Electric Current & DC Circuitsphys.ufl.edu/~avery/course/2061/2061C_circuits_from2054.pdfChapter 31:...

1 PHY2061: Chapter 31

Electric Current & DC Circuits


Chapter 31: DC Circuits 1.  (Nature of electrical current) 2.  EMF and circuits 3.  (Microscopic nature of current) 4.  (Ohm’s law) 5.  (Conductivity, resistivity and resistance) 6.  Kirchhoff’s rules 7.  Series and parallel circuits 8.  Analysis of single-loop and multi-loop circuits 9.  Power in electric circuits 10. RC circuits

3

EMF Device

è Moves charge carriers from low to high potential

è Converts energy to electrical energy

è Maintains potential difference across terminals

PHY2061: Chapter 31

4

Some EMF Devices

PHY2061: Chapter 31

Type Energy source Battery à Electrolytic reaction Generator à Changing magnetic field Fuel cell à Oxidation of fuel Solar cell à Electromagnetic energy Thermopile à Nuclear decay

5

Battery è Two electrodes (different metals)

u Immersed in electrolyte (dilute acid) u One electrode develops + charge u Other electrode develops – charge

PHY2061: Chapter 31

6

Common dry cell battery

PHY2061: Chapter 31

7

Electric Current

è Connecting the terminals of a battery across device leads to an electric circuit in which current flows u Charge begins to flow: electric current u Units: 1 Coulomb/s = 1 Ampere (A)

è Symbol: + - V + - or

I = Δq

Δt

PHY2061: Chapter 31

8

Direction of current

è Current is defined to be the direction of movement of + charge u Current flows from positive to negative terminal u But in conductors, electrons are free and carry the charge u Current thus points in opposite direction from electron movement

In the wire, electrons move very slowly (0.1 mm/s). ~ 1 meter per 3 hours!!

+++ ---

- -

-

-

-

I

PHY2061: Chapter 31

9

Direction of current (2)

PHY2061: Chapter 31

è What if we have a medium with + and – charge carriers e.g., ions in a gas or liquid? u Positive ions move away from

positive terminal u Negative ions move toward

positive terminal u But both currents are in

same direction: away from positive terminal

10

Example of Electron Flow

è Consider a current of 1A. Find the number of electrons flowing past a point per second

è So, in one second, number of electrons passing the point is

1 A 1 coulomb / secqt

Δ = ⇒Δ

1819

1 coulomb 6.2 10 electrons1.6 10e

N −= = ××

PHY2061: Chapter 31

11

Current and Electron Drift Speed

è Consider a material where current (electrons) is flowing u Let ne = # free charge carriers / m3 u Let q = charge per charge carrier u Let A = cross sectional area of material

è Total charge ΔQ in volume element moving past a point

è If charges moving with average drift speed vd, then Δx = vd Δt

è Thus, current can be written in terms of macroscopic and microscopic quantities

( )eQ n A x qΔ = Δ

( )e dQ n Av t qΔ = Δ

e dQi n qAvt

Δ= =Δ

- -

-

-

-

xΔ

A

using V A xΔ = Δ

PHY2061: Chapter 31

12

Example of Drift Speed

è 10A flowing through a copper wire of diameter 2mm u Density of Cu = 8.92 g/cm3 = 8920 kg/m3

u 1 free electron per Cu atom! u Atomic mass ACu = 63.5 ⇒

è Find drift speed vd using

u e is electron charge

u Area A:

u Still need ne = density of electrons (#/m3). Use Cu properties

e di n eAv=

A = πr2 = 3.14 × 10−3( )2 = 3.14 ×10−6 m2

191.6 10e −= ×

ne =

ρCumCu

×1= 8.92×103

1.05×10−25= 8.5×1028 / m3

mCu =

63.5×10−3

6.02×1023= 1.05×10−25kg

PHY2061: Chapter 31

13

Example of Drift Speed (cont.)

