28-1 (SJP, Phys 1120)
Electrical Circuits: Most electrical phenomena (everything from light bulbs to
toaster ovens to computer screens to lightning bolts..) involve the flow of
current. Flow happens when you have a source of potential difference, and a
circuit, a closed path for current to flow. (Since charge is conserved, charges
ultimately have to go around in a circuit for the flow to be sustainable!)
Let's start by thinking about sources of potential difference.
HRW call any such device an "emf" device.
You could think of "emf" as a funny spelling of "oomph". It's what provides the
oomph, the required potential difference to drive current. Examples: generators,
solar panels, fuel cells, or batteries.
Batteries will be our "typical" emf devices:
Batteries: An example of an EMF device
Zinc ions (+ charged) get pulled off by
chemistry (we won’t go into the details!) into
acid bath, leaving behind a residual “-” charge
the Zn rod (terminal, electrode).
Meanwhile, electrons (- charged) are pulled off
carbon rod into the acid bath, leaving a residual
charge on the C terminal.
Battery, or electric cell.
VA
VB
Zinc
Carbon
Acid
the
on
the
+
That means the carbon side is now at a higher potential, VA > VB. This
potential builds up, but if VA gets too high, the acid can’t pull electrons off any
more (the electrostatic attraction of e-’s back onto the + carbon rod will equal
the chemical attraction of the e-’s into the acid) So you reach an equilibrium
with
ΔV = VAB = VA - VB = some fixed value depending on the chemicals.
People usually drop the Δ, and just talk about “V”, the battery’s voltage. (Too
bad, remember they really mean the difference in voltage between the two
electrodes.) Some people will call this the EMF of the battery, using a curly E
that I don't have in my fonts. EMF=work/charge, the electrical potential
difference created.
If there's no circuit, the two terminals are at some potential difference, but no
current flows. If you connect something across them, current will flow, but the
IDEAL battery maintains a constant voltage difference between the terminals!
The EMF device provides energy to the charges moving around the circuit.
28-2 (SJP, Phys 1120)
In diagrams, we use a symbol for batteries:
The “+” and “-“ are often left off: the longer line always
represents the “+” side.
It’s a little like the symbol for capacitors, except the lines are different length.
Capacitors and batteries have some common aspects, but they are still very
different. Capacitors don’t spontaneously build up a ΔV, like batteries do, and
they don’t always have the same value of ΔV.
You might expect that the “+” charges at the top of the C electrode would want
to go over to the “-” post. They are attracted. The +’s would drop in energy,
ΔPE = qΔV: they’d like that, like rolling down a hill. They can’t go through the
acid, though: the chemical reactions are stopping them. But what if you let them
go some other way, outside of the acid? E.g.:
ideal wire
Symbol for ideal
chunk of
material, e.g.
metal.
Battery
wire:
Symbol for chunk of material that
allows current
through:
ideal wire
Now we’ve provided an outside path, a conducting
path, or circuit, for charges to flow from the + to sides of the battery like they want to. There is a
current flowing continuously through the circuit..
This is a simple electric circuit.
I
This is NOT like discharging a capacitor (where the flow is quick, and then stops
when the capacitor is discharged). The battery keeps maintaining a constant
voltage difference, the current is continuous.
28-3 (SJP, Phys 1120)
Example: A bike light’s battery drives 2A of current through the bulb. How
much charge has flowed in one hour?
Answer: I = Q/t, so Q = I*t = (2 A)* (3600 sec) = 7200 C
This corresponds to 7200 C / (1.6E19 C/electron) = 5E22 electrons have flowed
through the bulb! (Sounds like a lot, although 2 A really isn’t an unusual current.
Electrons are small.)
In that last example:
Current (I) is the same everywhere along this
circuit. That means
= I through the wires =
= I through the “chunk of material” =
I
= I through the battery =
D
C
= I passing by point A
IA = IB = IC = ID.
Also, VA= VB (because, there is no voltage change along ideal wires!) Be very
clear: VA refers to the value of voltage at point A. It's a number. It doesn't refer
to a difference. Similarly, VC= VD (again, because there is no voltage change, or
voltage difference along ideal wires.)
