long and has a resistance of 1 Ω. Find

a. the cross-sectional area of the tungsten wire.

Solution: Write the expression relating R, A, , and l, solved for A, then substitute in the numbers and calculate the area.

(V-7 #3)

b. the resistance of the iron wire.

Solution: Use the same expression in (a) but solve it for R.

(V-7 #4)

c. the current in each wire.

Solution: The charge flowing through the iron wire per unit time better be the same as the charges flowing per unit time through the tungsten wire. In other words, the current in both segments are the same since the wires are connected in series. Using the “sum-of-the-voltages around a closed loop” idea from the lab exercise and Ohm’s law:

Solving for I and substituting:

(V-7 #5)

V-8


d. ∆V across each section of wire.

Solution: Since R is known for both sections of wire and the current is known, use Ohm’s law and calculate ∆V across each section.

(V-7 #6)

(V-7 #7)

e. the -field within each wire.

Solution: Write the expression relating the magnitudes of , , I, and A:

see p. V-6 (V-8 #1)

Thus the -field in the iron is:

(V-8 #2)

and in tungsten:

(V-8 #3)


ENERGY & POWERConsider the circuit at the right consisting of an energy source, in this case a battery, and a box with “some stuff” in it (maybe a motor, resistor, another battery, etc. or a combination of these) that is an energy consumer or an “energy sink.” The conventional current enters the box at A and exits at B. No matter what the circuit looks like inside the box, we know that since it is a “sink” that VA > VB. In other words a (+) charge at B has less potential energy than it had at A by an amount q∆VAB. Also since our circuit is isolated, we know that its loss in potential energy in the box must equal its gain in potential energy (q∆V) in the battery, thus ∆V must equal ∆VAB. We can write then, that the loss in potential energy by a little hunk of charge dq as it moves through the box is

POWER is the rate at which energy is gained, lost, transferred, or whatever. Thus

V-9


Similarly, the energy supplied per unit time by the battery is P = I∆V. As you recall, this is the relation for power developed in your lab exercise.Generally then, the rate at which energy is supplied to of lost by charges moving through a circuit element (a battery, resistor, motor, diode, etc.) is the product of the current through the element and the potential difference across the element, or

If and only if a circuit element obeys Ohm’s law ∆V = RI, the following expressions for power are also valid:

The MKSA unit for power is the J/s or watt. These units are valid for the electrical case also:

Sample Exercise V-5: A 100 W light bulb is designed to operate at 115 V. Assuming it obeys Ohm’s law, find

a. the current in the bulb under operating conditions

Solution: Write the expression relating power, current, and voltage, plug in the numbers and calculate the answer:

(V-9 #1)

b. the resistance of the filament

Solution: To avoid using the answer to (a), write the expression involving power, ∆V, and R, plug in the numbers and calculate the answer:

(V-9 #2)

c. the energy radiated in 1 hour

Since the power output of the bulb is 100 W this means it radiates ______ joules (V-9 #3)

energy in the form of heat and light in each second. Thus, the energy radiated in 1 hour is:

(V-9 #4)

d. the cost at 5 cents per if the bulb is left on overnight (about 10 hrs).

The is an energy unit used inthe electric power industry. First let’s find the conversion factor between and

joules:

V-10


(V-9 #5)

V-11


d. (cont.) In (c) we found that we used 0.36 MJ or ________ when we let the bulb burn for 1 hr. Therefore, if we leave it on for 10 hrs we use ________ and at 5 cents per , the cost will be ________. (V-10 #1)


In the following sections of this unit we will apply the concepts developed in the lab exercise and the first part of this unit specifically to dealing with direct-current (D.C.) circuits. For simplicity, when the word “current” is used we will always be referring to conventional current, that is, the flow of (+) charge. Thus, since circuit elements like resistors, motors, batteries being “charged”, etc. are energy consuming elements, the (+) charge always moves through them moving from higher to lower electric potential. Since a discharging battery, solar cell, D.C. generator, etc. are all energy supplying devices (+) charge moves through them going from lower to higher electric potential since the charges gain energy by passing though them.