è Solve for electron drift speed vd

è Thus vd is 0.24 mm/sec: ~1 hour to move 1 m

è But electrons actually move ~ 106 m/s in material! u This is ~ 4 ×109 times larger than drift speed u So what’s going on?

vd =i

neeA= 10

8.5×1028( ) 1.6 ×10−19( ) 3.14 ×10−6( ) = 2.4 ×10−4 m/s

PHY2061: Chapter 31

14

Drift Speed vs Individual Electron Speeds

è Resolution of paradox u Individual electrons move at very high speeds, but in random directions u Current depends only on the net velocity of a large number of electrons

è Many examples of this in real life (as opposed to physics life) u Passengers moving around on a cruise ship: vd = ship velocity u Air molecules bouncing around inside elevator: vd = elevator velocity u People moving around on planet earth: vd = earth velocity

PHY2061: Chapter 31

15

Electrons in the Wire

è If the electrons move so slowly through the wire, why does the light go on right away when we flip a switch? u (1) Household wires have almost no resistance u (2) The electric field inside the wire travels much faster u (3) Light switches do not involve currents u (4) None of the above

PHY2061: Chapter 31

16

Electrons in the Wire

è If the electrons move so slowly through the wire, why does the light go on right away when we flip a switch? u (1) Household wires have almost no resistance u (2) The electric field inside the wire travels much faster u (3) Light switches do not involve currents u (4) None of the above

PHY2061: Chapter 31

Electric field moves at a fraction of the velocity of light, immediately moving electrons which fill the entire circuit. This is like a hose full of water when you turn on the faucet. The pressure wave moves much more rapidly than the water itself.

17

Electrons in the Wire, Part 2

è Okay, okay, okay. So the electric field in a wire travels quickly. But, didn’t we just learn that E = 0 inside a conductor? u (1) True, it can’t be the electric field after all!! u (2) The electric field travels along the outside of the conductor u (3) E = 0 inside the conductor applies only to static charges u (4) None of the above

PHY2061: Chapter 31

18

Electrons in the Wire, Part 2

è Okay, okay, okay. So the electric field in a wire travels quickly. But, didn’t we just learn that E = 0 inside a conductor? u (1) True, it can’t be the electric field after all!! u (2) The electric field travels along the outside of the conductor u (3) E = 0 inside the conductor applies only to static charges u (4) None of the above

EMF source constantly replenishes E field. But this takes energy provided by the emf source.

PHY2061: Chapter 31

19

Ohm’s Law (Current vs Voltage)

PHY2061: Chapter 31

Obey’s Ohm’s law (almost all materials)

Violates Ohm’s law (semiconductor)

20

Ohm’s Law

è Ohm’s law is an empirical observation that the current in a material is proportional to the applied E field: u  Current vs applied voltage difference V

è It is not a “law” in the sense of other physical laws u Only approximately true u Violated completely by semiconductors and superconductors u But it is observed in a diverse range of materials over a very wide range of

applied voltage differences u It can also be understood in terms of an atomic level description of current u (Later slide)

PHY2061: Chapter 31

i ∝V

21

Microscopic Understanding of Ohm’s Law

è Consider conductor with applied electric field E è Electrons within conductor

u Move at high speed ~ 106 m/s, but very small drift velocity vd u Undergo collisions: mean collision time τ u Drift speed & mean collision time related by acceleration

u Where L = length of material and we used ΔV = EL

PHY2061: Chapter 31

vd = aτ =

Eeme

⎛

⎝⎜⎞

⎠⎟τ = ΔVe

Lme

⎛

⎝⎜⎞

⎠⎟τ ΔV = EL( )

22

Microscopic Understanding (2)

è We can now write the current as

è So resistance is a product of two factors u Resistivity: Property of the material u Length/area: Geometrical dimensions of material

è Ohm’s law: Collision time almost unaffected by E field u Why è Individual velocities are huge compared to drift velocity u Collision time is ~constant, so resistivity is almost independent of ΔV u But resistivity is temperature dependent

PHY2061: Chapter 31

i = Aneevd = Anee( ) eΔVτLme

⎛

⎝⎜⎞

⎠⎟≡ A

L1ρΔV ≡ 1

RΔV

ρ =

mee2neτ

= "resistivity" R = ρ L

A

23

Collision Time Example (Copper)

è Copper properties u ne = 8.5 × 1028 / m3 (calculated earlier) u me = 9.1 × 10−31 kg (electron mass) u ρ =1.67 × 10−8 Ω·m (Cu resistivity) u Calculate collision time

è Interpretation u Typical electron velocity is ~106 m/s à collision distance ~ 25nm u Copper atom spacing = (1 / ne)1/3 = 0.23 nm u Average collision distance ~210 atoms

PHY2061: Chapter 31

τ = me / e2neρ = 25×10

−15 sec = 25fsec

24

Table of Resistivities

PHY2061: Chapter 31

25

Temperature Dependence of Resistivity

è Resistivity increases with temperature for most materials

u Where values are typically tabulated at 20o C (see table) u Not true for semiconductors, which behave very differently

