This week. 10/1/2013 Physics 214 Fall

This week

Electrical Circuits

Series or parallel that’s the question.

Current, Power and Energy

Why does my laptop battery die?

Transmission of power to your home

Why do we have big transmission towers?

Household Appliances

Why do fuses blow?

Electrical power and energy

In order to separate charge we have to do work and energy is stored. The simplest example is a battery. The quantity we call voltage is related to the energy stored

∆V = ∆PE/q (joules/coulomb = volt) When positive charge moves to the

negatively charged plate or vice versa then the stored energy is released. In order for this to happen the two sides of the battery must be joined by a conductor so that the charge can move easily. This is a simple circuit.

Electrical circuits

All circuits are basically the same.

There is an external source of energy which produces a voltage.

In a charged battery there is separation of charge. When the

circuit is made positive charge flows from high to low voltage or negative charge flows from low to high

voltage releasing the stored energy.

Normally it is electrons which flow.

As they move they “collide” with the atoms of the wire and lose some of their energy in the form of heat.

There is resistance to the flow.

Current and resistance

The rate of flow of charge determines how much energy is released per unit time

I = q/t amperes (coulombs/sec)

The direction of I is the flow of positive charge or opposite to the flow of negative charge.

OHM’s Law R = ∆V/I ohms

Every part of a circuit has resistance including an internal resistance in the battery. The higher the resistance the lower is the current for a given voltage difference

Electromotive force

The electromotive force of a battery ,

ε

, is the voltage difference between the two terminals when no current is

being drawn.

When it is connected to a simple circuit

I = ε ^/(R

^{+ R}

⁾

A voltage difference is the energy

stored or the energy released per unit charge. So if charge +q goes from

high to low voltage then the energy released is q

ε

Series circuit

If we add more light bulbs in the circuit in series the total resistance increases and the current will be reduced. The current is the same in all parts of the circuit and the voltage difference across each light is the same

ε ^{= I(R}

+ R + R + R)

The voltage difference across any light = IR

Parallel circuit

In the circuit shown the voltage difference across each light is the same and the total current is the sum of the three currents

I = I₁ + I₂ + I₃ and since a current = ∆V /R

∆V_AB /R_circuit = ∆V_AB /R₁ + ∆V_AB /R₂ + ∆V_AB /R₃ and 1/R_circuit = 1/R₁ + 1/R₂ + 1/R₃

That is the total resistance of the circuit is smaller than any of the single resistances. It is also true that if one bulb fails the other two will stay lit at the same brightness.

The current q at A

divides into three and then recombines at B

What is voltage?

We have seen that the definition of voltage is

∆V = ∆PE/q when a charge q is moved in an electrical force field.

So energy is stored as potential energy as a positive charge is moved in the opposite direction to E or a negative charge is moved in the same direction as E.

If we move a positive charge toward a positive charge potential energy and ∆V increase or if we move a negative charge away from a positive charge.

Just as in the gravitational field there are only differences in PE.

So normally we use the term ∆V. But very often for circuits we choose one point, usually the negative terminal, to be zero and then instead of ∆V we just use V.

When charge is free to move, that is positive charge moving to a lower voltage or negative charge moving to a higher voltage the PE will transform into KE just like dropping something. In a simple circuit with resistance this KE is turned into heat and light so

there is a voltage drop across every element in the circuit.

Voltage drop

If we have a circuit with many different resistors then there is a voltage drop across each resistor and there is also a voltage drop for the whole circuit.

Current only flows if there is a voltage difference and in a time t charge q

passes through the resistor.

I = q/t and ∆V = IR Case1

I = 6/60 = 1/10

∆V₁₅ = 15/10 ∆V₂₀ = 20/10 ∆V₂₅ = 25/10 Case 2

I = 12/8 I₂₄ = 12/24 Case 3

I = V_AB/5.5 ∆V₃₁ = 3I/2 ∆V₃ = 3I ∆V₃₃ = 3I/3

Series plus parallel

R₁ R₂ R₃

A B

I

I

R₄

V

V

= I

R

V

= I

R

V

= I

R

I = I

+ I

+ I

A

B

R_parallel R₄

B V

V

= IR

V = V

+ V

V = I(R

+ R

)

Measuring current and Voltage

It is often very important to know the current in a circuit or the

voltage difference between two points.