EMF & INTERNAL RESISTANCEConsider a (+) charge at terminal B of the battery. This charge is repelled by other (+) charges at B and hence has some electric potential energy at B. Due to this repulsive force it travels down through the external circuit (aka the “load”) therein losing electric potential energy and eventually arrives at terminal A, the (-) terminal of the battery. This upsets the potential difference across the battery. The battery is designed to use chemical potential energy to maintain a constant ∆V across its terminals so some of this chemical energy is transferred to our (+) charge and it is pulled away from the negative charges at A and forced through the battery against the repulsive charge at B back to point B. In being forced from A to B, its electric potential energy increases. The energy supplied per unit charge in this way is termed the “electromotive force” or “emf” of the battery (or generator, or solar cell, etc.). The MKSA unit for energy per unit charge is the joule per coulomb or volt. Admittedly emf is not a force and is nothing more than the difference in potential (or voltage) across the terminals of the battery. So why the jazzy name? Why not just call it the voltage across the battery? The only answer that can be given is: TRADITION! The funny name “emf” and symbol E (read “script eee”) won’t go away so we have to define it and use it.

Suppose we measure the voltage across a battery when there is no external circuit attached to it. Now measure it with an external circuit connected. When this is done it will be found that with the external load the voltage across the battery will be slightly less than when there is no load. The emf (E) is defined as the “open circuit” or “no load” voltage across the battery. This decrease in terminal voltage when a load is connected stems from the fact that every electrical energy source (aka battery, generator, etc.) has some (although usually small) resistance within the source itself. This resistance is termed the internal resistance of the source and is given the symbol r. Therefore, when a load is attached to a battery and a current flows through the battery, the voltage across the terminals will be

V = E - Ir

V-12


where Ir is the voltage across the internal resistance r. As you would expect, when I = 0, ∆V = E.In circuit diagrams the internal resistance is designated as follows:

where E = 5 V and the internal resistance of the battery r = 0.5 Ω.

Now that we are through with the preliminaries let’s look at some simple circuits.

D.C. CIRCUIT ANALYSIS.In the lab exercise on D.C. Circuits there were several general rules that, hopefully, you were able to discover which will now be formalized and given fancy names.

I. Kirchhoff’s First Rule [let’s call it K-1] -

K-1: The sum of the currents into a junction equals the sum of the currents out of the junction.

K-1 is nothing more than a statement of the principle of conservation of charge. That is, that charge can neither be created or destroyed. For example, suppose there are five wires all connected at a point forming a “junction”. If the sum I1 + I2 + I3 is less than the sum I4 + I5

(that is, if the charge/time coming in is less than the charge/time going out), then somehow charge is being created at the junction, which is hogwash. If I1 + I2 + I3 is greater than I4 + I5 then somehow charge is being destroyed at the junction, which is also baloney. If charge is to be conserved, I1 + I2 + I3 must equal I4 + I5.

II. Kirchhoff’s Second Rule [let’s call it K-2] -

K-2: The sum of the potential increases equals the sum of potential decreases around any closed loop in a circuit.

As discussed in the lab exercise, this rule is nothing more than a statement of the principle of conservation of energy. For example, suppose a charge Q starts at point A and we follow it clockwise around the loop and back to A. As it moves through the battery it has its electric potential energy increased by Q∆VAB = Q(E – Ir), its potential energy decreases by Q∆VCD

upon passing through R1, and its potential energy decreases again by Q∆VEF upon passing through R2. Since the potential at A is still the same, the potential energy of Q is the same as it arrives at A as when it left. Thus, the potential energy it gained must be equal to the amount lost going around the loop. That is

or

V-13


this is just K-2!Now let’s outline a consistent way of working circuit problems using K-1 and K-2 to minimize sign and other goof-ups.

Sample Exercise V-6 – Find the currents in R1, R2, and R3

given that R1 = 8 Ω, R2 = 4 Ω, and R3 = 7 Ω.

Step 1* - Label the polarity of all sources of emf. That is all circuit elements that can supply energy to the charges flowing in the circuit. (Notice this has been done on the two batteries.)

Step 2 – Indicate a current in each branch of the circuit. If you can guess the direction the current should go ahead of time, choose it. If not, just assume some direction. If the direction you assume is wrong, the numerical value for that current will come out negative, indicating the actual current is in the opposite direction. (Notice I’ve done this on the drawing.)

Step 3 – According to the directions chosen for the currents, label each current element that reduces the energy of the charges, with a (+) on the side having the higher potential, and a (-) on the side having the lower potential. Conventional current flows from higher to lower potential, thus, the side the current enters is (+). The energy consumers are resistors in this problem, but they could be motors, light bulbs, etc. . (Notice I’ve done this on the drawing.)

Step 4 – Decide a direction for traveling around each loop to add up the voltage. Notice in the drawing I have chosen counterclockwise in both loop 1 & 2. The choice is arbitrary. I could have chosen them both the other way or one each way. I could also have labeled a third loop going from A through R1 to B, through R3 and back to A, but it will turn out that we won’t need this loop. The main thing in choosing each loop is that every branch in the circuit is included in at least one loop.