è Example: find resistivity of copper at 100o C u α = 4.05 x 10–3 / oC u ΔT = 80

u So resistivity increases by 32%, a substantial amount

PHY2061: Chapter 31

ρ T( ) = ρ0 1+αΔT( )

ρ 100( ) = 1.67 ×10−8 1+ 4.05×10−3( )×80⎡⎣⎢ ⎤⎦⎥= 1.67 ×10−8 1.324( )= 2.21×10−8


Reading Quiz for Chapter 18

è When resistors are connected in series u (1) the current in each resistor is different u (2) the current in each resistor is the same u (3) the voltage in each resistor is the same


Series Circuit

è Simple series situation u 1 battery and two resistors R1 and R2 u Crucial: common current I

è Loop starting at (a): u 0 + E – V1 – V2= 0 u E – IR1 – IR2 = 0 u E = IR1 + IR2 ≡ IRs u Rs = R1 + R2

è So in series circuit battery sees an equivalent resistance = Rs u From POV of battery, Rs behaves in all ways (current, power) identically

to series combination of R1, R2 u Works for any number of resistors u Rs = R1 + R2 + R3 + R4 + …

R1 R2

Ea

E


Resistors in series

è Battery EMF = 12 V, 3 identical resistors. What is the potential difference across each resistor? u 12 V u 0 V u 3 V u 4 V

R R R

30

Resistors in series

è Battery EMF = 12 V, 3 identical resistors. What is the potential difference across each resistor? u 12 V u 0 V u 3 V u 4 V

PHY2061: Chapter 31

Current is the same, so voltage across each is the same

R R R


Resistors in series è If the light bulbs are all the same in each of these two circuits,

which circuit has the higher current? u (1) circuit A u (2) circuit B u (3) both the same

A B


Resistors in series è If the light bulbs are all the same in each of these two circuits,

which circuit has the higher current? u (1) circuit A u (2) circuit B u (3) both the same

A B

A has twice the current as B


Real EMF Sources: Internal Resistance

è Real batteries have small internal resistance u Lowers effective potential delivered to circuit

CC

aVbV r

R

i = E

r + R

Veff =Vb −Va =E − ir

=E − Er + R

r

Veff =

E1+ r / R

This is the voltage measured across the terminals!

E


Internal Resistance Example

è Loss of voltage is highly dependent on load

è E = 12V, r = 0.1Ω, R = 100Ω u Loss of 0.01V



è E = 12V, r = 0.1Ω, R = 0.5Ω u Loss of 2.0V

eff 12/1.01 11.9V V= =

eff 12 /1.1 10.9V V= =

Veff = 12 / 1.2 = 10.0V

eff 12 /1.001 11.99V V= =

Veff =

E1+ r / R

= 121+ 0.1/ R


Resistors in Parallel è Current splits into several branches.

è ΔV is the same across each branch u  ΔV = V1 = V2

è Total current is conserved u  I = I1 + I2

a

d

I

R1 R2

I1 I2 ΔV

ΔVRp

= ΔVR1

+ ΔVR2

1Rp

= 1R1

+ 1R2

I

a

d

I

I ΔV

I

Rp

Rp = equivalent parallel resistance. Works for any number of resistors

36

è As more resistors R are added in parallel to the circuit, how does total resistance between points P and Q change? u (a) increases u (b) remains the same u (c) decreases

PHY2061: Chapter 31

Resistors in Parallel

37

è As more resistors R are added in parallel to the circuit, how does total resistance between points P and Q change? u (a) increases u (b) remains the same u (c) decreases

PHY2061: Chapter 31


Req =1

1R+ 1

R+ 1

R+


è If the voltage between P & Q is held constant, and more resistors are added, what happens to the current through each resistor? u (a) increases u (b) remains the same u (c) decreases



è If the voltage between P & Q is held constant, and more resistors are added, what happens to the current through each resistor? u (a) increases u (b) remains the same u (c) decreases


i = ΔV

R For each branch


Resistance Example

è Find net resistance of the circuit connected to the battery. Each resistance has R = 3 kΩ

1

2

3 4

56

12

A

B

CNote that in the following figures that kΩ displays as kW for some reason.


