electricity circuits 2 simple circuit analysis

Electricity Circuits 2

Simple Circuit Analysis

Introductory Notes

Facts About Our Circuit Model1. Batteries have a positive and negative terminal. How do we label the

terminals of the battery in the circuit opposite?The long line is the positive and the short fat line is the negative.

2. Batteries will cause an electric current to flow around a circuit if it has a fully connected path to follow. What direction will an electric current flow in the circuit opposite? Electric current flows from positive to negative.

3. Electric current is the movement of charge around a circuit and represents the number of charges per second passing through each component in the circuit? Instead of talking about individual charges in an electric current we use a special unit called Coulombs . Why do we need this special unit and what does it mean? Because a typical current represents millions of millions of millions of charges per second it is much easier to talk about “groups” of charge called Coulombs and represent current as Coulombs per second.1 Coulomb is just over 6 million million million individual charges

4. Electric currents are measured in amps which represents the number of Coulombs per second passing through each part of the circuit.What does it mean if a current of 2 amps is measured in the circuit opposite?

There are 2 Coulombs of charge flowing through the battery and resistor every second.

12V+–

C = 2C/s2 A

Part A: Analysis Without Formulae

Part A: Analysis Without FormulaeQuestion 1

A 9.0V battery gives 9J of energy to each coulomb that passes through it

?

G1 G2

9V 0J 9J

Q2a Fill in the virtual energies and the ammeter and voltmeter readings in the circuit below.

4A 4 A?R1

Energy+ 0J 6J

0J

0J

0J

0J

0J

0J

0J 0J 0J

2J 2J

6J 6J 6J

6J

6J

6J

6J

6J 6JR2

2J

V

Different Resistances

4V V2V

V6V

A1 A2

V = voltage increase

= voltage dropV

Q2b Fill in the virtual energies and the ammeter and voltmeter readings in the circuit below.

3A 3 A?R1

Energy+ 0J 8J

0J

0J

0J

0J

0J

0J

0J 0J 0J

3J 3J

8J 8J 8J

8J

8J

8J

8J

8J 8JR2

3J

V

Different Resistances

5V V3V

V8V

A1 A2


= voltage dropV


State the potential difference from the battery configurations below:

Vout = 12V

Vout = 6V

?

?

Q4 Fill in the virtual energies and the voltmeter readings in the circuit below.

V12V

G2 G3

9V

6J 0J

18J

?V18V

0J 9J 9J

9V

V6V

18J 12J

G1THESE ARE identical Globes?V0V


= voltage dropV

?V0V

Part A: Analysis Without Formulae5. The unit used to measure electric charge in circuits

is called the Coulomb

6. A current of 3.0A through a battery means that 3.0 C of charge passes through the battery each second

7. If 2.0A flows through a resistor for 5.0 seconds, then 10 C of charge would have passed through the resistor

8. If 18C of charge passed through a resistor in 3.0seconds then the current through the resistor is 6.0 A

?????

??

??

??


State whether the globes in the circuits below are in series or in parallel.

Series Parallel Parallel


In the circuit below the currents in ammeter 2 and ammeter 4 are A2=2.0A and A4 =5.0A.

(a) A3 = 2.0 A(b) A1 = 3.0 A

2A

5A

3A

??

??


In the circuit below the potential difference across B and C (VBC) = 4.0V.

a) Each coulomb passing through the battery drains 15 J of energy from it.

b) Each coulomb passing through R1 gives off 4.0 J of energy.

c) VAB = 0 Vd) VCD = 11 V

??

??

??

??

V4V V11V

V0V


If the variable resistance below is increased

a) The reading on V1 will increase.

b) The reading on V2 will decrease.

??

??

Question 13(a) When only switch 1 is closed

globes 1 and 2 would be on.

??

(b) If switch 1 is shut and then switch 2 is closed globe 1 will go out and globe 2 will get brighter.

?? ??

Short circuiting occurs when a very low resistance path is put across a component. In such circumstances the current is diverted away from the component that has been short circuited. Because the short circuited components have essentially been removed from the circuit the overall resistance will have decreased and hence the current will increase.

Part A: Analysis Without FormulaeQuestion 14For the Circuit:

(a) When the switch is open an observer would notice that all the globes would be on.

(b) When the switch is shut the changes observed would be that.

globe 1 will get brighter globe 2 will go outglobe 3 will go out

??

??

??

??


(a) Use a red pen to show how to connect a wire to make globes 3 and 4 go out.

(b) When this wire is connected globes 1, 2 and 5 will get brighter??

Question 16The symbol in the diagrams below represents a fuse which is designed to blow out if there is too much current flowing through it.

Circuit 1 Circuit 2 Circuit 3

YX

A

YX

A

YX

A

(a) Which circuit(s) could damage the ammeter if there was a short circuit from X to Y.

With the resistance out of the circuit there will be a high current and the ammeter could be damaged.

No current will flow through the ammeter it is safe.

With the resistance out of the circuit there will be a high current but the fuse will blow protecting the ammeter.

Circuit 1

Question 16The symbol in the diagrams below represents a fuse which is designed to blow out if there is too much current flowing through it.

Circuit 1 Circuit 2 Circuit 3

YX

A

YX

A

YX

A

(b) Which circuit will protect the ammeter by having the fuse blow out if there was a short circuit from X to Y.