A
B
However, VA-VD = “V of battery” (which we REALLY should call ΔV, but
people rarely do) is fixed, V>0.
Look at the picture and convince yourself that this means
VB-VC= VA-VD=V (of battery) (which we should really call ΔV!)
The order of those terms matters: VB is higher than VC.
28-4 (SJP, Phys 1120)
There are several analogies that might help you think intuitively about V, I, and
R in circuits.
Analogy #1: Voltage tells about electrical potential energy, so think of
gravitational potential energy instead, as the analogue.
• Think of flowing charges as people sliding around at a ski area.
• Think of batteries (which lift charges up to high voltage) as chair lifts that lift
people up to high (gravitational) potential.
• Think of resistors (which allow current flow, but eat up energy) as bumpy
mogul runs, which let people ski past, but slow you down.
Flat, smooth,
top of hill.
Switch
(gate) open
Ski Lift
Flat, smooth,
bottom of hill.
Moguls!
Steep downhill.
I added a new circuit element here, a
switch.
As shown, it’s “don’t pass”,
Unfortunately, this is called an “open
switch”, (but that means the run is closed,
no current can flow. )
As shown, we have an “open circuit”, the gate is forbidding flow: no flow of
skiers, no current. (People will build up briefly at the top, but the lift operators
will frantically call down and say “hold up, no more people!” and there’ll soon
be no flow of skiers anywhere)
When you close the switch (which unfortunately means “open the run”) skiers
(current) flows. In steady state, equal numbers of
skiers
Switch
go UP the lift every hour as go DOWN the run
every
(gate) closed
hour.
Ski Lift
The ski lift is a pump, a battery, giving skiers
potential energy, keeping the current flowing. If
dies, the flow of skiers halts.
the lift
The lift raises your potential energy. You’re not allowed to ride the lift DOWN,
you have to ski.
The “flat smooth” parts are ideal wires. Skiers can freely wander forwards or
backwards at the top (or bottom) with no change in energy.
28-5 (SJP, Phys 1120)
Lots of bumps in the mogul means lots of resistance. People ski slowly. Only a
small number of people can go down the hill every hour. So, only an equal small
number of people can go up. (They better not let any more people on the lift than
come down, otherwise there’s a buildup of people, which is forbidden at this ski
area!)
The resistance here is high, the current is small.
If the ski run is smooth and easy => LOW resistance => lots of people will go
down in a short time. There’s a large current of skiers. The lift has to bring lots
of people up, the steady state current is high.
If the run ices up (no resistance at all) we have big trouble. People get hurt, the
lift goes crazy trying to bring everyone up fast enough, because they’re back
down at the bottom essentially instantly. The lift will quickly break down, it’ll
fry. It’s a “short circuit”!
+
If lift #1 raises you V1 feet, and lift 2 raises you V2
you’ve gone up a total of V1+V2 feet.
(We call this “batteries in series”. It’s like the
example problem a few pages back.)
feet,
V2
-
+
V1+V2
V1
-
On the way down, you lose EXACTLY the same height (potential energy,
voltage) as you gained from the lift on the way up. That’s just conservation of
energy.
Inside the mogul run (R), people (+ charges) flow DOWNHILL, from high
voltage to low voltage.
Inside the lift (battery), people (+ charges) are lifted UPHILL, from low voltage
to high.
We’ll come back to this analogy later, when we have more complicated circuits.
It’s not perfect, but sometimes it can help you physically picture what’s going
on.
28-6 (SJP, Phys 1120)
Analogy #2: Water flow. (Physicists really like this analogy, but frankly my
intuition about pipes and pumps and pressure isn’t much better than my intuition
about circuits! But here it is for you to think about, anyway.)
Think of current as “gallons of water/sec”
Think of voltage as “water pressure”.
Think of wires as pipes, and resistors as constrictions in pipes.
Pressure
P1
constricted
pipe
Pressure
P2
is like
R
Voltage
V1
Voltage
V2
If P1 = P2 (no pressure DIFFERENCE), then there is equal force on the water in
the middle. There’s no flow. Similarly, if V1=V2, there’s no difference in
voltage across the resistor, and I = ΔV/R = 0.
