Ch. 20 – Electric Circuits

20 1 Electromotive Force20.1 – Electromotive Force

Every electronic device depends on circuits.

Electrical energy is transferred from a power source such as aElectrical energy is transferred from a power source, such as a battery, to a device, say a light bulb.

Light bulbA diagram of this circuit

ld l k lik th f ll iConducting Light bulb

+ Light

would look like the following:

+-

Battery

symbol

gwires

Battery

symbol

Inside a battery, a chemical reaction separates positive and negative charges, creating a potential difference.

This potential difference is equivalent to the battery’s- This potential difference is equivalent to the battery s voltage, or emf (ε) (electromotive force).

This is not really a “force” but a potential.

Because of the emf of the battery, an electric field is produced within and parallel to the wires.Because of the emf of the battery, an electric field is produced within and parallel to the wires.This creates a force on the charges in the wire and moves them around the circuit.

This flow of charge in a conductor is called electrical current (I).

A measure of the current is how much charge passes a certain point in a given time:

qI ΔtqIΔ

=Electrical Current

Units?[ ] [ ]AAmpere

sC

timeCharge

==⎥⎦⎤

⎢⎣⎡=⎥⎦

⎤⎢⎣⎡

If the current only moves in one direction, like with batteries, it’s called Direct Current (DC).

If the current moves in both directions, like in your house, it’s called Alternating Current (AC).

f f

Light bulb

+

I

Electric current is due to the flow of moving electrons, but we will use the positive conventional current in the circuit diagrams.

Battery I

-

I

e So I shows the direction of “positive” charge flow from high potential to low potential.

20.2 – Ohm’s Law

The flow of electric current is very analogous to the flow of water through a pipe:

The battery pushing the current acts like the water pump pushing the water.

The voltage of the battery is analogous to the pump pressure – the higher the pump th f t I h th t th h Th th l b ttpressure, the faster I can push the water through. Thus, the larger my battery

voltage, the greater my current.

IV ∝

Let’s make this an equality: IRV = This is Ohm’s Law.

The proportionality constant, R, is called the electrical resistance.

[ ] [ ]⎤⎡⎤⎡ VVoltsVUnits? [ ] [ ]Ω==⎥⎦⎤

⎢⎣⎡=⎥

⎦

⎤⎢⎣

⎡= Ohm

AV

AmpVolts

IVR

Define Resistor: A component of an electrical circuit that offers resistance to the flow of electric current.

Symbol for resistors:

RResistor,

+ -

ε = VStraight lines have ε = VStraight lines have essentially zero resistance

20.3 - Resistivity

The electrical resistance of a conductor depends on its shape:

-Longer wires have more resistance

-Fatter wires have less resistance

Thus, .ALR ∝

Length

Let’s make this an equality:ALR ρ=

ACross-sectional area

A

The proportionality constant, ρ, is the electrical resistivity.

Units? [ ]mmm

2

⋅Ω→⎥⎦

⎤⎢⎣

⎡ ⋅Ω=

LARρ

Resistivity is an intrinsic property of materials, like density:

Every piece of copper has the same resistivity, but the resistance of any one piece depends on its size and shape.

ρ, R ρ, Rρ, R ρ,

A copper wire with a circular cross section has a resistance R What o ld the resistance of the ire be if the radi s of

Clicker Question 20-1

R. What would the resistance of the wire be if the radius of the wire was reduced by a factor of 2?

1. R2. 2R3. 4R4. R/25. R/4

Temperature Dependence of Resistivity

The resistance of most materials changes with temperature.g p

For good conductors (metals) the resistance decreases with decreasing temperature.

R Conductors R Insulators

T T

For insulators (poor conductors) the resistance increases with decreasing temperature.

[ ]For many materials, we find that: [ ])(1 oo TTRR −+= αR = Resistance at temperature T

t lF0>αRo = Resistance at temperature To α is the temperature coefficient of resistivity

insulatorsFor

metalsFor

0 0

<>

αα

20.4 Electrical PowerOur standard definition of power is: .

TimeWorkPower =Time

So what would electrical power be?

WFrom the definition of potential: qVW

qWV =⇒=

qVThus,

tqVP = IVP =⇒

We can write this different ways using Ohm’s Law, V=IR:

VPRIP2

2 ==R

PRIP ==

*So we have 3 ways of calculating electrical power depending on what other quantities are known.

Electrical Energy

Work and Energy have the same units (Joules)Work and Energy have the same units (Joules).

Thus TimePowerEnergy ×=Electrical companies, like Entergy, measure your monthly energy use this Thus, TimePowerEnergy ×= y y gyway, in units of kilowatt⋅hours (kWh).

For example, if you used an average power of 1500 W for 31 days (744 hours), your energy consumption would be:

kWh1116)h744)(kW51( ==E kWh 1116)h744)(kW5.1( ==E

At a cost of roughly $0.13/kWh, this would be a monthly bill of $145.

J10603kWh1 6×= J 1060.3kWh1 ×

Two light bulbs, A and B, are connected to a 120-V outlet (a constant lt ) Li ht b lb A i t d t 60 W d li ht b lb B i t d

Example:

voltage source). Light bulb A is rated at 60 W and light bulb B is rated at 100 W. Which bulb draws the most current from the outlet?