Step 5 – Starting anywhere, go around each loop in the direction chosen, and apply K-2. Start at point A in our diagram and apply K-2 to loop 1 & 2.

K-2: Loop 1: ___________________ = ___________________ (Eqn. I)

Loop 2: ___________________ = ___________________ (Eqn. II) (V-12 #1)

Step 6 - Equate the sum of the currents entering a junction with the sum of the currents exiting. (In other words, apply K-1.) Apply K-1 to the junction at point A and point B.

K-1: Point A: ___________________ = ___________________ (Eqn. III)

Point B: ___________________ = ___________________ (V-12 #2)

V-14


With these 6 steps we’re finished with the physics of the problem. The most common mistakes are made in step 5 due to sign goofs; that is why it is so important to complete the labeling outlined in steps 1 through 4.

All that is left to do now involves an orgy of algebra. However, students of your caliber can solve simultaneous equations in your head - right? Let’s do it and find I1, I2, and I3.Substitute the known values into equations I, II, and III:

I. ________________ = ________________

II. ________________ = ________________

III. ________________ = ________________ (V-13 #1)

So with these three equations involving only three unknowns, solve them for I1, I2, and I3:

I1 = ________________ , I2 = ________________ , I3 = ________________ (V-13 #2)

Notice that I1 came out (–). This means that we guessed wrong as to the direction of I1, but no biggie, now we know its correct direction. You see, if you set it up right, the system is self-correcting!

Another thing – Suppose I1 did pass through battery #1 going (+) to (–). That just means the charges making up the current are losing potential energy since they are going from higher to lower potential. Energy is being transferred to the battery and is stored as chemical potential energy in the battery. This is what “charging” a battery means. It means merely that current is being forced to flow backwards through the battery.

Let’s use the values of I1, I2, I3 and the given quantities to calculate some other stuff.

Find:a. the potential difference across R1.

V1 = ( _____ )( _____ ) = ( __________ )( __________ ) = __________ (V-13 #3) Equation Substitution Answer

b. the rate at which battery #1 is converting chemical potential energy into electrical energy and heat.

P = ( _____ )( _____ ) = ( __________ )( __________ ) = __________ (V-13 #4) Equation Substitution Answer

c. the potential difference between the terminals of battery #2.

VBatt. #2 = ( _____ ) – ( _____ ) = ( ______ ) – ( ______ )( ______ ) = ________ (V-13 #5)

Equation Substitution Answer

V-15


d. the energy radiated as heat in R2 and R3 in 10 minutes.

E2 = ( _____ )( _____ ) = ( _____ )( _____ )( _____ ) = ( _____ )( _____ )( _____ ) Equation Equation Substitution

= __________ (V-14 #1) Answer

E3 = ( _____ )( _____ ) = ( _____ )( _____ )( _____ ) = ( _____ )( _____ )( _____ ) Equation Equation Substitution

= __________ (V-14 #2) Answer

For further examples see Serway & Beichner pages 879-882.

At this point try problems 12 & 13 at the back of this unit.

V-16


EQUIVALENT RESISTANCESResistors in Series - Suppose you have three resistors connected in series with one another and a battery as shown. Using K-2 and Ohm’s law:

E = V1 + V2 + V3 = IR1 + IR2 + IR3

orE = I(R1 + R2 + R3)

Note that the same current would flow if the three resistors were replaced by an equivalent resistance, Req where

Req = R1 + R2 + R3

The same argument can be made for any number of resistors connected in series. Req for the n resistors in the sketch at the right is given by:

Resistors in Parallel - Consider the three resistors connected in parallel with a battery as shown. From applying K-2 (and from your lab experiment) it follows that ∆V1 = ∆V2 = ∆V3 = E. From K-1 and Ohm’s Law:

or

I = ENote again that the same current will flow from the battery if the resistors are replaced by an equivalent resistance Req given by

If we have n resistors all connected in parallel the Req is given by


V-17


ELECTRICAL INSTRUMENTSWhen a meter (either a voltmeter or ammeter) is introduced into a circuit to make a measurement, it can’t help but change, ever so slightly, the quantity being measured. This is because the meter has some resistance and thus introducing it effects the voltages and currents in the circuit. It should be evident then, that the better the meter, the less it disturbs the circuit.

The Galvanometer (symbol: )

The galvanometer is a device that is employed as part of both analog voltmeters and ammeters. It consists of basically a coil of wire that is immersed in the magnetic field of a permanent magnet. When a current is passed through the coil a magnetic force causes the coil to rotate through an angle proportional to the current in the coil. Attached to the axle of the coil is a pointer or a needle whose deflection angle is also proportional to the current. So if one of these things is constructed and calibrated we have a device that measures current.