On the other hand, if P1>P2, you have high pressure on one side (like, e.g. the
faucet), and a constriction (e.g. the hose) and low pressure on the far side (e.g.
the air outside the hose), and water will flow. Just like current through a resistor
with a voltage drop.
So a circuit looks like this:
big pipe
high
pressure
pump
constricted
pipe
low
pressure
High V
is
like
I
Low V
28-7 (SJP, Phys 1120)
Many real world electronic devices are just collections of wires, resistors,
capacitors, and batteries, forming circuits that do something (flashlights,
toasters, blowdryers, radios, amplifiers,...) It's important to understand (and
predict) the currents and voltages in such circuits. E.g., consider first a simple
“flashlight circuit”:
I
R1
V
The “R” here might represent the resistance of the
flashlight bulb.
Here, V=IR1, or I = V*(1/R1)
Circuit 1
Now consider a slightly more complicated circuit:
These resistors are in series.
There will be a voltage drop V1= I*R1 across the first
resistor, and V2 = I*R2 across the second.
The TOTAL voltage drop from top to bottom is V =
V1+V2 = I*(R1+R2)
R1
V
R2
The resistances simply add up!
In other words, this circuit
Circuit 2
is essentially equivalent to
the following simpler circuit:
V
Requiv =
R1 + R2
Similarly, for many resistors in series: Requiv = R1 + R2 + R3 + K
R1
R2
R3
Requiv = R1+R2+R3
=
I
I
Important: the current I is the SAME through each of these series resistors.
(What goes in must come out: conservation of charge!
This is not an approximation of any kind, it’s exactly true)
However, that doesn’t mean the current I in circuit 1 is the same as the current in
circuit 2. Those are different circuits....
28-8 (SJP, Phys 1120)
Here’s a different circuit. We say R1 and R2 are “in parallel”:
This time, the current I is NOT necessarily the same
through R1 and R2.
I
I2
V
R1
I
The current divides up: I1 goes left, I2 goes right.
I1
Circuit 3
R2
(Conservation of current, however, does tell us that
I = I1 + I2, can you see that?)
It also says I at the bottom (going into the battery)
is exactly the same as I at the top (leaving the
battery)
Note: the voltage across R1 is exactly the same as the voltage across R2! This
is an important point, stare at the picture and try to understand why. Think of
this as two different ski runs. Both have the same top and bottom (the same
height, the same voltage), but they have different resistances, so different
number of skiers/hour. (Different currents through each resistor)
Or, you might think of water flowing
through pipes:
Here, the difference in pressure (like
voltage difference) is exactly the same
for both pipes (pressure at the top of
either is identical, pressure at the
bottom of either is identical, so the
difference across either is identical) but
the current through each will be
different.
The total current is just the sum of the
two currents, I = I1+I2
fixed, high
pressure
total flow, I
pump
constricted
pipe 1
pipe
2
I1
fixed, low
pressure
I2
28-9 (SJP, Phys 1120)
The previous parallel circuit (#3) is essentially equivalent
to the following simpler circuit:
V
Requiv
In this situation (resistors “in parallel”) I claim
1
I
Requivalent
R3
R1
R2
=
1
1
1
+
+
+K
R1 R2 R3
We can prove it mathematically
(if you’re interested):
It comes from the fact that I = V / R equiv ,
but conservation of charges (current) says
I = I1 + I2 + I3 + K =
V V
V
+
+
+K
R1 R2 R3
1

1
1
= V * +
+
+ K = V / Requiv
 R1 R2 R3

Examples of equivalent resistances:
100 Ω
=
50 Ω
1
Requivalent
=
1
1
+
= .02 Ω −1 ,
100Ω 100Ω
i.e. R
100 Ω
equivalent
2Ω
=
0.67 Ω
1
Requivalent
=
i.e. R
1Ω
1
1
+
= 1.5 Ω −1 ,
2Ω 1Ω
equivalent
2Ω
=
0 Ω => (short circuit!)