Resistance Example (1+2)

è Combine #1 & #2 in series u R12 = 3 + 3 = 6 kΩ

1

2

3 4

56

12

A

B

C

6kW

34

56

12

A

B

C

Note that kΩ displays as kW.


Resistance Example (1+2+3)

è Combine 6kΩ & #3 in parallel u R123 = 6 || 3 = 2 kΩ

6kW

34

56

12

A

B

C

2kW 4

56

12

A

B

C


Resistance Example (1+2+3+4)

è Combine 2kΩ & #4 in series u R1234 = 2 + 3 = 5 kΩ

2kW 4

56

12

A

B

C

5kW

56

12

A C


5kW

56

12

A C

Resistance Example (1+2+3+4+5)

è Combine 5kΩ & #5 in parallel u R12345 = 5 || 3 = 1.875 kΩ

1.875kW6

12

A C


Resistance Example (1+2+3+4+5+6)

è Combine 1.875kΩ & #6 in series u R123456 = 1.875 + 3 = 4.875 kΩ

1.875kW6

12

A C 4.875kW

12


Circuits

è If the light bulbs are all the same in each of these two circuits, which circuit has the higher current through the battery? u (a) circuit A u (b) circuit B u (c) both the same

A

B


Circuits

è If the light bulbs are all the same in each of these two circuits, which circuit has the higher current through the battery? u (a) circuit A u (b) circuit B u (c) both the same

A

B

B draws twice the current as A


Circuits

è In which circuit is each light bulb brighter? u (a) circuit A u (b) circuit B u (c) both the same

A

B


Circuits

è In which circuit is each light bulb brighter? u (a) circuit A u (b) circuit B u (c) both the same

A

B

Current through each branch is unchanged (V / R)

50

Household Circuits: Parallel Paths! è All devices are added in parallel

u Each power outlet is 115 V, independently of others

u Enough power must be supplied from connection to power company

è Overload: too many devices that require a lot of current can draw more current than wires can handle. u Overheating of wires u Fire hazard!

PHY2061: Chapter 31


Circuit Problem (1)

è The light bulbs in the circuit are identical. What happens when the switch is closed? u a) both bulbs go out u b) the intensity of both bulbs increases u c) the intensity of both bulbs decreases u d) nothing changes

(a) (b)

12V

12V


Circuit Problem (1)

è The light bulbs in the circuit are identical. What happens when the switch is closed? u a) both bulbs go out u b) the intensity of both bulbs increases u c) the intensity of both bulbs decreases u d) nothing changes

Before switch closed: Va = 12V because of battery. Vb =12 because equal resistance divides 24V in half. After switch closed: Nothing changes since (a) and (b) are still at same potential.

(a) (b)

12V

12V


Circuit Problem (2)

è The light bulbs in the circuit shown below are identical. When the switch is closed, what happens to the intensity of the light bulbs? u a) bulb A increases u b) bulb A decreases u c) bulb B increases u d) bulb B decreases u e) nothing changes

(b) (a)

12V

12V

12V


Circuit Problem (2)

è The light bulbs in the circuit shown below are identical. When the switch is closed, what happens to the intensity of the light bulbs? u a) bulb A increases u b) bulb A decreases u c) bulb B increases u d) bulb B decreases u e) nothing changes

(b) (a)

Before switch closed: Va = 12V because of battery. Vb =12 because equal resistance divides 24V in half. After switch closed: Nothing changes since (a) and (b) are still at same potential.

12V

12V 12V


(a) (b)

Circuit Problem (3)

è The bulbs A and B are identical. What happens when the switch is closed? u a) nothing happens u b) A gets brighter, B dimmer u c) B gets brighter, A dimmer u d) both go out

24V

12V


(a) (b)

Circuit Problem (3)

è The bulbs A and B are identical. What happens when the switch is closed? u a) nothing happens u b) A gets brighter, B dimmer u c) B gets brighter, A dimmer u d) both go out

Before: Va = 24, Vb = 18. Bulb A and bulb B both have 18V across them. After: Va = 24, Vb = 24 (forced by the batteries). Bulb A has 12V across it and bulb B has 24V across it.

