Download “In parallel” means A] both resistors necessarily have the same

“Over the weekend, I reviewed
the exam and figured out what
concepts I don’t understand.”
A] true
B] false
1 point for either answer.
A battery is connected to a metal wire.
What is the direction of the electric field at point P, in the wire?
A] Up
B] Down
C] There is no field, as it is in a conductor
A battery is connected to a metal wire. The wire thins down to
half the diameter along the right side of the circuit.
What is the magnitude of the E field at Q, compared to at P?
A] same
B] larger
C] smaller
A current of 1 ampere flows at point P. The direction of
the current is
A] up
B] down
A current of 1 ampere flows at point P. The current is
carried by electrons, which have negative charge. What
direction do the electrons move?
A] up
B] down
A current of 1 ampere flows at point P. What is the
direction of the current inside the battery?
A] up
B] down
C] there is no current inside the battery, it just flows from
+ to - outside the battery.
Assume wires have zero resistance
Assume wires have zero resistance
A] same
B] 3R
C] 9R
D] R/3
E] R/9
We can draw a graph of voltage around a circuit.
This is analogous to height.
For each resistor, V=IR. V is the voltage drop across
that resistor.
Let’s figure out the effective resistance of two resistors
in series.
Resistors dissipate power.
The electric field does work on the charges as
they drop through the resistor. dW = V dq
So P = V I
The work done by the field does not go into
accelerating the charges, as they move (on
average) at a steady drift velocity.
It goes into HEAT.
A] A is brighter
B] B is brighter
C] same
A] bulb A on left is brighter B] bulb A on right is brighter C] same
Batteries are pumps.
They pump charges up to a potential Vbatt
(above the - terminal)
An ideal battery will provide as much current as is needed
to keep the potential difference = Vbatt
The potential provided by the battery drives current around
a circuit (and through the battery itself.)
We call a potential that drives current an
Electromotive force or EMF.
V = IR
What happens to the brightness of bulb
B when the switch is closed?
A] it gets brighter
B] it gets dimmer
C] it is unchanged
In series, R1 + R2 = Ref
We also found that the power dissipated in a resistor is
P = IV & P = I2R
& P = V2/R
A 60 W incandescent bulb dissipates 60W
A 100 W incandescent bulb dissipates 100W
If an incandescent bulb dissipates more power, it
will be brighter.
Which has a higher resistance, a 100 W bulb or a
60 W bulb?
A] 60 W
B] 100 W
C] they have the same resistance
Here’s a DIFFERENT question
If R1 > R2, which bulb is brighter?
A] R1
B] R2
C] Same
Here, the more resistive bulb is brighter. (If you
put a 60W bulb and a 100W bulb IN SERIES
the 60W bulb will be brighter!)
Bulbs in your home are IN PARALLEL
What is the effective resistance of two resistors
in parallel?
“In parallel” means
A] both resistors necessarily have the same current
B] both resistors necessarily have the same potential drop
C] both resistors are placed side-by-side
Are R1 and R2 in parallel, in series, or neither?
A] parallel
B] series
C] neither
1
1
1
 
Ref R1 R2

What happens to the current I as we add R5?
A] it goes down if R5 is larger than R4
B] it goes down regardless of R5
C] it goes up regardless of R5
For resistors in series, resistances add.
For capacitors in parallel, capacitances add.
V=IR
But
V = Q/C
What happens to the
brightness of bulb A when
the switch is closed?
A] it gets brighter
B] it gets dimmer
C] it is unchanged
B&C in parallel have a lower resistance than B alone.
So the total current goes up, and A gets brighter.
Real(ish) Batteries
Our ideal batteries provide “as much current as needed”
at the battery potential.
When you draw a lot of current out of a real battery, the
potential drops a little.
So a better model of a real battery includes an “internal
resistance.”
Ammeters and Voltmeters
Measuring current and voltage (drop)
To measure current, you must send the current
THROUGH the meter.
To measure voltage, you must connect your meter
ACROSS the voltage drop.
A good voltmeter should have a
A] very high
B] very low
internal resistance?
Answer: very high. It’s connected in parallel,
and we don’t want it to change the circuit.
Similarly, a good ammeter should have a very
low internal resistance.
A very nice worksheet on simple circuits and lightbulbs has
been posted. It is a great source of exam questions!
Written homework due next Friday :
Chapter 26: 46, 69, 73
Kirchoff’s Rules
Current is “conserved”
(current into a junction = current out) (did this last time)
Voltage changes around a loop sum to zero.
(Change = Voltage After element - Voltage Before Element)
Voltage change across an ideal battery is the EMF, (or EMF, if you go the other way).
Voltage change across a resistor is IR (or -IR, if you go
the other way!)
A negative voltage change across an element
means the voltage goes down, when traversing the
element in the specified direction.
There is no right direction or wrong direction to
apply Kirchoff’s voltage rule. YOU choose a
direction.
Use +IR for the voltage change across a resistor if
you traverse it OPPOSITE TO THE CURRENT.
(“paddling upstream”)
Use -IR for the voltage change across a resistor if
you traverse it WITH THE CURRENT.
(“paddling downstream”)
BUT BEFORE YOU ANALYZE A CIRCUIT, YOU DON’T
ACTUALLY KNOW THE DIRECTION OF THE CURRENT!
It doesn’t matter. YOU choose the current!
Here’s an example: I have deliberately (and obviously) drawn the
current going the wrong way. Let’s solve for I1
Starting at point A and going clockwise, what are the voltage
changes around the loop?
First: What is the voltage change
across the ideal battery?