0Ω
1
Requivalent
=
= 1 / (.02 Ω −1 ) = 50Ω
= 1 / (1.5 Ω−1 ) = 0.67Ω
1
1
+
= (0.5 + ∞) Ω −1 = ∞ Ω −1 ,
2Ω 0Ω
i.e. Requivalent = 1 / (∞ Ω−1 ) = 0. Ω
(The last is a short circuit, 0 resistance. All the current is happy to flow through
the 0 Ω side!)
Note that R_Equiv always comes out less than any of the individual parallel R's.
That means, if there are two (or more) ways for the current to go, there is LESS
overall resistance to flow.
(More ways for current to flow makes it easier for the current to flow. More ski
runs at a resort means you can get more people skiing: more current, less overall
resistance.)
28-10 (SJP, Phys 1120)
You can have both parallel
and series together. To find the
“single equivalent resistor”,
just "break it apart" bit by bit:
1Ω
3Ω
3Ω
0.67 Ω
=
3.67 Ω
=
2Ω
Drawing circuits requires some abstraction. The real situation may look
different, but be equivalent. The fact that wires can have funny shapes makes
things especially confusing. So, e.g. the following two circuits are exactly the
same:
V
V
R
Straight lines (even with bends)
always represent ideal wires.
So you need to think hard in lots of
these pictures!
=
R
E.g. here's another pair of circuits which
are exactly the same. (I labeled some
points, to help guide your eye). Study
this and convince yourself that it's two
different representations of the SAME
CIRCUIT.
a
R1
c
d
R1
R2
b
R3
R3 =
c
b
a
R2
d
R4
If I asked for the "equivalent resistance"
between a and b in the above two, the one on the left
looks scary. But the one on the right makes it apparent
that it's not so bad, just combine R3 and R4 (which are
in series) to get this picture:
R4
R3+R4
R1
R2
Now you could combine "R3+R4" with "R2", which
are manifestly in parallel (dotted circle), and add R1 to
get the equivalent resistance. (It's not as hard as the 1st picture implied!)
28-11 (SJP, Phys 1120)
Here's yet another example of identical circuits. ALL the pictures represent
exactly the same situation! Again, study them all, and convince yourself that
there's no difference!
R1
R2
V
R1
R2
R3
=
R3
V
R3
V
R1
=
R2
R1
V
R2
All those circuits are just 3 resistors in parallel. I personally prefer the first
drawing, because it makes it visually obvious that the voltage V across all 3
resistors is exactly the same. (Like 3 parallel runs down the ski hill) By the way,
this setup is pretty much how houses are wired up for appliances: here's one
MORE drawing, which is pretty much equivalent to all the ones above:
fuse
I
V
Plug
Plug
Plug
R1
Stereo
R2
Toaster
R3
Light Bulb
(The funny symbol on the left represents not a battery but an alternating current
V=120 V AC. The fuse shuts off current to all 3 appliances if I exceeds about
15-20 A)
28-12 (SJP, Phys 1120)
Another example. Here's a more complicated circuit, let's find the currents I1, I2,
and I3. And, let's find Vab, the voltage difference between points a and b.
R1=30 Ω
Look at R2 and R3. They're in
parallel.
a
I1
R2 = 20 Ω
10V
R3 = 20 Ω
So we can combine them into
an effective single resistor.
b
The circuit is equivalent to this one:
R1=30 Ω
a
I know what R/ / is because
I1
R // = 10 Ω
10V
b
1
1
1
1
1
=
+
=
+
= 0.1Ω−1 ,
R/ / R2 R3 20Ω 20Ω
i.e. R/ / = 10Ω.
But now you can see we have two resistors in series. Thus, this circuit in turn is
effectively equivalent to the next one:
Life is easy at this point!
I1
R Equiv
=30+10 Ω
10V
= 40 Ω
b
Use V = I1 Requiv ,
giving
I1 =(10 V)/(40 Ohm) = 0.25 A.
This tells us I1 (the same in any of the diagrams!)
To find Vab, go back a step to the "equivalent circuit" just above.
In that picture, you can see that
Vab = I1* R/ / = 0.25 A * 10 Ohm = 2.5 V.
To find I2 and I3, you go back yet another step, to the very top-most picture.