A hand held meter is very useful to test batteries or a circuit.

An ammeter is a device inserted into a circuit. The resistance of an ammeter is very small so as to

minimize the effect on the circuit.

A voltmeter is attached in parallel and V is found by measuring the current and V = I_meterR_meter. The resistance has to be much larger than the circuit resistance so that the current is very small and does not disturb the main circuit.

Both meters actually

measure current

Power

Electromotive force or voltage difference between two points is the difference in potential energy/unit charge.

So the energy delivered if charge q is transferred is

energy = ∆Vq

power = ∆Vq/t = ∆VI watts

For any voltage difference ∆V and current I.

power = ∆VI

For circuits that obey OHM’s law

∆V = I R

P = ∆VI = I

R = ∆V

/R watts

The power used appears as heat or light

Practical Unit

1 kilowatt

Energy Unit 1 kilowatt hour = 1000 x 3600 joules

Summary Chapter 13

I = q/t amperes (coulombs/sec)

OHM’s Law R = ∆V/I ohms

I = ε ^/(R

^{+ R}

⁾

P = ∆VI = I

R = (∆V)

/R watts

Transmission

In the distribution of electric power the goal is to deliver to the user as large a fraction as possible of the generated power.

Practical cables have a specific resistance so the power losses

will be I²R_cable and we need I to be as small as possible.

But we also need the delivered power

P = V

I

to be as high as possible, therefore, the electrical power is distributed at very high voltage and low current.

The voltage is reduced from 250,000volts to 220volts for

households by using a transformer.

The current increases by the same

factor since for an ideal transformer no power is lost. Transformers are the

dominant reason electrical

transmission is alternating current

V_source

R_cable i

i

R_user V_user

V_user = V_source – IR_cable

P_user = iV_user = iV_source – i²R_cable

Household appliances

Household circuits are wired in parallel so that when more than one appliance is plugged in each sees the same voltage and can get the required current.

As we plug in more and more appliances the current in the circuit increases and the I²R losses could cause a fire. This is why we have fuses and why major appliances use 220 volts and many parts of the world use 220 volts for all household use.

As many people turn on appliances (air conditioners) the grid has to supply more power by increasing the current.

P

= iV

= iV

– i

R

This results in a higher fraction of the power being lost in the cable

In cases of very heavy load the power station reduces the

transmission voltage resulting in a “brown out” and in extreme cases there are rolling blackouts.

Mini Review

∆V

I = q/t In order for a current to flow there must

be a voltage difference. The voltage difference is the amount of potential energy/unit charge. As the charge, usually electrons move along a conductor they lose energy

continuously so the voltage along the conductor decreases continuously. For a simple circuit we ignore losses in the conductor but there is a voltage drop across every resistor.

Energy liberated is qV. So the power is qV/t = IV For a resistor V = IR and power = I²R = VI = V²/R

Simple circuits

The current is the same everywhere R = R₁ +R₂ + R₃

V = V₁ + V₂ +V₃

V

V is the same across all resistors V = IR_circuit

1/R_circuit = 1/R₁ + 1/R₂ + 1/R₃ I₁ = V/R₁ I₂ = V/R₂ I₃ = V/R₃

R

R₂

R = R₁ + R₂ + R₃ V_ab = IR

Questions Chapter 13

Q3

Q4

The current is the same

No. electric current is the flow of charge I = q/t

Q6

Є

A

B

•

•

No. The voltage drop from A to B is stiil the same so the current through the bulb does not change.

Q7

Open Switch

1.5 V 1.5 V

(a) ^(b)

In a) the battery is not connected. In b) the bulb is lit

Q11

voltmeter is connected across the terminals. Can the battery still be used to light a bulb?

Q12

A battery has internal resistance so although one can measure an open circuit voltage when connected to a circuit the voltage will drop and the current flow will be very low.

The voltage will be less because of the voltage drop due to the internal resistance

Q13

A. Which of the two resistors, if either, has the greater current flowing through it? Explain.

B. Which of the two resistors, if either, has the greatest voltage difference across it? Explain.

Є

R₁

R₂

The current is the same in

both. Since V = IR the greatest voltage drop will be across R₂

Q14

I₃ = I₁ + I₂ so I₃ is the largest Є

R₃

R₁ R₂

Q15

Q16

V = IR and both I and R are the largest for R₃

Є

R₃

R₁ R₂ The resistance of the

circuit will increase so the current

through R₃ will decrease.