Since the coil is just a long hunk of wire wound up, the must have some resistance

(remember R where is the length of the wire). In this section we will use values for a typical galvanometer: 1 µA will cause full scale deflection of the pointer and the galvanometer has an internal resistance RG = 1 kΩ. This means that when 1 µA is flowing thorough thegalvanometer the ∆V across it is

∆V = IR = (10-6 A)(103 Ω) = 10-3 V = 1 mV.

When discussing voltmeters and ammeters we will let this represent the galvanometer, where R6 is its internal resistance.

The Ammeter (symbol: )

As stated above, a galvanometer is a current measuring device, in other words an ammeter.

So basically, = . But a given galvanometer only measures currents within a limited range; for example the one we will use for this discussion only measure currents between 0 and 1 µA. Suppose we need to measure currents as high as 1 mA. To use our galvanometer we must fix our circuit so that most of the current is detoured or “shunted” around the galvanometer so that no more than 1 µA flows though it.

This is done by connecting a “shunt” resistance, RS, in parallel with the galvanometer like so as shown at the right. The proper choice of RS will insure that the current through the galvanometer is less than (or = to) 1 µA.

Consider the following ammeter. Since the galvanometer and the shunt resistor are in parallel, then ∆VG = ∆VS and from K-1, I = IS + IG and therefore IS = I – IG. Applying Ohm’s Law:

∆VG = IGRG and ∆VS = Is Rs = (I – IG)Rs.

Using these and solving for RS for a desired current range is given by:

V-18


For example, suppose we wish to measure currents up to a maximum of 1 mA (that is, I < 1 mA). Full scale deflection on our galvanometer is 1 µA (that is, IG < 1 µA) and its RG is 1 kΩ. To do this we must connect an RS in parallel with the galvanometer where

Hopefully, this is as you would expect. RS should be much less than RG if you want most of the current to pass around the galvanometer.

Now we have a 0 - 1 mA ammeter where: =

and full scale deflection on the meter now means 1 mA. To convert the meter to a 0 – 1 A, 0 – 5 A or whatever range you wish, just calculate and insert the proper RS in parallel to the galvanometer.

The Voltmeter – (symbol: )

As mentioned before, when out sample galvanometer reads 1 µA a difference in potential of 1mV is across its terminals. When it reads 0.5 µA, 0.5 mV is across its terminals, etc. This is true because ∆V I. Therefore, if we just change the units on the face of the meter our galvanometer is also a 0 – 1 mV voltmeter, but all by itself it is limited to that range. As you recall from your lab exercise, a voltmeter is connected across a circuit element. If the ∆V across R at the right is less than or equal to 1 mV then we just connect our galvanometer as shown in parallel to R (i.e.,across R). If, as is usually the case, ∆V > 1 mV, we must add a resistor in series, RV, with the galvanometer so that no more than 1 mV is across its terminals. The value of the resistance RV is given by:

solving for RV:

Using our sample galvanometer, suppose we want full scale deflection to mean 1 V. This will be when 1 mV is across the galvanometer and the current through it is IG = 1 µA. If the maximum value of ∆V then is to be 1 V, then RV must be

Now we have a 0 - 1 V voltmeter where: and full scale deflection on

the meter means 1 V. To convert the meter to 0 – 5 V, 1 – 12 V, or whatever range you desire, just calculate and insert the proper value of RV in series with the galvanometer.

V-19


Sample Exercise V-7 - A certain meter movement has an internal resistance, RG, of 100 Ω and requires a current of 200 µA for full scale deflection.

a. Make a sketch showing how you would use this meter to make a voltmeter with a 0 – 10 V range, giving the values of any resistances used.

b. Make another sketch showing how you would use the meter to make a 0 – 5 A ammeter, giving the values of any resistances used.

Solution:a. To use the meter as a voltmeter, a resistor must be connected in ___________ with

(series or parallel?)

the galvanometer so that it looks like this (complete drawing) (V-18 #1)

The value of the resistor that was added in the sketch above is given by:

RV = _____________ = _____________ = _____________ (V-18 #2)

(Equation) (Substitution) (Answer)

b. To use the meter as an ammeter, a resistor must be connected in ___________ with (series or parallel?)

the galvanometer so that it looks like this (complete drawing) (V-18 #3)

The value of the resistor that was added in the sketch is given by:

RS = _____________ = _____________ = _____________ (V-18 #4)

(Equation) (Substitution) (Answer)

At this point try problems 18 and 19 at the back of this unit.