(Note! Vab is the same across resistors 2 and 3)
Vab = I2 R2 ⇒ I2 = Vab / R2 =2.5V / 20Ω = 0.125A
Vab = I3 R3 ⇒ I3 = Vab / R3 =2.5V / 20Ω = 0.125 A
(Also notice: I2+I3 = .125+.125 = .25A = I1, which makes sense!)
28-13 (SJP, Phys 1120)
There are devices to measure V, I, and R in circuits.
A
I
An ammeter: a device that measures the current running through
itself.
(An ideal ammeter has zero internal resistance, so it won't
hinder the current it is trying to measure...)
a
A voltmeter: a device that measures the voltage difference
across its terminals:
V
b
Vab = Va - Vb.
(An ideal voltmeter has infinite internal resistance, so it won't
alter the voltage it is trying to measure. It won't suck any
current into itself...)
Examples of simple circuits with volt- and ammeters in place:
Good
A
I
V
R
To measure the current "I" flowing through
the resister R in this circuit, we must place
the ammeter in series, as shown.
If you put the ammeter in parallel, like this,
it's very bad. You'll blow a fuse, burn out
your battery, make sparks... Why?
Bad!!
I’
V
R
A
Remember, the ammeter has zero resistance you have short-circuited the battery.
(The current is happy to flow through the ammeter, with R=0, so V=I'R, but if
R=0, and V is finite, I' will be VERY big!)
28-14 (SJP, Phys 1120)
Good
To measure the voltage "V" across the resistor, you
must place the voltmeter in parallel, as shown. The
voltmeter will read off "V" for you
I
V
V
R
Wrong!
This circuit is bad. Remember, voltmeters have
very LARGE internal resistance. In this circuit, no
current can flow! (V=IR, R is huge, so I =0).
V
R
V
There will be no current, no voltage drop across the meter, it will simple read
zero. Nothing will "burn out" like in the previous bad example, but you won't get
the reading you were interested in!
Note: real batteries always have some small, unavoidable internal resistance "r"
in them. We can usually neglect it, but in real life, if e.g. you short circuit a
battery, the current is large but never infinite due to the small but finite "r".
A real battery can be "modeled" in the following way:
r (tiny)
V (real)
(“Real” battery)
=
V
Remember, our text writes an ideal
voltage source as “EMF” (with a
curly E) instead of V.
(“ideal” battery)
If you ever need to think about circuits with real batteries instead of ideal ones,
just add in a small “internal resistance” r next to the battery. As batteries get old,
the internal resistance gets larger (and so they put out less current)
Have you ever noticed that if you start your car with the lights on, the lights
dim? That’s because of the internal resistance of the car battery- as current
flows, there is some voltage drop across “r”, which means less voltage for the
light bulbs...
28-15 (SJP, Phys 1120)
Finding R_equiv piece by piece, like we've done, usually works fine. But
sometimes, circuits get a little too complicated. E.g.
or
V
V
(2 batteries!)
You can still find the currents and voltages throughout, using ideas we call
"Kirchoff's rules". (There's nothing especially new here, we've already
intuitively seen and used the rules!)
Kirchoff's first rule is a statement of conservation of current (charge): "whatever
current goes in, must come out".
I2
I1
a
I3
At this junction "a" (which might be part of a bigger
circuit),
I1
=
I2 + I3
Current Entering = Current Exiting
Just be careful to watch the arrows. You will be drawing
arrows for currents, and they might point either way. E.g. in this picture:
I2
I1
a
I1 + I3
=
I2
Current Entering = Current Exiting
(Note the direction of I3's arrow)
I3
Example: Look at the junction labeled "a".
Here, I = I1 + I2 + I3
(I enters, the other three exit)
(Look back at the discussion about household
wiring. We had this same diagram.
I is the total current drawn by your house. I1, I2,
and I3 are currents in the individual appliances.)
a
I
V
I1
I2
I3
28-16 (SJP, Phys 1120)
Kirchoff's second rule says "The sum of voltage changes around any closed loop
is always zero". (This is just conservation of energy.) Think of our ski-lift
analogy, where voltage <-> height. As you move around the ski area, you go up
and down, but if you make a loop (ending up right where you started), the total
sum of all rises plus all drops will add to zero!