B] C] 0 
A]
What is the voltage change across the top resistor, going clockwise?
A] I1R
B] -I1R
C] 0
We are swimming upstream, according to our choice of current direction, so
change in voltage = +IR.
Adding all the voltage changes together gives zero. Solving for I1 = -EMF/(2R).
The negative sign means the current actually flows clockwise!
What is the direction of the
current through the 3V battery?
A] up
B] down
C] Ibatt = 0
D] it depends on R
Keep in mind: current CAN flow through a battery either direction.
If it flows from the - end to the + end, the battery is discharging:
chemical energy is changing into electrical energy (electrical power
added P= IV).
If it flows the other way, the battery is charging: in this case,
something else in the circuit is adding power to the battery; electrical
power lost P=IV)
What is the current in the top branch?
A] 10 A, to the right
B] 10 A, to the left
C] 2 A, to the right
D] 2 A, to the left
E] cannot determine w/o R
What is the current in the top branch?
Answer: 2A, to the left
Let’s apply the voltage loop law to the
outer loop, going anticlockwise.
What is the voltage change across the 3
ohm resistor?
A] 18 V
B] -18 V
C] 2 V
D] - 2 V
What is the current in the top branch?
Answer: 2A, to the left
Let’s apply the voltage loop law to the
outer loop, going anticlockwise.
What is the voltage change across the 3
ohm resistor? Answer -18 V. We are
paddling DOWNstream.
Continuing around the outer loop, what is
the voltage change across R?
A] -2V
B] -10V
C] +14V
What is the current in the top branch?
Answer: 2A, to the left
Let’s apply the voltage loop law to the
outer loop, going anticlockwise.
What is the voltage change across the 3
ohm resistor? Answer -18 V. We are
paddling DOWNstream.
Continuing around the outer loop, what is
the voltage change across R?
Answer: -10V.
What is the resistance R?
A] 5 ohms
B] -5 ohms
C] 10 ohms
D] 28 ohms
Using V=IR, R = 5 ohms.
(What happened to the minus sign???)
Traversing the bottom loop anticlockwise
starting at x, what is the voltage change
through the 6 ohm resistor?
A] 10 V
B] -10 V
C] 24 V
D] -24 V
E] cannot determine without knowing
EMF
Using V=IR, R = 5 ohms.
(What happened to the minus sign???)
Traversing the bottom loop anticlockwise
starting at x, what is the voltage change
through the 6 ohm resistor?
Answer -24V. We drop 24 V in the 6 ohm
R and 18 V in the 3 ohm R, so the
unknown emf must be:
42 V.
NOTE: if the battery had been drawn the
other way, we could correctly conclude
that it was drawn backwards! (when
solving, we would find emf = -42V)
Any circuit with resistors & batteries can in principle be solved
with Kirchoff’s loop & current rules.
With multiple loops, you will need to solve simultaneous linear
equations.
Your FIRST EFFORT should be to simply, WHERE
POSSIBLE, by replacing series and parallel resistors with Req
Your SECOND EFFORT should be to look for any loop that is
solvable on its own.
Your THIRD and last EFFORT is to write down N equations in
N unknowns, when 1 & 2 have been taken as far as possible.
What is I , and I5 ?
All R = 20 ohms, and emf = 10V
Are any resistors in series?
A] 2 & 4 are
B] 1 & 3 are
C] 1 & 2 are
D] both A& B
E] none are in series
Are any resistors in parallel?
A] 4 & 5 are
B] 2 & 3 are
C] 1 & 4 are
D] both B&C
E] none are in parallel
We redraw the circuit, replacing the parallel resistors with Refs
We see that R5 is now in parallel with the 2 series Refs.
The total Ref = R/2 = 10 ohms. So I= 1 A and I5 = 0.5 A.