1. A1. A2. B

IVP =

If P is bigger, than I will be bigger for the same V.

20.5 Alternating Current

When the voltage provided changes periodically with time!g p g p y

0 sin 2V V ftπ= 0 sin 2I I ftπ=/I V R=

0 0 /I V R= Peak values 60f Hz= frequency

Sinusoidal function

20 0 sin 2P IV I V ftπ= =Power:

Mean Value of Power:0 0

1rms rmsP I V I V= ≡0 02 rms rmsV V

0

2rmsII = 0

2rmsVV = root-mean square value

2 2

20.6 Series Circuits

The first way is to wire them together in series:

The same current runs through twoThe same current runs through two components connected in series.

V1 and V2 are called voltage drops.

*We speak of currents running through resistors, and voltages drops across resistors.

Thus, the current through resistor R1 is I, and the voltage drop across R1 is V1.

How would we find the net resistance (equivalent resistance, Req) for resistors connected in series?

For resistors connected in series, the sum of the voltage drops across all the resistors must equal the battery voltagemust equal the battery voltage.

Th VVV +Thus, 21 VVV +=

B t f Oh ’ L IRIRIR RRRBut from Ohm’s Law:21 IRIRIReq += 21 RRReq +=⇒

Thus, for resistors wired up in series, the equivalent resistance is:, p , q

⋅⋅⋅+++= 321 RRRReq

i.e. you just add them!!!

What is the current I in the following circuit?

Clicker Question 20-4

g4 Ω 2 Ω

12 V

1. 0.50 A2 2 00 A2. 2.00 A3. 10.0 A4. 60.0 A

20.7 Parallel Circuits

For resistors connected in parallel the voltage

I1

For resistors connected in parallel, the voltage drop across each resistor is the same.

I

I2

Thus, .21 VVV ==

The current through each might be different. It splits: I = I1 + I2.

From Ohm’s Law:

21 IIV

IVReq +

==2

2

1

1RV

RV

V+

=21

11

1

RR +=

21 RR 21 RR

Thus, ⋅⋅⋅+++=321

1111RRRReq

for resistors in parallel.

Given the circuit below, a larger current flows through hi h i t ?

Clicker Question 20- 5

which resistor?=2 Ω

=4 Ω

+ -

I

1. R1

+ -V = 12 V

1. R1

2. R2

3. The current through each is the same.

What is the current I in the following circuit?

20.8 Series and Parallel CircuitsNow let’s hook resistors up both in series and in parallel in the same circuit!

What is the current I in the following circuit?

We need to find the I I

equivalent resistance!

I I

Thus, Ω= 240R, Ω240eqR

VI = A1024==

II eqRI A1.0

240

What is the equivalent resistance between points A d B?

Clicker Question 20- 6

A and B?

8.00 Ω

1. 2 Ω2. 4 Ω3. 6 Ω3. 6 Ω4. 8 Ω

Three identical light bulbs are connected to a battery as shown in the diagram. What happens when bulb #3 burns g pp

out?

1. Bulbs 1 and 2 both get brighter2. Bulb 1 gets brighter and 2 stays

the same3. Bulb 1 gets dimmer and 2 gets

b i htbrighter4. Bulb 1 gets dimmer and 2 stays

the same

#1

#2

#3

+ -

C1

C2

21 QQQTot += VCVC 21 += VCC )( 21 += VCeq=

+ -

V

Thus, ⋅⋅⋅+++= 321 CCCCeq

To find the equivalent capacitance forTo find the equivalent capacitance for capacitors wired in parallel, you just add them!

Now consider two capacitors wired in series:

C1 C2 The voltage across each will now, in general, be different, but the charge on each of the plates must be the same.

+ -

V

but t e c a ge o eac o t e p ates ust be t e sa e

21 VVV +=V1 V2

21 CQ

CQ += ( )

21

11CCQ +=

( ) 1V ( ) VCVQ eqCC =+=⇒ −11121

Thus, ⋅⋅⋅+++= 1111CCCC

For capacitors wired in series, the +++321 CCCCeq reciprocal of the equivalent

capacitance is just the sum of the reciprocal capacitances.*Notice that these rules are just the

opposite for resistors in combination.

What is the equivalent capacitance between points A and B?Clicker Question 20- 7

1. 0.5 μF2 1 2 μF2. 1.2 μF3. 2.0 μF4. 6 μF

Clicker Question 20-7 (cont.)Let’s first number the capacitors:

1 2

3 4

5 6

Capacitors 2, 4, and 6 are in series, thus:

642246

1111CCCC ++= 8

1121

241 ++= 4

1=

Now the circuit looks like this:

F0.4246 μ=⇒C1

3246

4.0 μF

5

μ

Now capacitor 3 is in parallel with capacitor 246: 24632346 CCC += 844 =+=

Now the circuit looks like this:

1

23468.0 μF

Now capacitors 1,2346, and 5 are in series:

5

523461123456

1111CCCC ++= 2

161

81

51 ≈++= F 0.2123456 μ==⇒ CCeq523461123456 CCCC 685 q