Example: Consider the following circuit (by now familiar!)
Look at the loop labeled #1.
You can start wherever you
want, let's begin at the bottom
and imagine "walking around"
the loop in the direction
Loop #4
shown.
(around the outside)
We first go UP the battery
(voltage change +V), and then we go DOWN the resistor R1, with voltage
change -I1*R1. It’s a drop: think hard about every minus sign here:
V
R1
Loop
I1
#1
R2
Loop
I2
#2
Loop
I3
#3
R3
+V − I1 R1 = 0 (Or V = I1 R1 , which if you think about it makes sense!
Resistor R1 has a voltage V across it, after all)
Alternatively, you could go around the loop labeled #4. (That's going around the
whole circuit, basically). Now you go up the battery, across the top (no voltage
drop there), and then down resistor R3. So Kirchoff's second rule says
+V − I3 R3 = 0
(Or V = I3 R3 , again makes sense)
Another alternative: Loop #2. Thinking of this in the "skier analogy", loop #2 is
for backcountry skiers. You ski UP R1, and then back down R2. Kirchoff's
second rule would say
+I1 R1 − I2 R2 = 0
(Or I1 R1 = I2 R2 , both also =V, again makes sense)
Notice the signs. The I1 R1 term is +, you're going UP the hill there (going
against the current), so your voltage is increasing....
28-17 (SJP, Phys 1120)
Kirchoff's rules are fairly intuitive, except the signs can be tricky! Unfortunately
it's critical that you get them right. Currents always have arrows (directions)
associated with them, and so do the loops.
R
I
If you are going around a loop and encounter a resister with a current
like this, then IF you are heading RIGHT -->, you're traveling WITH
the current, the voltage DROPS across R, so the change is -IR.
On the other hand, if you happen to be going around the loop the other way, <--,
(so you're "fighting the current"), you're going UP the hill, change in voltage is
+IR.
Batteries can also be confusing:
If you are going around a loop and encounter a battery like in
this picture, then IF you are heading RIGHT -->, you're
V
travelling from the low V side to the high V side of the battery,
the change is +V.
On the other hand, if you are going around the loop the other way, (<--), then
you're going DOWN the hill (down the battery) and the change in voltage is -V.
(Think of the ski lift analogy)
low side
high side
Example: Let's figure out the currents in each of the resistors here.
I3
R1= 5 Ω
V1=
10 V
I1
I2
I4
R2
= 50 Ω
V2=
25 V
The first step is to draw arrows for
currents, and label them.
The direction you initially pick for the arrow is arbitrary!
It doesn't matter. If you solve for, say, I1 in the picture above, and if it happens
to come out negative, that just means the real current is heading the OPPOSITE
way from the arrow you drew. (So, it'd be left, in the picture above.)
28-18 (SJP, Phys 1120)
Before proceeding, try to avoid making up too many different symbols/labels for
currents. E.g., in this example, look over on the left. I4 is entering the battery
(labeled V1) and I1 is leaving. But Kirchoff's rule #1 says "what goes in, must
come out". That's true everywhere, including batteries. So I4=I1. Why make up
a different label? Just replace I4 with I1! There are then only 3 unknowns,
rather than 4...
Here's the circuit again, with currents (and a few points) labeled.
R1
I1
V2
R2
I2
V1
What we're after is I1, I2, and I3.
I3
a
b
Look at junction "a", we have
I1 + I3 = I2
(Eq'n 1)
(current in = current out: watch the arrows!)
That's ONE equation, for three unknowns! We
need two more.
It turns out, in this case, that no more "current junctions" (i.e. Kirchoff's first
rule) will help. We could TRY, look e.g. at point b.
I2
I1
b
At point b, Kirchoff's first rule says
I2
= I1 + I3.
I3 (Current in = Current out)
But that's the same equation we had before!
So we use Kirchoff's second rule for our other two equations.
Here's the circuit one more time, with some loops labeled:
R1
I3
Let's follow Loop #1.
a
I1
V1
R2
Loop
#1
I2
Loop
#2
b
+V1 − I1 R1 − I2 R2 = 0
V2
We can start wherever we want,
e.g. at point b, and go around.