V-20


RC CIRCUITSIn the previous sections we dealt with circuit elements that were resistors, light bulbs, batteries, etc., where the currents did not vary with time. In this section we will introduce the capacitor as a circuit element along with the resistor, which will lead to the concept of time varying currents.

Let’s consider the circuit at the right qualitatively.

Case 1 – Attach a wire between A and B, then close switch S. Charge flows into the plates “super quick,” so that, if we assume resistanceless connecting wires, the difference in potential across the capacitor, ∆VC, “instantaneously” becomes equal to the emf, E, across the battery. The battery has given up some energy and it has been stored in the capacitor

where Ustored = C(∆VC)2. The charge Q stored on the capacitor is Q = C∆VC = CE.

Case 2 – Insert a resistor between A and B, then close switch S. Using K-2, at any time after the switch is closed

E = ∆VR + ∆VC

The instant the switch is closed ∆VC = 0 since there isn’t any charge on C. Thus the initial current will be the same as if C wasn’t even in the circuit! That is,

I = ∆VR/R = E/R at t ~ 0.

As time goes on, ∆VC increases (‘cause charge is flowing onto C) and ∆VR decreases (‘cause ∆VR + ∆VC = a constant = E). This means that I = ∆VR/R must be decreasing. That’s logical ‘cause the charge flowing onto C is being repelled by those already there. As I approaches zero ∆VR becomes zero and ∆VC E. The circuit now acts as it R were not even in the circuit and the situation is static.

Case 3 – Insert an even larger resistor between A and B and close the switch. This time the initial current is less than in case 2 since

I = ∆VR/R = E/R at t ~ 0

and R(case 3) > R(case 2). Since the rate that charge is moving toward C is less, it will take longer to charge C to where ∆VC ≈ E.

The graphs below illustrate qualitatively how ∆VC and ∆VR change with time.

V-21


Try your hand at qualitatively sketching the graphs of the charge Q being stored on the capacitor as a function of time (hint: ∆VC = Q/C) and the current as a function of time (hint: ∆VR = IR).

(V-20 #1)

Now let’s quickly look at discharging a capacitor through a resistor. Suppose that capacitor C in the diagram has been charged to a difference in potential (∆VC)0 then connected into the circuit as shown. When switch S is closed, the charge on C will flow off the capacitor discharging it. Conventional current in the circuit will flow ____________ around the loop.

(clockwise/counter-CW) (V-20 #2)

Applying K-2 to the loop, at any time after the switch is closed

VC = VR

Since ∆VC = (∆VC)0 initially, the initial current will be

(V-20 #3)

Since ∆VC = Q/C as the charge on the capacitor decreases, ∆VC ____________ and therefore (decreases/increases) (V-20 #4)

∆VR ______________This being true and since I = ∆VR/R, then I also _____________ with (decreases/increases) (V-20 #5) (decreases/increases) (V-20 #6)

time until after a long time, I = _______, and ∆VR becomes _______. (V-20 #7)

On the graphs below, sketch roughly how the quantities vary with time for a discharging capacitor.

(V-20 #8)

How would each change if R was made larger or smaller?

V-22


Now that we’ve got a feeling for what happens, let’s derive the equations for charging and discharging the capacitor through a resistor.

Consider the setup at the right. When switch S is switched to A the capacitor C is charged through the resistor R. After C is charged, if we switch S to point B, the capacitor will discharge through R.

Charging C – At t = 0 the switch S is switched to A and the charge on C is initially zero. Generally at any given t, if we apply K-2 we get

E = VR + VC = IR + Q/Cor

E =

Rearranging:

(CE – Q)/RC

Let’s solve this for Q as a function of time:

Integrating this should be old stuff,

use substitution: u = CE – Q du = - dQ

An orgy of algebra:

Solving the last expression for the charge Q as a function of time:

Recall from

Algrbra II:

If

Since: VR = IR, I = dQ/dt, and VC = Q/C

V-23


The current flowing in the circuit as a function of time as C is charging is found from

resulting in the relation:

∆VR as a function of time as C is charging comes from Ohm’s law:

∆VC as a function of time as C is charging comes from the relation

Notice these equations for Q(t), I(t), ∆VR(t), and ∆VC(t) give all the initial and final values which we arrived at in our qualitative description of the charging of a capacitor:

Initially (that is, when t ~ 0) since e-t/RC = e0 = 1,

I(t = 0) = Q(t = 0) = CE(1 - 1) = 0∆VR(t = 0) = E(1) = E ∆VC(t = 0) = E(1 - 1) = 0

After a long time (that is, when t ~ ∞) since e-∞/RC = 0,

I(t ~ ∞) = Q(t ~ ∞) = CE(1 - 0) = CE

∆VR(t ~ ∞) = E(0) =0 ∆VC(t ~ ∞) = E(1 - 0) = EAll these values should correspond to the limiting values of these quantities on the graphs at the bottom of page V-20. Check ‘em out!