24V

12V

57

Using Kirchhoff’s Rules (1)

è Junction rule (conservation of current/charge) u Current into junction = sum of currents out of it

I I2

I1

I3

1 2 3I I I I= + +

9 V22 W

15 W

6 V

I3

I1

I2 I1 = I2 + I3

PHY2061: Chapter 31


Using Kirchhoff’s Rules (2) è Loop rule (conservation of energy)

u Algebraic sum of voltages around a closed loop is 0

3 1 2 2 2 3 1 2 0I R I R E I R E− − − − + =

1 2

1 1 1 1 1 2 2 2 0I R E I R E I R− − − + + =

1

2

I3 = I1 + I2


Problem Solving Using Kirchhoff’s Rules

è Label the current in each branch of the circuit u Choice of direction is arbitrary u Signs will work out in the end (if you are careful!!) u Apply the junction rule at each junction u Keep track of sign of currents entering and leaving

è Apply loop rule to each loop (follow in one direction only) u Resistor: loop direction same as current direction: ⇒ –iR u Resistor: loop direction opposite current direction: ⇒ +iR u Battery: loop direction in “normal” direction: ⇒ +V u Battery: loop direction in “opposite” direction: ⇒ –V

è Solve equations simultaneously u You need as many equations as you have unknowns

60

+

+ 22Ω

15Ω

6 V

9 V

I1

I3

I2

PHY2061: Chapter 31

Kirchhoff’s Rules: Example

è Determine the magnitudes and directions of the currents through the two resistors in the figure below. u Take two loops, 1 and 2, as shown on next slide

1

2

61

+

+ 22Ω

15Ω

6 V

9 V

I1

I3

I2

PHY2061: Chapter 31

Kirchhoff’s Rules: Example

è Determine the magnitudes and directions of the currents through the two resistors in the figure below. u Take two loops, 1 and 2, as shown

+6−15I3 = 0−22I2 + 9+15I3 = 0

I3 = 6 / 15 = +0.40I2 = 15 / 22 = +0.68I1 = I2 + I3 = +1.08

1

2

1

2

I1 = I2 + I3

Currents are all +, so are in directions shown


Circuit Quiz

1W

3W

2W

2V4V

6V

I1 I2 I31W

è Which of the equations is valid for the circuit shown below? u a) 2 − I1 − 2I2 = 0 u b) 2 − 2I1 − 2I2 − 4I3 = 0 u c) −2 − I1 − 4 − 2I2 = 0 u d) I3 − 2I2 − 4I3 = 0 u e) 2 − 2I1 − 2I2 − 4I3 = 0


Resistors

è Current flows through a light bulb. If a wire is now connected across the bulb as shown, what happens? u (a) All the current continues to flow through the bulb u (b) Current splits 50-50 into wire and bulb u (c) Almost all the current flows through the wire u (d) None of the above


Resistors

è Current flows through a light bulb. If a wire is now connected across the bulb as shown, what happens? u (a) All the current continues to flow through the bulb u (b) Current splits 50-50 into wire and bulb u (c) Almost all the current flows through the wire u (d) None of the above

Current is maintained by external voltage source. The wire “shunt” has almost no resistance and it is in parallel with a bulb having resistance. Therefore all the current follows the zero (or extremely low) resistance path.

66

Circuit Example (1)

è Consider the network of resistors shown below. The switch is initially open, then closed. u What happens to the current within and voltage across R1? u What happens to the current within and voltage across R2, R3, R4? u What happens to the total power output of the battery?

PHY2061: Chapter 31

+

V

R1

R2 R3 R4

67

Circuit Example (2)

è Consider the network of resistors shown below. The switch is initially open, then closed. u What happens to the current and voltage through R1?

PHY2061: Chapter 31

+

V

R1

R2 R3 R4

After switch closes: Ø  More resistors in parallel Ø  Parallel resistance decreases Ø  Current through R1 increases Ø  Voltage across R1 increases

68

Circuit Example (3)

è Consider the network of resistors shown below. The switch is initially open, then closed. u What happens to the current and voltage in R2, R3, R4?

PHY2061: Chapter 31

+

V

R1

R2 R3 R4

Ø  Voltage across R1 increases Ø  Voltages across R2, R3, R4 decrease Ø  Current in R2 increases (from 0) Ø  Currents in R3 & R4 decrease

69

Circuit Example (4)

è Consider the network of resistors shown below. The switch is initially open, then closed. u What happens to the total power output of battery?

PHY2061: Chapter 31

+

V

R1

R2 R3 R4

After switch closes: Ø  Same emf, more current Ø  Power = Emf * i à increases

70

Circuit Example (5)

è Consider the network of resistors shown below. The switch is initially open, then closed. u Let R1 = R2 = R3 = R4 = 90Ω and V = 45 V. u Initial total resistance is 90Ω + 45Ω = 135Ω, final is 90Ω + 30Ω = 120Ω u Find current through each resistor before and after closing the switch.

