V-1
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
This unit assumes that you have completed the laboratory exercise entitled “E-V.1
DIRECT CURRENT ELECTRIC CIRCUITS”. If you have not completed it, you will do well
to complete it before continuing.
Unit Objectives
When 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 crosssectional 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 resistors 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 resistor 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 resistor 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 resistor, 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-2
Unit V – OHM'S LAW AND D.C. CIRCUITS
In the laboratory exercise entitled “E-V.1 DIRECT CURRENT ELECTRIC CIRCUITS” you were
introduced to the following ideas as they apply to DC 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.
CURRENT
Suppose 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)
For answers & solutions see “In-Text
Fill-in Answers” beginning on page V-26
V-3
Unit V – OHM'S LAW AND D.C. CIRCUITS
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.
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 charges
less 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.
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
= ____________ 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-4
Unit V – OHM'S LAW AND D.C. CIRCUITS
Back to our hunk of wire. If there is an -field exerting a force on
the electrons equal to e 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 l represent the distance an electron travels at speed vd in
time t. Thus, l = 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 nlA = 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 (e), or
The current is the charge passing through D-D’ per unit time. In this case, charge QT 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.
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 1 A.
Solution: Solve the equation above for vd.
vd = ____________
(V-4 #1)
where I = ___ A, e = ____________ C and A = ___ mm2 = ______ m2
(V-4 #2)
If we can find n, then we can calculate vd.
Notice that unit-wise
all this reduces to
electrons/volume
_____ , ( atoms
____________ ,
( electrons
atom ) =
mole ) =
for copper = ____________ , \ ( mass
____________ ,
( mass
mole )
mole ) =
mass
(V-4 #3)
( volume
) for copper = ____________
From givens and previous courses:
Now multiply these together and make sure your final answer is in MKSA units.
n = _________________ electrons/cm3 = _________________ electrons/m3
(V-4 #4)
V-5
Unit V – OHM'S LAW AND D.C. CIRCUITS
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 crosssectional area A. Since the -field is constant within the wire, it
follows that
Now I = nevdA and since (the number of electrons/vol) and e 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
Unit V – OHM'S LAW AND D.C. CIRCUITS
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 sample’s 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 >10 5
“non-conductors” and those in the broad mid-range are “semi-conductors.
RESISTIVITY AT
ROOM TEMP.
MATERIAL
Silver
Copper
Aluminum
Tungsten
Iron
Manganin (Cu 84%,Mn 12%,Ni 4%)
Constantan (Cu 60%, Ni 40 %
Nichrome
Graphite
Carbon
Germanium
Silicon
Rock (Granite)
Wood (Maple)
Mica
Quartz (Fused)
1.47 x 10-8
1.72 x 10-8
2.83 x 10-8
5.51 x 10-8
10 x 10-8
44 x 10-8
44.1 x 10-8
100 x 10-8
8 x 10-6
3.5 x 10-5
0.43
2.6 x 103
105 - 107
4 x 10 11
9 x 1013
5 x 1016
V-7
Unit V – OHM'S LAW AND D.C. CIRCUITS
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 R 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)
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
Unit V – OHM'S LAW AND D.C. CIRCUITS
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
Thus the
(V-8 #1)
-field in the iron is:
(V-8 #2)
and in tungsten:
(V-8 #3)
At this point try problems 4 through 7 at the back of this unit.
ENERGY & POWER
Consider 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
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.
V-9
Unit V – OHM'S LAW AND D.C. CIRCUITS
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:
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)
of 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
The
if the bulb is left on overnight (about 10 hrs).
is an energy unit {
In the electric power industry. First let’s find the conversion factor between
joules:
} used
and
(V-9 #5)
V-10
Unit V – OHM'S LAW AND D.C. CIRCUITS
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 would be ________.
(V-10 #1)
At this point try problems 8 through 11 at the back of this unit.
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 RESISTANCE
Consider 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
where Ir is the voltage across the internal resistance r. As you would expect, when I = 0,
∆V = E.