V-24


Discharging C – Now suppose after we charge the capacitor until the potential difference across it is E and the charge it has stored is Q = CE, we switch S from A to B in the diagram. The capacitor will now begin discharging through the resistor. If we apply K-2 to the loop the instant S is switched to B, we get

∆VR = ∆VC

or

IR = Q/C

Since the charge on the capacitor is decreasing with time, then I = -dQ/dt (negative since Q is decreasing) and

Putting everything involving Q on the left-hand side of the = sign and everything else on the right

____________ = ____________ (V-23 #1)

Setting up the integral with the limits of integration being from t = 0 to t and Q = Q0 (at t = 0) to Q (at t):

.

Integrating we get:

______________ = _____________(V-23 #2)

To solve the last expression for Q as a function of time, remember from the box at the bottom of page V-22 that if ln(x) = y then x = ey. The result is:

where at t = 0, Q0 = CE.

Since I = dQ/dt, then the magnitude of the current as a function of time is given by:

I =

Since ∆VR = IR, then ∆VR as a function of time is

∆VR =

Since ∆VC = Q/C, then ∆VC as a function of time is

∆VC =(V-23 #3)

V-25


In the same way that we did for the charging circuit, find the following:

I (at t = 0) = ____________ Q (at t = 0) = ____________

∆VR (at t = 0) = ____________ ∆VC (at t = 0) = ____________ (V-24 #1)

After a long time (t ~ ∞)

I (at t = 0) = ____________ Q (at t = 0) = ____________

∆VR (at t = 0) = ____________ ∆VC (at t = 0) = ____________ (V-24 #2)

Again these should correspond to the limiting values of these quantities on your graphs at the bottom of page V-20.

The RC Time Constant (Greek letter tau) A useful parameter for describing RC circuits is the time required for the exponential term e -t/RC to become e-1. This occurs when t = RC. This time is defined as “” the RC time constant, where = RC. It is a measure of the speed with which the current and voltage change in an RC circuit. For example, in discharging a capacitor, in one time constant (when t = = RC) the charge becomes

Q = Q0(e-1) = Q0/e 0.37Q0

In other words, in one , Q has decreased to ~37% of it’s original value.Similarly, in charging a capacitor, in 1 the charge becomes

Q = Q0(1 – e-1) 0.63Q0

or 63% of its maximum value.

Label the values of ∆VR, ∆VC, Q, and I at one on your charging and discharging graphs on pages V-19 and V-20. (V-24 #3)

Sample Exercise V-8 – How many must elapse before a capacitor in an RC circuit is charged within 1% of its maximum charge?

Solution: Q = CE(1 – e-t/RC) = Qf(1 – e-t/RC)

thus, Q/Qf = 0.99 = (1 – e-n/RC) e-nRC = 0.01

Taking natural logs (i.e., ln) of both sides of the last expression:

- n/RC = - 4.6 n = 4.6RC

Hence it takes 4.6 to charge the capacitor to within 1% of its maximum charge.


– End Unit V –In-Text Fill-In Answers

Page Fill-In # Fill-In Answer

V-2 1 Coulomb, 6.25 x 1018

V-26


V-3 1 e- 1.6 x 10-19 C, p+ 1.6 x 10-19 C

2 (1.6 x 10-19 C)(3.1 x 1018 electrons) = 0.5 C

3

4 (1.6 x 10-19 C)(1.1 x 1018 protons) = 0.18 C

5

6 I- + I+ = 0.5 A + 0.18 A = 0.68 A to the right

7 The force on both the e- and the p+ is the same since F = qE and the magnitude of their charge q is the same and they are in the same electric field E. Since the e- is less massive it will have a greater acceleration thus more e- will move across A-A’ per second.