With the resistance out of the circuit there will be a high current and the ammeter could be damaged.

No current will flow through the ammeter it is safe.

With the resistance out of the circuit there will be a high current but the fuse will blow protecting the ammeter.

Circuit 3


The heater in an electric hot water service has two switch sensors, one which goes on when there is enough water in the heater and one which goes on when the water is below a certain temperature. The heater only goes on when both switches are on. Using a battery, two switches and a light, design a model of a hot water circuit that turns on a light only when both switches are on.

S1

S2 OFFON

Part A: Analysis Without FormulaeQuestion 18The alarm in a burglar system has two switch sensors, and when either switch detects an intruder the alarm goes off. Using a battery, two switches and a light, design a model of a burglar alarm that turns a light on when either of the switches is turned on.

S1 S2

OFFON

Part B: Ohm’s Law

Ohm’s Law1. In 1827 Georg Ohm published his experimental results that the

electric current (I) in a circuit depends on the Voltage (V) of a battery and the Electrical Resistance (R) in a Circuit? What rule did he find for electric current and what name is it given?The rule is I and it is called Ohm’s Law.

2. What is the meaning and units of the symbols in Ohm’s Law?

?? ??

Symbol Meaning Unit

V = Voltage or Potential Difference= Difference in coulomb energies

Volts (V)

I = Electric Current= Coulombs Per Second Flowing Through A Point

Amps (A)

R = Electrical Resistance = opposition to electrical current flow

Ohms()

??

??

??

??

??

??

Ohm’s Law3. What electrical current will flow in the circuit?

4. What does the current in the previous question actually mean?There are 6 Coulombs of charge flowing through the circuit every second

5. Work out in your head the current that would flow if another

2.0 resistor was added to the circuit in question 3 as shown above?

using I in your head I = 3Amps

I = ? V = 12V R = 2.0I I I = 6.0A

12V 2Ω+–

2Ω

6.0A

C

6 C/s

3.0A

3 C/s

Ohm’s Law6. Ohm’s Law can be written in three forms and each rule can be

worked out from the triangle below. What are the 3 rules?

V= IRI R

V I R

Ohm’s Law

Part C: A Quirk In History

A Quirk In History1. How many coulombs of charge are flowing through

the battery every second in the circuit opposite ?Current = 2.0A so 2 coulombs per second

2. What particles are REALLY being driven around the circuit by the battery?electrons

3. What does a Coulomb represent when talking about current in an electric wire?

1 Coulomb = 6.25 1018 electrons.4. What direction are the electrons flowing?

Since electrons have a negative charge they actually flow from negative to positive. THE OPPOSITE DIRECTION TO THE CURRENT!!!!!

5. Why is there a contradiction between the direction of the current and the actual direction of charge flow?

18V 7Ω+–

I = ? V = 18V R = 9.0I I I = 2.0A

VI R

2Ω

2.0A

C

2 C/s

Ohm’s Law

?

Physicists wrote the direction of the current into circuit theory before they knew the existence of electrons. By that time the current direction was too entrenched in theory.

e–

e–

Part D:Working With Ohm’s Law

Ohm’s Law & The Power FormulaeV = Voltage or Potential Difference (measured in Volts-

V) = J/C gained by a Coulomb passing through a battery OR J/C lost from a Coulomb passing through a

resistor or device

I = Electrical Current (measured in Amps- A) = C/s passing through a device

R = Resistance (measured in Ohms- ) = opposition to electrical current flow

P = Power (measured in Watts- W) = J/s dissapated in a device = J/s drain on a battery

VI R

P = VI = I2R = =

V = IR

I =

R =

Working With Ohm’s LawQuestion 1

(a) What will be the electrical potential difference of the battery in the circuit below? V = ? I = 0.20A R = 10

V = IRV = 0.2 10V = 2.0V

VI R

P = VI = I2R = =


(b) What will be the power dissapated in the resistor below?

P = ? I = 0.20A R = 10P = I2RP = 0.22 10P = 0.40 W

VI R

P = VI = I2R = =


(a) What will be the electrical resistance of the resistor in the circuit below?

VI R

P = VI = I2R = =

R = ? V = 9.0V I = 0.10A R RR = 90


(b) What will be power drain on the battery below?

VI R

P = VI = I2R = =

P = ? V = 9.0V I = 0.10A P = VI P = 9 0.1R = 0.90W


(a) What will be the electrical current flowing through the circuit below?

I = ? V = 12V R = 20

I = 0.60A

VI R

P = VI = I2R = =


(b) What will be the power dissipated in the resistor below?

P = ? V = 12V R = 20

P P P = 7.2W

VI R

P = VI = I2R = =


For the circuit.a) Total resistance =b) Work out the total

current

I = ? V = 12V R = 150

I = 0.080A

VI R

P = VI = I2R = =

150


For the circuit.c) Work out V50

V = ? I = 0.080A R = 50V = IRV = 0.08 50V = 4.0V V

I RP = VI = I2R = =

0.080A

electricity circuits 2 simple circuit analysis

Documents

j 12jg1these

j 9vv6v

j 6jr2

j 8jr2

j 9jq2a

j of energy

circuit opposite

circuit modelbatteries