V-11
Unit V – OHM'S LAW AND D.C. CIRCUITS
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
is less than the sum
(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
is greater than
then
somehow charge is being destroyed at the junction, which is also baloney. If charge is to be
conserved,
must equal
.
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
this is just K-2!
V-12
Unit V – OHM'S LAW AND D.C. CIRCUITS
Now let’s outline a consistent way of working circuit problems using K-1 and K-2 to minimize
sign and other goof-ups.
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-13
Unit V – OHM'S LAW AND D.C. CIRCUITS
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 selfcorrecting!
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 = ( _____ )( _____ ) = ( __________ )( __________ ) = __________
Equation
Substitution
(V-13 #3)
Answer
b. the rate at which battery #1 is converting chemical potential energy into electrical
energy and heat.
P = ( _____ )( _____ ) = ( __________ )( __________ ) = __________
Equation
Substitution
(V-13 #4)
Answer
c. the potential difference between the terminals of battery #2.
VBatt. #2 = ( _____ ) – ( _____ ) = ( ______ ) – ( ______ )( ______ ) = ________
Equation
Substitution
Answer
(V-13 #5)
V-14
Unit V – OHM'S LAW AND D.C. CIRCUITS
d. the energy radiated as heat in R2 and R3 in 10 minutes.
E2 = ( _____ )( _____ ) = ( _____ )( _____ )( _____ ) = ( _____ )( _____ )( _____ )
Equation
Equation
Substitution
= __________
(V-14 #1)
Answer
E3 = ( _____ )( _____ ) = ( _____ )( _____ )( _____ ) = ( _____ )( _____ )( _____ )
Equation
Equation
= __________
Answer
For further examples see Serway & Beichner pages 879-882.
At this point try problems 12 & 13 at the back of this unit.
Substitution
(V-14 #2)
V-15
Unit V – OHM'S LAW AND D.C. CIRCUITS
EQUIVALENT RESISTANCES
Resistors 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
or
E = 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=E
Note 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
At this point try problems 14 through 17 at the back of this unit.
V-16
Unit V – OHM'S LAW AND D.C. CIRCUITS
ELECTRICAL INSTRUMENTS
When 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 µ 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 the
galvanometer 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 Þ
galvanometer, where R6 is its internal resistance.
The Ammeter (symbol:
represent the
)
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 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,
and
therefore
. Applying Ohm’s Law:
Using these and solving for RS for a desired current range is given by:
V-17
Unit V – OHM'S LAW AND D.C. CIRCUITS
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,
< 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 our sample galvanometer reads 1 µA a difference in potential of
1 mV is across its terminals. When it reads 0.5 µA, 0.5 mV is across its terminals, etc. This is
true because ∆V µ 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-18
Unit V – OHM'S LAW AND D.C. CIRCUITS
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 the galvanometer so that it looks like this Þ
(series or parallel?)
Ü
Ü (complete drawing)
(V-18 #1)
The value of the resistor that was added in the sketch above is given by:
RV = _____________ = _____________ = _____________
(Equation)
(Substitution)
(V-18 #2)
(Answer)
b. To use the meter as an ammeter, a resistor must be connected in ___________
(series or parallel?)
Ü (complete drawing)
with the galvanometer so that it looks like this Þ
(V-18 #3)
The value of the resistor that was added in the sketch is given by:
RS = _____________ = _____________ = _____________
(Equation)
(Substitution)
(Answer)
At this point try problems 18 and 19 at the back of this unit.
(V-18 #4)
V-19
Unit V – OHM'S LAW AND D.C. CIRCUITS
RC CIRCUITS
In 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-20
Unit V – OHM'S LAW AND D.C. CIRCUITS
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)
time until after a long time, I = _______, and ∆VR becomes _______.