V-4 1

2 I = 1 A, qe = 1.6 x 10-19 C and A = 1 mm2 = 10-6 m2

3 = 1, = 6 x 1023 (Avogadro’s #),

for copper = 63.5 g/mole, =1/63.5 = 0.016 mole/g,

for copper = 9 g/cm3

4

V-5 1 vd = 7.3 x 10-5 m/s = 0.073 mm/s

V-7 1

2

3

In-Text Fill-In Answers

V-27



V-7 4

5

6

7

V-8 1

2

3

V-91

2

3 100 J

4

5

V-10 1 0.36 MJ = 0.1 , 10 hrs we use 1 , cost: 5¢

V-12 1 Loop 1: E1+ I1r1 + I1R1 = I2R2

Loop 2: I2R2 + I3R3 + I3r2 = E2

V-28


2 Point A: I1 + I2 = I3

Point B: I3 = I1 + I2

V-29




V-13 1 I. 6V + I1(1) + I1(8) = I2(4) 6 = 4I2 – 9I1

II. I2(4) + I3(7) + I3(1) = 10V 5 = 2I2 + 4I3

III. I1 + I2 = I3

2 I1 = –0.23 A, I2 = 0.99 A, I3 = 0.76 A

3 a. V1 = (I1)(R1) = (0.23 A)(8) = 1.8 V

4 b. P = (I1)(E1) = (0.23 A)(6 V) = 1.4 W

5 c. VBatt. #2 = (E2) – (I3r2) = (10 V) – (0.76 A)(1) = 9.3 V

V-14 1 d. Conversion: 10 min x 60 s/min = 600 s E2 = (P)(t) = (I2)2(R2)(t) = (0.99 A)2( 4 )( 600 s ) = 2352 J

~ 2.4 kJ

2 d. E3 = P)(t) = (I3)2(R3)(t) = (0.76 A)2(7)( 600 s ) = 2426 J ~ 2.4 kJ

V-18 1 Series,

2

3 parallel,

4

V-30




V-20 1

2 I flows CCW

3

4 VC decreases

5 VR decreases

6 I decreases

7 After a long time (i.e., t = ) I = 0 and VR = 0

8

V-23 1

2

3 , ,

V-31




V-24 1 I = E/R, Q = Q0 = CE, VR = VC = E2 I = Q = ∆VR = ∆VC = 0

3

V-32


End of Unit Problems

1. A circular dielectric disk has a uniform surface charge density 1 µC/m2. If the disk has a radius of 10 cm, at what angular speed must the disk be rotated to produce a current of 50 nA across a straight line extending from the center of the disk to the rim? [Ans: 10 rad/s]

2. The total charge that has passed any given cross-section of wire varies with time according to the relation

were Qo equals 30 µC and B equals 1 second.

a. Find the current in the wire is a function of time. [Ans: ]

b. What is the ratio of the current at 1 second to the current at 0 seconds? [Ans: ]c. Sketch a graph of the current versus time.

3. The current in a wire varies according to the relation

where t is in seconds and the constant 120 has units of rad/s.

a. Sketch a graph of the current as a function of time out to t = 1/60 s.

How many net coulombs of charge pass a cross-section of the wire inb. 1/4th of a cycle? [Ans: 1/(6) C]c. 1/2 of a cycle? [Ans: 1/(3) C]d. 1 full cycle? [Ans: 0]

Note that all you were doing in calculating b, c, and d is finding the area between the graph and the t axis from t = 0 to t.

4. A 10 m length of #18 copper wire is carrying a current of 5 A. If the diameter of the wire were doubled, with the same current, how would the drift speed change? Support your conclusion with a short physical argument. [Ans: 1/4 of the original value]

5. A potential difference, V, is applied to a copper wire diameter d and length L. What will be the effect on the drift speed of the electrons in the wire of (a) doubling V, (b) doubling L, and (c) doubling d? [Ans: double, 1/2, 1/4]

6. A wire with a resistance of 6 is drawn out through a die so that its new length is three times its original length. Find the resistance of the longer wire, assuming that the resistivity and density of the material are not changed during the drawing process. [Ans: 54]

V-33



7. A wire 4 m long and 0.5 mm in diameter has a resistance of 1.1 . If a difference of potential of 20 V is applied between its ends find:

a. the current in the wire. [Ans: 18.2 A]b. the resistivity of the material. [Ans: 5.4 x 10-8 ]c. the electric field in the wire. [Ans: 5 V/m]

8. A 50 resistor is rated at 0.5 W. (Meaning the maximum power it can dissipate without melting is 0.5 W).

a. What is the maximum voltage that can be applied across the resistor? [Ans: 5 V]b. What is the maximum current allowable through the resistor? [Ans: 0.1 A]

9. A coil of wire dissipates energy at the rate of 5 W when a difference in potential of 200 V is applied to it. A second coil made of the same wire dissipates 15 W when the same potential difference is across it. What is the ratio of the length of the wire in the second coil to that in the first? [Ans: 1:3]

10. Find the cost of electricity heating 40 gallons (about 150 liters) of water from 20°C to 90°C see if the power company charges 5¢ per . (recall that about 4200 J will raise 1 L of water 1°C) [Ans: 61¢]