Q18

voltmeter. Which of the following statements is correct? Explain.

A. The voltmeter is in the correct position for measuring the voltage difference across R.

B. No current will flow through the meter, so it will have no effect.

C. The meter will draw a large current.

Є R

+ +

V

The voltmeter is in the correct position. Current will flow

through the meter but the current will be very small because the resistance is very high.

Q19

ammeter. Which of these statements is correct? Comment on each.

A. The meter is in the correct position for measuring the current through R.

B. No current will flow through the meter, so it will have no effect.

C. The meter will draw a significant current from the battery.

Є R

+ +

A

The meter is in the wrong position. A large current will flow because the meter resistance is very low.

Q21

Q22

Q23

Power is the rate at which energy is used. Your electrical bill is for the total energy you use.

P = I²R so the power increases by a factor of 4

Yes because increasing the resistance lowers the current

Q29

First if one failed then all the appliances will not function. Secondly the current will be determined by the whole string of appliances by the

voltage drop from end to end and will depend on the number of

appliances. Each appliance requires the same voltage drop and this is what happens when they are connected in parallel.

Ch 13 E 14

24  resistor has voltage difference 3V across leads.

a) What is the current through the resistor?

b) What is the power dissipated in resistor?

a) V = IR

I = V/R = 3/24 = 0.125A

b) P = IV = V

/R = (3)

/24 = 0.375W

3V 24 

Ch 13 E 16

A toaster draws current = 7A on a 110-V AC line

a) What is the power consumption of the toaster?

b) What is the resistance of the heating element

in the toaster?

110 V R

→ I = 7A

a) P = IV = 7.110 = 770 W

b) V = IR, R = V/I = 110/7 = 15.7 

Ch 13 CP 2

Three 30- light bulbs connected in PARALLEL to 1.5

V battery with negligible internal resistance.

a) What is the current through the battery?

b) What is the current through each bulb?

c) If one bulb burns out, does the brightness of the

other bulbs change?

1.5 V R R R R = 30 

Ch 13 CP2 cont.

a) 1/R_p = 1/R₁ + 1/R₂ + 1/R₃ = 1/30 + 1/30 + 1/30 = 1/10 R_p = 10 

V = I_tR_p , I_t = V/R_p = 1.5/10 = 0.15 A

b) V = IR, I = V/R = 1.5/30 = 0.05 A

Notice that total current, I_t, through the battery is the sum of currents through each bulb.

I_t = 3(I) = 0.15 A

c) Brightness of remaining two bulbs do not change. Instead, load on the battery is reduced. Each remaining bulb still feels 1.5 V.

So, each remaining bulb still draws 0.05 A of current. Since, P=IV, each remaining bulb still outputs same power. This is benefit of hooking circuits in parallel.

CH 13 CP 4

A R

R R

R R

R B R = 3

a) What is the resistance of each parallel combination?

b) What is the total resistance between A and B?

c) 6V voltage difference b/w A and B. What is the current through the entire circuit?

d) What is the current through each resistor in three-resistor parallel combination?

a) Two-resistor parallel combination

I/R_P2 = 1/R + 1/R = 1/3 + 1/3 = 2/3, R_P2 = 3/2  b) Three resistor parallel combination

I/R_P3 = 1/R + 1/R + 1/R = 1/3 + 1/3 + 1/3 = 1, R_P3 = 1 

CH 13 CP 4 cont

c) V = I R_T, I = V/R_T = 6/(11/2) = 12/11 = 1.09 A

A R

R R

R R

R B R = 3

C

A

9/2  C 1 

B A

3/2  3  C 1 

B

CH 13 CP 4

After accounting for AC part of circuit we have left

V=1.09 V R R

B C

R

This answer could have been easily noted as follows: We calculated 1.09 A flowing from A to B. That means, at point C, there is 1.09 A. Now this

current must make it to point B, but there are 3 different paths. Since each path is of equal resistance, the current will equally choose all three paths.

thus any one path has 1.09/3 A = 0.37 A of current d) V_AC = I R_AC = (1.09) (9/2) = 4.91 V

V_CB = V – V_AC = 1.09 V I = V/R = 1.09/3 = 0.37 A