PHY2061: Chapter 31

+

V

R1

R2 R3 R4

Before Ø I1 = 45/135 = 1/3 Ø I2 = 0 Ø ΔV1 = I1R1 = 30, ΔV2 = 0 Ø ΔV3 = ΔV4 = 45 – 30 = 15 Ø I3 = I4 = 15/90=1/6

After Ø I1 = 45/120 = 3/8 Ø ΔV1 = I1R1 = 33.75 Ø ΔV2 = ΔV3 = ΔV4 = 11.25 Ø I2 = I3 = I4 = 1/8


Wheatstone Bridge

è An ammeter A is connected between points a and b in the circuit below. What is the current through the ammeter? u a) I / 2 u b) I / 4 u c) zero u d) need more information

V

R R

R R I

a

b


Wheatstone Bridge


Before ammeter is added: •  The top branch divides the voltage

evenly, so Va = V/2. •  The bottom branch also divides

the voltage evenly, so Vb = V/2. •  Thus Va = Vb and current is 0

across ammeter. V

R R

R R

a

b I


Wheatstone Bridge


2R 2R

V

R R a

b I


Wheatstone Bridge


2R 2R

Same analysis. Before ammeter is added: •  The top branch divides the voltage

evenly, so Va = V/2. •  The bottom branch also divides

the voltage evenly, so Vb = V/2. •  So although branch currents are

different, Va = Vb and current is 0 across ammeter.

V

R R

I

a

b


Res-Monster Maze

All batteries are 4V All resistors are 4Ω

Find current in R Hint: follow the batteries

R


Res-Monster Maze

All batteries are 4V All resistors are 4Ω

0V

4V

8V

12V

8V

0V

IR = 2 R

Find current in R Hint: follow the batteries

77

Power in Electric Circuits

è Work done by emf source raising charge q to E volts

è Power delivered to circuit

è Power dissipated by resistor

è In any circuit: power delivered = power dissipated PHY2061: Chapter 31

ΔU =Eq

P = dU

dt=E

dqdt

=E i

P =Vi = i2R

78

Power Example

è Let E = 24 V, R1 = 4Ω, R2 = 8Ω

è Power delivered

è Power dissipated by R1, R2 (as heat)

è Total power delivered = Total power dissipated = 48 W

PHY2061: Chapter 31

R1 R2

Ea

E

Pin =Ei = 24× 2 = 48 W

P1 = i2R1 = 2

2 × 4 = 16 W

P2 = i2R2 = 2

2 ×8 = 32 W

i =E / RT = 24 / 4+8( ) = 2A

79

Heating from Internal Resistance

è E = 12V, r = 0.1Ω (same as done previously)

è Previously, we calculated the voltage across the terminals from voltage drop across r

è This time we explore power dissipation u Battery heats up from power dissipation by internal resistance

PHY2061: Chapter 31

CCaVbV r

R

i = E

r + R= 12

0.1+ R


Heating From Internal Resistance

è Heating of EMF source: P = i2r u Heating is extremely dependent on load

è E = 12V, r = 0.1Ω u R = 100Ω i = 0.12 A P = 0.0014 W u R = 10Ω i = 1.19 A P = 0.14 W u R = 1.0Ω i = 10.9 A P = 11.9 W u R = 0.5Ω i = 20 A P = 40 W

P = i2r = E

r + R⎛⎝⎜

⎞⎠⎟

2

r = 14.4

R + 0.1( )2


Resistors in series è In which case is each light bulb brighter?

u circuit A u circuit B u both the same

A B


Resistors in series è In which case is each light bulb brighter?

u circuit A u circuit B u both the same

A B

B has ½ the current of A. Power dissipated by each bulb in B is ¼ the power dissipated by the bulb in A (i2R).


Light Bulbs

è A three-way light bulb contains two filaments that can be connected to the 120 V either individually or in parallel. u A three-way light bulb can produce 50 W, 100 W or 150 W, at the usual

household voltage of 120 V. u What are the resistances of the filaments that can give the three wattages

quoted above?


Light Bulbs

è A three-way light bulb contains two filaments that can be connected to the 120 V either individually or in parallel. u A three-way light bulb can produce 50 W, 100 W or 150 W, at the usual

household voltage of 120 V. u What are the resistances of the filaments that can give the three wattages

quoted above?