11. A good (but expensive) car battery is rated at 480 (meaning the total charge it can store is 480 ). Suppose while the car is parked, the two 80 W headlights are left on accidentally. Assuming the terminal voltage remains constant at 12 V, determine the number of hours that elapse before half the charge stored in the battery is used up. [Ans: 18 hrs]

12. In the drawing at the right assume the batteries have no internal resistance.

a. Find the current through the three branches of the circuit. [Ans: left 0.67 A ccw, center 0.33 A up, right 0.33 A ccw]

b. Find the difference in potential between points a and b. [Ans: 3.3 V]

13. In the circuit at the right, find:

a. the potential difference between points a and b. [Ans: 6.4 V]

b. the potential difference between points c and b. [Ans: 8 V]

c. the potential difference between points a and c. [Ans: 1.6 V]

d. If points a and c are connected, find the potential difference between points a and b. [Ans: 6.9 V]

V-34

Unit V – OHM'S LAW AND D.C. CIRCUITSV-35



14. Find the equivalent resistance for each of the networks below.

[Ans: (a) Req = 25/4 , (b) Req = 21, (c) Req = 14/3 ]d. If battery having an emf of 12 V is connected between points A and B in diagram (b)

above, what will I1, I2, and I3 be if the internal resistance of the battery is 1 ? [Ans: I1 = 0.54 A, I2 = 0.41 A, I3 = 0.14 A]

e. Which power is dissipated in the 10 resistor? . . . in the 24 resistor? [Ans: 2.9 W, 0.47 W]

15. Consider the circuit at the right.

a. What is the equivalent resistance for the resistance network? [Ans: 48/5

b. What is the current through the battery? [Ans: 2 Ac. If points A and B are connected with a wire having zero

resistance, what will the current through this wire be? [Ans: 5 A

16. Given the circuit at the right.

a. What s the potential difference across the resistor? [Ans: 30 V]b. What is the voltage across the motor? [Ans: 70 V]c. How much energy is given up by the charges in the motor during

each second? Ans: 210 J]d. If there is no heat generated in the motor as the charges passed

through it, how far can it lift the 10 kg mass in three seconds? [Ans: 6.3 m]

17. Given the circuit at the right, find:

a. the potential difference across the capacitor. [Ans: 10 V]b. the charge on the capacitor. [Ans: 20 C]c. the power dissipated in the 5 resistor. [Ans: 20 W]

18. A diagram of the internal wiring of a multi-range ammeter is shown at the right. When each of the three terminals is used, the full-scale reading of the meter is 10 A, 1 A, or 0.1 A. The galvanometer gives a full-scale deflection when a current of 10 mA flows through it and its internal resistance is 25 . Find the values of the resistors Ra, Rb, and Rc. [Ans: Ra = 0.0278 , Rb = 0.25 , Rc = 2.5 ]

V-36



19. Diagram of the internal wiring of a multi-range voltmeter is shown at the right. When each of the terminals is used the full-scale reading of the voltmeter is 3 V, 15 V, and 150 V. The galvanometer gives a full-scale deflection when a current of 1 mA flows through it and it’s internal resistance is 15 . Find the values of the resistors Ra, Rb, and Rc and the overall resistance of the meter on each of its ranges. [Ans: Ra = 2985 , Rb = 12 k, Rc = 135 k, overall: 3 k, 15 k, 150 k]

20. In the circuit at the right switch S is switched to position a for two hours then switched to position b.

a. 12 seconds after being switched to position b, what is the charge on the capacitor? [Ans: 22 C]

b. Two hours later S is moved back to position a. What is the charge on the capacitor 12 seconds after S is switched to a? [Ans: 38 C]

c. What is the current 2.5 after S is moved to position a? [Ans: 0.4 A]

21. Given the circuit at the right, what is the current in the battery 5 seconds after switch S is closed? (Hint: think Req

and Ceq) [Ans: 29 A]

22. A 3 M resistor and a 1 F capacitor are connected in a single loop circuit to a battery having an E of 4V. At 1 second after the connection is made, what are the rates at whicha. the charge of the capacitor is increasing? [Ans: 1 C/s]b. energy is being stored in the capacitor? [Ans: 1.1 W]c. thermal energy is appearing in the resistor? [Ans: 2.8 W]d. energy is being supplied to the circuit by the battery? [Ans: 3.8 W]

23. Consider the circuit at the right and suppose that at t = 0 the switch S is closed. Without writing down any of the circuit equations calculate the following:a. The initial current through the battery. [Ans: 0.2 A]b. The steady current through the battery a long time after the

switch has been closed. [Ans: 0.067 A]c. The final charge on the capacitor. [Ans: 67 C]

V-37

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