“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] UpB] DownC] 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] sameB] largerC] smaller

A current of 1 ampere flows at point P. The direction of the current is A] upB] 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] upB] down

A current of 1 ampere flows at point P. What is the direction of the current inside the battery?

A] upB] downC] 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] sameA] 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 isP = IV & P = I2R & P = V2/R

A 60 W incandescent bulb dissipates 60WA 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 WB] 100 WC] 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 resistorsin parallel?

“In parallel” means

A] both resistors necessarily have the same currentB] both resistors necessarily have the same potential dropC] both resistors are placed side-by-side

Are R1 and R2 in parallel, in series, or neither?

A] parallelB] seriesC] neither

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1Ref= 1R1+ 1R2

For resistors in series, resistances add.For capacitors in parallel, capacitances add.

V=IR But V = Q/C

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

What happens to the brightness of bulb A when the switch is closed?

A] it gets brighterB] it gets dimmerC] 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 highB] 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?

A]

B] -

C] 0

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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!

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 direction of the current through the 3V battery?

A] up

B] down

C] Ibatt = 0

D] it depends on R

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 ohmsC] 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 VB] -10 VC] 24 VD] -24 VE] 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 areB] 1 & 3 areC] 1 & 2 areD] both A& BE] none are in series

Are any resistors in parallel?

A] 4 & 5 areB] 2 & 3 areC] 1 & 4 areD] both B&CE] 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.