Watch all the signs!
(Eq'n 2)
We climbed UP the battery V1, it's +. We went DOWN both of the resisters R1
and R2 (i.e. we went WITH the current), so they're both voltage drops, or
negative. Putting in numbers, we have
+10V − I1 (5Ω) − I2 (50 Ω) = 0
(Eq'n 2 with numbers)
We need one more equation. Just pick another loop (any one will do). E.g., let's
go around loop #2, starting this time at point a.
−V2 + I2 R2 = 0
(Eq'n 3)
28-19 (SJP, Phys 1120)
As always, the minus signs are important to understand. Look at the picture
again: this time, when we pass battery V2, we are going DOWN the battery
(from the "high" side to the "low" side indicated by the long and short horizontal
lines on the battery symbol, respectively.) That's a voltage DROP. It's like we
took the ski lift labeled V2 DOWN the mountain! That's why V2 got a minus
sign above.
Similarly, now we're climbing up the resistor R2. (We're fighting the current I2)
That means we're going uphill: the voltage change is plus, so that's why we have
+I2*R2.
Putting in numbers,
−25V + I2 (50 Ω) = 0 , or (solving for I2),
I2 = 25V/50 Ohm = 0.5 A
We can plug this value for I2 back into Eq'n (2), giving
10V - I1*5 Ohm - (0.5 A)*50 Ohm =0, or (solving for I1)
I1 = [10 V - 0.5A*(50 Ohm)] / (5 Ohm) = -3 A.
Finally, Equation 1 said I1 + I3 = I2, or (-3A) + I3 = 0.5 A,
so I3 = 3.5A.
I1 came out negative. So we guessed wrong: current I1 is really heading left. It's
going INTO the left-hand battery. That's a little unusual, the big 25V battery on
the right is pumping current INTO the 10 V battery on the left, it's being
"charged". Only specially designed batteries can be effectively recharged, for
most normal ones this won't work. (You might damage the battery on the left.)
What does "charging" mean? Well... batteries can only put out so much total
charge = current*time, and then the chemicals inside die. So for the short term,
recharging is irrelevant, (the circuit diagram tells you what happens, regardless
of whether the battery is rechargeable or not) but after a long time, the voltage
(and current) of a normal battery will become zero, unless you can recharge it!
28-20 (SJP, Phys 1120)
RC Circuits:
Putting capacitors into circuits is a bit of a funny business. E.g.,
V
C
The battery on the left will charge up the capacitor on the right.
Charges move onto the top plate, and off the bottom plate, until
finally Q=CV has built up on both plates. Then, no more charges
will move.
So there's briefly a current (as charge flows from the
battery to the capacitor), but then the current stops.
I (briefly)
V
+Q
++++
---- C
Now consider a circuit with no battery at all, just a
capacitor and a resistor. Let's begin with the capacitor all charged up
(like we just described), and the switch open, as shown
here.
Q0
++++
---- C
R
Let's suppose we start with charge Q0, and voltage V0,
across the capacitor.
(V0 = Q0/C, of course)
At time t=0, we close the switch. What happens? The capacitor is now able to
discharge! At first, a large current, I, will flow. (There's a voltage across the
resistor at this point, and V=IR) This situation is similar
to a simple "battery and bulb" circuit (like way back on P.
I
19-1), but capacitors aren't batteries.
+++
The capacitor runs down, quickly! Current doesn't
CONTINUE to flow, like it does with a battery.
--- C
R
So the current STARTS off strong (I0 = V0/R), but it will decay away with time.
28-21 (SJP, Phys 1120)
At any instant in time, we have the following relations:
a
V = IR
V = Q/C
I
(Q)
+++
--- C
R
V=voltage difference
= Va-Vb
b
The voltage "V" is the SAME voltage across the capacitor and the resistor.
(Look at the picture, V=Va-Vb in the picture, it’s the difference in potential
between top and bottom)
What is I? It's the charges flowing/second. Where do the charges come from?
They were stored on the capacitor plate!
So I = -dQ/dt.
