How do we describe charge in motion?• Mass falls to lower P.E. in a g-field•Charge falls to lower P.E. in an E-field.•Electric Current (I) : Charge in motion

–Transient Current: charges move until E-field is cancelled in a conductor.–Electric Circuit:

•Charge “falls” from high P.E. to low P.E.•Charge is then “pumped” back up to high P.E. to fall again.

–Definition: I = ΔQ/Δt = amount of charge passing a point per second. Units: coulombs/second or amperes (“amps”) (A)

Current Direction and Electron Flow• What is ΔP.E. if a +e charge crosses from a plate at +1

volt to a plate at 0 volts?

• What is ΔP.E. if a -e charge crosses from a plate at 0 volts to a plate at +1 volt?

• Energetically: – A +e charge falling one way is the same as a -e charge falling

the other way.

• “Conventional Current”: defined as the perceived flow of (+) charge.

• “Electron Flow”: The actual direction of electron flow. (NYS Regents uses this)

How does a Battery (or Electric Cell) Supply Power to a Circuit?

• An electric cell converts chemical energy into electrical energy.

• Ex: Zinc Copper E-Cell– Zn2+ ions dissolve into H2SO4 leaving Zn

terminal with (-) charge and acid with (+) charge.

– e- are pulled into the (+) charged acid from the Cu terminal (becomes +).

– If in circuit, e- are pulled from wire to the Cu terminal creating a current.

– More Zn gets dissolved as cell is used up.

• Zn metallic bonding (P.E.) is converted to e- current (K.E.).

• voltages for Zn-Cu and Pb-Cu

At Zn: Zn(s) Zn2+ +2 e-

At Cu: 2H+ + 2e- H2(g)

Electric Circuit/Hydraulic CircuitThe two circuits are analogous using similar devices.

Hydraulic Electrical

(quantity) (unit)

Symbol

(meter also)

Pump

(Pressure)

Battery or Cell

(Voltage) (V)

Turbine

(Resistance)

Resistor

(Resistance) (Ω)

Water flow rate

(Electrical current) (A) or

(Electron flow)

Ohm’s Law• How are Voltage, Current, and Resistance related

in a circuit?• Ohm’s Law: R = V/I or V = IR

– V = Voltage between two points.– R = resistance between the two points.– I = current through the path between the two points.

• R is a measure of a material’s resistance to allowing current to flow through it.

• When we draw circuits, we assume the wires to have no resistance compared to the resistors.

• Ohm’s Law always applies between two points.

Current in a Very Simple Circuit•How does the current vary in this circuit?

•Where is it greatest?•Where is it least?

•MUTUAL REPULSION OF ELECTRONS:

• PREVENTS CHARGE BUILD-UP AT ANY POINT• CURRENT IS UNIFORM EVERYWHERE IN THIS “CURRENT LOOP”

Examples (Ohm’s Law)

1. Current of .10 A is measured to flow through a lamp connected by short wires to a 12.0 volt source. What is the resistance of the lamp?

2. How can we find the resistance of a resistor using:

a) A voltage source

b) A voltmeter

c) An ammeter ?

Experimentally using Ohm’s Law

Ohm’s Law Lab

Resistivity in a Conductor

• Wires and conductors do actually have resistance in them.

• For a wire with: – A = cross-sectional area (m2)– l = length (m)– ρ = resistivity (material property) (Ω·m)– R = ρl/A

• Why is R proportional to l ?– Variable Resistor: coil of long wire; can adjust l.

• Why does R vary inversely with A ?

Example (resistivity)

• Suppose we hook up an 8Ω speaker to a stereo system using 6 meters of copper wire with a cross-sectional area of 10-5m. The resistivity of copper is: ρ = 1.68×10-8Ω·m.

What is the resistance of the wire used?

Is this “a lot of resistance”? (Why or why not?)

Resistivity depends on Temperature

• In metals: – ρ increases with T (why?)

• In semiconductors: – ρ decreases with T

– semiconductors conduct more easily when heated (why?)

How much Power is Delivered by an Electrical Circuit?

• Suppose an amount of charge ΔQ falls across a voltage V between two points.

• How much work is done on the charge? W = V ΔQ

• What is the definition of the power? The rate at which work is done P = W/Δt = V ΔQ/ Δt = V I

• From Ohm’s Law: P = VI and V = IR gives:

P = I2R = V2/R (applies only across resistors)

Example (Ohm’s Law and Power)

1. A light bulb whose resistance is 240Ω is connected to a 120 V source.

• What is the current through the bulb?