Use P = V2 / R. Two filaments: Ø 50W R1 = 1202 / 50 = 288 Ω Ø 100W R2 = 1202 / 100 = 144 Ω


Problem

è What is the maximum number of 100 W light bulbs you can connect in parallel in a 100 V circuit before tripping a 20 A circuit breaker? u (a) 1 u (b) 5 u (c) 10 u (d) 20 u (e) 100


Problem

è What is the maximum number of 100 W light bulbs you can connect in parallel in a 100 V circuit before tripping a 20 A circuit breaker? u (a) 1 u (b) 5 u (c) 10 u (d) 20 u (e) 100

Each bulb draws a current of 1A. Thus only 20 bulbs are allowed before the circuit breaker is tripped.


Circuits

è Two light bulbs A and B are connected in series to a constant voltage source. When a wire is connected across B, what will happen to bulb A? u (a) burns more brightly than before u (b) burns as brightly as before u (c) burns more dimly than before u (d) goes out


Circuits

è Two light bulbs A and B are connected in series to a constant voltage source. When a wire is connected across B, what will happen to bulb A? u (a) burns more brightly than before u (b) burns as brightly as before u (c) burns more dimly than before u (d) goes out

The wire shunt effectively eliminates the second resistance, hence increasing the current in the circuit by 2x. The first bulb burns 4x brighter (i2R).


Circuits

è The light bulbs in the circuits below are identical. Which configuration produces more light? u (a) circuit I u (b) circuit II u (c) both the same


Circuits

è The light bulbs in the circuits below are identical. Which configuration produces more light? u (a) circuit I u (b) circuit II u (c) both the same

Circuit II has ½ current of each branch of circuit I, so each bulb is ¼ as bright. The total power in circuit I is thus 4x that of circuit II.


Circuits

è The three light bulbs in the circuit are identical. What is the brightness of bulb B compared to bulb A? u a) 4 times as much u b) twice as much u c) the same u d) half as much u e) 1/4 as much


Circuits

è The three light bulbs in the circuit are identical. What is the brightness of bulb B compared to bulb A? u a) 4 times as much u b) twice as much u c) the same u d) half as much u e) 1/4 as much

Use P = I2R. Thus 2x current in A means it is 4x brighter.


Light Bulb Problem

è Two light bulbs operate at 120 V, one with a power rating of 25W and the other with a power rating of 100W. Which one has the greater resistance? u (a) the one with 25 W u (b) the one with 100 W u (c) both have the same resistance


Light Bulb Problem

è Two light bulbs operate at 120 V, one with a power rating of 25W and the other with a power rating of 100W. Which one has the greater resistance? u (a) the one with 25 W u (b) the one with 100 W u (c) both have the same resistance

P = i2R = V2/R, so 100W bulb has ¼ the resistance of the 25W bulb.


Light Bulb Problem

è Which carries the greater current? u (a) the one with 25 W u (b) the one with 100 W u (c) both have the same current


Light Bulb Problem

è Which carries the greater current? u (a) the one with 25 W u (b) the one with 100 W u (c) both have the same current

P = i2R = V2/R, so 100W bulb has ¼ the resistance of the 25W bulb and carries 4x the current.

97

Dimmer

è Imagine a dimmer composed of a variable resistance put in series with a light. When you rotate the knob of the dimmer, what is being changed in the electric circuit? u (a) emf voltage u (b) resistance u (c) current u (d) both (a) and (b) u (e) both (b) and (c)

PHY2061: Chapter 31

98

Dimmer

è Imagine a dimmer composed of a variable resistance put in series with a light. When you rotate the knob of the dimmer, what is being changed in the electric circuit? u (a) emf voltage u (b) resistance u (c) current u (d) both (a) and (b) u (e) both (b) and (c)

Turning the knob increases the total resistance and thus lowers the current. Note: real dimmers are not made this way because of the dangerous amount of heat generated by the variable resistor.

PHY2061: Chapter 31


Power lines

è At large distances, the resistance of power lines becomes significant. To transmit maximum power, is it better to transmit high V, low I or high I, low V? u (a) high V, low I u (b) low V, high I u (c) makes no difference


Power lines

è At large distances, the resistance of power lines becomes significant. To transmit maximum power, is it better to transmit high V, low I or high I, low V? u (a) high V, low I u (b) low V, high I u (c) makes no difference

Power loss is i2R, so want to minimize current.