(decreases/increases) (V-20 #6)
(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-21
Unit V – OHM'S LAW AND D.C. CIRCUITS
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/C
or
Ü
Since: VR = IR, I = dQ/dt,
and VC = Q/C
E =
Rearranging:
(CE – Q)/RC
Let’s solve this for Q as a function of time:
Ü
Integrating this should be old stuff,
Ü use substitution: u = CE – Q \ du = - dQ
An orgy of algebra:
Recall from Algebra II:
If
Solving the last expression for the charge Q as a function of time:
V-22
Unit V – OHM'S LAW AND D.C. CIRCUITS
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) = E
All 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-23
Unit V – OHM'S LAW AND D.C. CIRCUITS
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
is decreasing) and
(negative since Q
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 = Q 0 (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
, 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-24
Unit V – OHM'S LAW AND D.C. CIRCUITS
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)
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)
Þ
e-nRC = 0.01
Taking natural logs (i.e., ln) of both sides of the last expression:
- n/RC = - 4.6
Þ
n = 4.6RC
Hence it takes 4.6 to charge the capacitor to within 1% of its maximum charge.
At this point try problems 20 through 23 at the back of this unit.
– End Unit V –
V-25
Unit V – OHM'S LAW AND D.C. CIRCUITS
In-Text Fill-In Answers
Page
Fill-In #
Fill-In Answer
V-2
1
Coulomb, 6.25 x 1018
V-3
1
e- Þ 1.6 x 10-19 C, p+ Þ 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
7
V-4
= 0.5 A + 0.18 A = 0.68 A to the right
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.
1
2
3
I = 1 A, qe = 1.6 x 10-19 C and A = 1 mm2 = 10-6 m2
= 1,
= 6 x 1023 (Avogadro’s #),
for copper = 63.5 g/mole,
for copper = 9 g/cm3
4
V-5
1
V-7
1
2
3
vd = 7.3 x 10-5 m/s = 0.073 mm/s
=1/63.5 = 0.016 mole/g,
V-26
Unit V – OHM'S LAW AND D.C. CIRCUITS
In-Text Fill-In Answers
Page
Fill-In #
V-7
4
Fill-In Answer
5
6
7
V-8
1
2
3
V-9
1
2
3
100 J
4
5
V-10
1
0.36 MJ = 0.1
, 10 hrs we use 1
V-12
1
Loop 1: E1+ I1r1 + I1R1 = I2R2
Loop 2: I2R2 + I3R3 + I3r2 = E2
2
Point A: I1 + I2 = I3
Point B: I3 = I1 + I2
, cost: 5¢
V-27
Unit V – OHM'S LAW AND D.C. CIRCUITS
In-Text Fill-In Answers
Page
Fill-In #
V-13
1
I. 6V + I1(1) + I1(8) = I2(4) Þ 6 = 4I2 – 9I1
II. I2(4) + I3(7) + I3(1) = 10V Þ 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
1
d. Conversion: 10 min x 60 s/min = 600 s
E2 = (P)(t) = (I 2)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
1
Series,
V-14
V-18
Fill-In Answer
2
3
4
parallel,
V-28
Unit V – OHM'S LAW AND D.C. CIRCUITS
In-Text Fill-In Answers
Page
Fill-In #
V-20
1
2
Fill-In Answer
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-29
Unit V – OHM'S LAW AND D.C. CIRCUITS
In-Text Fill-In Answers
Page
Fill-In #
V-24
1
I = E/R, Q = Q0 = CE, VR = VC = E
2
I = Q = ∆VR = ∆VC = 0
3
Fill-In Answer
V-30
Unit V – OHM'S LAW AND D.C. CIRCUITS
End of Unit Problems
1. A circular dielectric disk has a uniform surface charge density 1 µC/m 2. 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 in
b. 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
]
V-31
Unit V – OHM'S LAW AND D.C. CIRCUITS
End of Unit Problems
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 Þ 0.67 A CCW, center Þ 0.33 A up, right Þ
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-32
Unit V – OHM'S LAW AND D.C. CIRCUITS
End of Unit Problems
14. Find the equivalent resistance for each of the networks below.
[Ans: (a) Req = 25/4 , (b) Req
, (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.
c. If points A and B are connected with a wire having zero
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-33
Unit V – OHM'S LAW AND D.C. CIRCUITS
End of Unit Problems
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 which
a. 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]