• What is the power used by the bulb?

• After operating for 10 minutes, how many joules of heat and light energy are produced?

Examples (cont’d)2. A lightning bolt delivers 109 J of energy across a

voltage of 5× 107 V over a period of .2s.

• What is the power delivered?

• What is the current?

• What is the amount of charge transferred?

The kilowatt-hour

• We pay the utility company for energy used every month.

• Convenient unit of energy: (power * time)– Kilowatt-hour (kWh)

• Typical charge by ConEd: $.18 to $.26 per kWh

• What would it cost to run a 100 watt light bulb for 10 hours at $.25 per kWh?– Cost = (.100 kW) (1 hour) ($.25/kWh) = $.25

How can you be killed by Electrical Shock?

• Electric Current (not voltage) causes you harm by– Burning you– Disrupting your normal electrical (nerve) bodily

functions

• Voltage causes current, resistance helps determine how large the resulting current will be, but it is the current that causes the damage.

Electrical Shock (cont’d)

Current (Amps) Effect on Body

.001 Can be felt.

.005-.010 Sensation to mildly painful

.010-.015 Muscle spasms and lack of control

.020-.070 Labored breathing to cessation of breathing

.100-.200 Ventricular fibrillation; heart damage

> .200 Heart can be clamped to a stop; prevents damage

1.0 Severe burns

DC and AC

• Direct Current (DC): – Current always flows in one direction.– Example: battery

• Alternating Current (AC):– Direction of current reverses back and forth.– Direction alternates with a certain frequency.– Sign of voltage changes with current. (“Polarity”

changes sign)– More dangerous shock hazard.– Example: Household Current

• 120V, 60 Hz (U.S.)

• 240V, 50 Hz (Europe)

Symbols in Electric Circuits

Series Circuit (A Voltage Divider)

• There is only one pathway through which current can flow.– Breaking the circuit anywhere stops all current.– Current is the same at every point.– Resistances add up to a higher net, total, (“or

equivalent”) resistance. (REQ = R1 + R2)

– Voltage is divided into smaller voltages across each resistor:

VT = V1 + V2

Series Circuit ExampleFor VT = 120 V, andR1= R2= R3= 200Ω: a. Find the total resistance

of the circuit.b. Find the current through

the circuit.c. Find the voltage across

each resistord. Find the power used by

each resistor and by the entire circuit.

Parallel Circuit (A Current Divider)

• Separate pathways (branches) for current to flow.

• Voltages across parallel branches are the same.

• Current divides to flow through different branches:– More current through low

resistance branches.– Less current through high

resistance branches.– Breaking one branch leaves

current unchanged in other branches.

• Net resistance decreases with more branches.

Parallel Circuit (A Current Divider)(closer look)

• Voltages across parallel branches are the same:

VT = V1 = V2 = V3

• Current divides to flow through different branches:

IT = I1 + I2 + I3

• Net resistance decreases with more branches.

1/REQ = 1/R1 + 1/R2 + 1/R3

Don’t forget to flip 1/REQ

Parallel Circuit Example

• For VT = 120 V, and

• R1= R2= R3= 200Ω:

a. Find the total resistance of the circuit.

b. Find the current through each branch and the total current drawn.

c. Find the voltage across each resistor

d. Find the power used by each resistor.

= 120V

Which will burn brighter?

At what cost?

Overloading a Household Circuit

• Are household appliances hooked up in series or in parallel? Why?

• When we add appliances, each with a load (resistance) R to a power line– How does REQ change?– How does voltage change?– How does current drawn change?– How does power change?

• If the wire is too thin it can heat up and cause a fire.• How do we prevent overloading a circuit?

– Fuse or Circuit Breaker.– How should a fuse be installed in the circuit?

Series-Parallel Combinations• Equivalent Resistance: For any two points

a and b in a circuit, with a configuration of resistors between them: – We can replace the configuration between

them with an equivalent resistance (REQ) without changing • The current (I) through a and b

• The voltage across a and b (Vb -Va)

II

Configuration of resistorsa b a REQ b

I

Examples

1. Between points a and b in a circuit are two resistors (R1=500 Ω and R2=700 Ω) in parallel with each other. In series with a and b are a resistor (R3=400 Ω) and a 12V source.

a. Find the equivalent resistance for this circuit.

b. Find the total current drawn from the battery.

c. Find the voltage across R3.

d. Find the voltage across points a and b.

e. Find the currents through each resistor.