101

E

è Charging a capacitor takes time in a real circuit u Resistance allows only a certain amount of current to flow u … so current takes time to charge a capacitor

è Example: assume uncharged capacitor initially u Close switch at t = 0 u Initial current is (no charge on capacitor)

è Current flows, charging capacitor u Generates capacitor potential of q / C

è Current decreases continuously as capacitor charges! u Goes to 0 when fully charged

PHY2061: Chapter 31

RC Circuits

i =E / R

i = E − q / C

R

102

Analysis of RC Circuits vs Time (1)

è Current and charge are related

è So can recast previous equation as “differential equation“

dqdt

+ qRC

= ER

/i dq dt=

i ≡ dq

dt= E − q / C

R

PHY2061: Chapter 31


Analysis of RC Circuits vs Time (2)

è Solve for charge q and current i vs time

è τ = RC is the “characteristic time” of circuit u Circuit behavior depends only on t / RC

q = CE 1− e−t / RC( ) = qmax 1− e−t / RC( )i = dq

dt= E

Re−t / RC = imaxe

−t / RC

1

2


q t( ) = qmax 1− e−t / RC( ) i t( ) = imaxe−t / RC

Charging Capacitor (Capacitor Initially Uncharged)


RC Example

è Let E = 12 volts, C = 6µF, R = 100kΩ u τ = RC = 0.6 s

è Charge vs time u Using units of µC and sec

è Current vs time

u Using units of mA and sec

è This circuit will fully charge in a few sec u Initial values: i = 0.12 mA q = 0 µC u Final values: i = 0 q = 72 µC

q = CE 1− e−t / RC( ) = 72 1− e−t /0.6( )

i = E

Re−t / RC = 0.12e−t /0.6

106

Exponential Behavior of Charge & Current è τ = RC is the “characteristic time” of any RC circuit

u Only t / RC is meaningful

è t = 1.0τ u Current falls to 37% of maximum value u Charge rises to 63% of maximum value

è t =2.0τ u Current falls to 13.5% of maximum value u Charge rises to 86.5% of maximum value

è t =3.0τ u Current falls to 5% of maximum value u Charge rises to 95% of maximum value

è t =5.0τ u Current falls to 0.7% of maximum value u Charge rises to 99.3% of maximum value

PHY2061: Chapter 31


Discharging a Capacitor

è Connect fully charged capacitor (voltage V) to a resistor at t = 0

è Solve for current i

0qiRC

− − =

0dq qdt RC

+ =

q = CVe−t / RC = qmaxe

−t / RC

i = dq

dt= −V

Re−t / RC = −imaxe

−t / RC

R C

S Kirchhoff loop rule

Charge vs t

Current vs t (i < 0 because discharge is in opposite direction)


Discharging Capacitor (Capacitor Initially Charged)

q t( ) = qmaxe−t / RC

i t( ) = −imaxe−t / RC


Same RC Example

è C = 6µF, R = 100kΩ u τ = RC = 0.6 s

è Charge capacitor to 12 volts, then pull out battery and let the RC circuit discharge

è Charge vs time u Using units of µC and sec

è Current vs time

u Using units of mA and sec

è This circuit will fully discharge in a few sec

q = CVe−t / RC = 72e−t /0.6

i = V

Re−t / RC = 0.12e−t /0.6

110

Application: Camera Flash (1)

è Basic idea of a flash circuit u Position 1: Capacitor is charged

slowly (10 – 30 sec) by battery in series with a large resistance R1. Low current required to maintain battery lifetime.

u Position 2: Capacitor discharges rapidly (~1/1000 sec) through a small resistance R2 & flash.

è Next slide shows charge on capacitor during charging and discharging cycles u Note slow charge & rapid discharge

PHY2061: Chapter 31

+

1 2

R1

R2C

111

Application: Camera Flash (2)

è Example shows the basic idea u C = 12 F, emf = 6V (Qmax = 72 C) u R1 = 1Ω (RC = 12 sec) u R2 = 0.05Ω (RC = 0.6 sec)

è Charge and discharge cycles u Charge over 6 RC times (72 s) u Discharge quickly (~1-2 s) u (In reality, R2 is chosen to make

discharge time < 1 ms)

è Defibrillator uses similar idea u Slow charge, extremely fast charge

discharge to jumpstart the heart

PHY2061: Chapter 31

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