(If charges leave the capacitor, they must flow through the resistor. The minus
sign says that Q is decreasing when current flows in the direction shown)
Putting this together with the equations above gives
(dQ/dt) = -I = -V/R = -Q/)RC)
In words, the RATE at which charges leave is proportional to the AMOUNT of
charge still sitting on the plate. (Q on the right side is of course changing all the
time, this equation holds at some instant)
Look at the units:
[RC]=[Ohm*Farad]=[V/I][Coul/V]=[Coul/I]=[Coul/(Coul/sec)]=sec
(This makes sense, dQ/dt SHOULD be charge/time)
We define the quantity τ ≡ RC .
(That's the Greek latter "tau")
Whenever you have something decaying, and the rate is proportional to how
much you have left, you get "exponential decay". The mathematical formula for
Q as a function of time is
Q(t) = Q0 e− t /( RC) = Q0 e −t /τ
There are MANY examples of "exponential
decay" in nature. Some examples: cooling objects (where the temperature decays
with time), or radiation (where the number of particles decays with time), or
drug concentrations (where the amount decays with time)...
28-22 (SJP, Phys 1120)
In the exponential decay formula, "e" is a number, e=2.718... It's a little like Pi.
It's also called the "base of natural logarithms".
e^something really means 2.718^something.
(You do NOT use the "E" or "EE" key on your calculator! Look for the e^x
button)
Since V=Q/C, we have (dividing that last equation through by C)
V(t) = V0 e− t / ( RC) = V0 e −t /τ .
At time t=0, you have e^(0) = 1, in other words V(0) = V0. (Duh)
If you wait till a later time, the formula tells you what the voltage is (or, the
earlier formula tells you the remaining charge on the plates)
Example: When t=tau, (i.e. if you wait a time t=RC seconds, which is also
called waiting "one time constant"), then you have
1
1
V(τ ) = V0 e −1 = V0 =
V0 ≈ .37V0 .
e
2.718
The voltage has dropped to almost a third of where it started.
If you wait TWO time constants, i.e. t=2RC = 2*tau, then
V(2τ ) = V0 e −2 =
1
V ≈ .14V0
e2 0
Here's a sketch of Voltage as a function of time:
+V0
Voltage (AC)
.37V0
.14V0
τ
2τ
3τ
time
Whenever you wait "tau" more seconds, the voltage (or similarly, the charge)
has decreased to 1/e = 37% less than what you just had.
28-23 (SJP, Phys 1120)
Example: Consider the following "RC" circuit:
C = 10 nF
+++
---
R =2 kΩ
Here, "tau"=RC=(10*10^-9 F)*(2000 Ohm)
= 2*E-5 sec.
The time constant of this circuit is
2E-5 sec. (pretty short)
Suppose you begin with Q0=50 microC of charge.
(So V0 = Q0/C = 50.E-6C / 10.E-9 F = 5000 V = 5kV)
The charge will begin to leak off (through R).
After 2E-5 sec (one time constant), you will have
1
Q(τ ) = Q0 e −1 = Q0 ≈ .37Q0 ≈ 18.5 µC left on the plates.
e
After 4E-5 sec (two time constants), you will have
Q0/e^2 = 6.8 micro Coulombs left.
You can plug ANY time into this formula. E.g., after one second,
you have Q(1sec) = Q0 e− (1sec)/ RC =Q0 e −1/ (2E −5) = 50 µC e−5E 4 ≈ 0
It's long gone!
TV and computer monitors usually have BIG capacitors in them to store up
charges. They also have big resistors across those capacitors, and often have
very high voltage power supplies (converting the 120V at the wall to much
higher voltages, many kV, inside). So Q0 = CV0 is large (lots of charge stored!)
and tau = RC is also large (the time constant is big, it takes a relatively long time
for that charge to leak back off the capacitors). If you open up a monitor and
start poking around in there, you're liable to get into big trouble.
C
+++
---
R
R(you)
You provide a LOWER resistance path for the
charges to flow through - the current goes
through YOU (with a smaller time constant, i.e.
quicker!)
The inside of TV's and computer monitors can be pretty dangerous, even when
they’re unplugged, because the capacitors in them can hold a lot of charge for a
long time (even after being disconnected to the external voltage source.)