STOP HERE

• BEYOND THIS SLIDE IS AP LEVEL

DC and AC (cont’d)

Voltage: V(t) = V0 sin (2πft) = V0 sin (ωt)Current: I(t) = V(t)/R = (V0/R) sin (ωt) = I0 sin (ωt)“R” = overall resistance of household circuit

•V0 and I0 are “peak” values of voltage and current

•Irms = I0/√2; Vrms = V0 /√2 where “rms” = root mean square

•Average Power: Pav = I2rms/R = ½ I2

0R

Examples (cont’d)

2. Req for a network of resistors:

Emf (“Electromotive Force”)

• A battery in an “open circuit” has an ideal voltage, we call “emf” or “ε”

• A real battery has an internal resistance (r)

• When connected to a circuit, r causes a drop from the ideal voltage to the actual voltage we measure V = ε – I r

• In your lab, try measuring the voltage across your batteries with the switch turned on and then with the switch turned off.

Kirchoff’s Rules(Rules by which to analyze complicated circuits)

• Applying the rules gives equations for unknown currents, resistances and/or voltages.

• Terms:– Junction (or node): point where three or more wires

meet.– Branch: line containing circuit elements in series;

connects two junctions.

1. Junction Rule: The sum of all currents entering a junction equals the sum of all currents leaving the junction.

2. Loop Rule: The sum of the voltages around any closed path is zero.

Applying Kirchoff’s Rules

1. Label the (+) and (-) signs for each battery.

2. Label the current in each branch and assign it (that is, guess) a direction. (If your guess is wrong, the current will come out (-)).

3. If there are “j” junctions, apply the junction rule at j-1 junctions: Σ Ii = 0.

4. If there are “b” branches, apply the loop rule to b-j+1 loops: Σ Vi = 0.

5. There will be a total of “b” equations and “b” unknowns. Solve them.

Applying the Loop Rule

• Follow each loop in one direction (clockwise or counterclockwise) (call this the “loop direction”).– When crossing a resistor

• the sign of the voltage is (-) if the loop direction is the same as the current direction

• the sign of the voltage is (+) if the loop direction is opposite to the current direction

– When crossing a battery• The sign of the voltage is (+) if the loop direction goes from (-)

to (+)

• The sign of the voltage is (-) if the loop direction goes from (+) to (-)

Example 1

For the circuit given in class:

1. Find I1, I2, I3 using Kirchoff’s Rules

2. Find the voltage and power used by the a. 30 Ω resistor.

b. 40 Ω resistor

3. How much power is lost to the internal resistance of the batteries? (The r=1Ω resistors represent these resistances).

Grounds, Shorts, Opens• A ground is a point (or points) specified in a circuit to

be at zero potential. Symbol:

• A short circuit exists between two points connected by

wire with no resistance between them. (Usually this occurs through damage or defect.)

• An open circuit exists between two points when they are separated by a gap so that no current can flow. (Usually this occurs through damage or defect.)

How does a Capacitor behave in a circuit?

• Charging a Capacitor across a resistor– Initially capacitor acts a short circuit.– Finally (when charged), capacitor acts an open circuit.

• Discharging a Capacitor– Initial current and voltage exist as charge spreads out

over circuit.– Finally: current and voltage across capacitor are zero.

How do Capacitors combine in Series and Parallel?

• Series– Voltage is divided by capacitors.– The charge on all the capacitors is the same.– They combine as: (1/CEQ = 1/C1 + 1/C2 + …)– Can think of the distance between plates as being

increased by adding capacitors in series.

• Parallel– Charge is divided by capacitors.– Voltage across each is the same.– They combine as CEQ = C1 + C2 + ….– Can think of the area of plates being increased by adding

capacitors in parallel.

Example• Two capacitors (C1= .1μF and C2 = .2 μF) are

connected in parallel with each other and in series with another capacitor (C3=.6 μF) and with a 12V voltage source.

a. Find the equivalent capacitance.

b. Find the total charge stored.

c. Find the voltage across C3 and across C1 and C2.

d. Find the charge stored on C1 and on C2.