Physics - D.C. Circuits

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D.C. CircuitsIn this section, we will learn to

draw circuit diagrams,

calculate the current, potential difference and resistance in a series and parallel circuit.

18.1Electric Circuits

Lamp

d.c. supply

Switch

1

23

4Conductors

18.1 Electric Circuits1. An electric circuit is a closed path in

which electric charges can flow from one terminal to another.

2. In every electric circuit, the following must be present:

• A e.m.f. source that drives charge and produces a current, e.g a battery.

18.1 Electric Circuits• Components that “feed” on current and do a job, e.g. a bell that makes a sound.

• Conductors that join the source and various components together, e.g. copper wires.

• Switches that break or complete a circuit.

18.2 Circuit DiagramsA circuit diagram is used to represent an electric circuit. Symbols are used to represent the devices or components used in electric circuits.

(Ref. textbook pg 321)

switch

18.2 Circuit Diagrams

resistor (fixed)

18.2 Circuit Diagrams

variable resistorrheostat

18.2 Circuit Diagrams

fuse

18.2 Circuit Diagrams

ammeter

A18.2 Circuit Diagrams

voltmeter

V18.2 Circuit Diagrams

galvanometerG

1. A galvanometer is used to measure small currents while an ammeter is used to measure greater currents.

2. A galvanometer must be connected in series to a circuit.

18.2 Circuit Diagrams

lamp

18.2 Circuit Diagrams

cell

18.2 Circuit Diagrams

bell/buzzer

18.2 Circuit Diagrams

ground/earth

18.2 Circuit Diagrams

power supply

240V240V

18.2 Circuit Diagrams

wires joinedat a point

18.2 Circuit Diagrams

wires crossed over each other but not joined

18.2 Circuit Diagrams

18.3 Series Circuit1. In a series circuit, there is ONLY

ONE path in which charge can flow.

AResistor

Lamp

Ammeter

CellI I

II

18.3 Series Circuit2. If there is a break anywhere in

the circuit, current will stop flowing.

AResistor Ammeter

Cell

No Current Flow

18.3 Series Circuit Let’s practice:

Draw a circuit diagram containing a d.c supply, a lamp, an ammeter, a switch and a variable resistor all connected in series. (J92 P2)

18.3 Series CircuitAnswer:

A

Lamp

Ammeter

d.c. supply

Switch

Variable resistor

18.4 Parallel Circuit1. In a parallel circuit, there is MORE

THAN ONE path in which charge can flow.

Lamps

Cell

AB

Current splits up at junction A and join together again at B

18.4 Parallel Circuit2. Even if one of the lamps is

removed, there will still be a flow of current in the other lamp.

Lamp

Cell

18.4 Parallel Circuit Let’s practice:

Draw a circuit diagram containing a d.c supply, two lamps connected in parallel and a switch that can turn off only one of the lamps.

18.4 Parallel CircuitAnswer:

Lamps

d.c. supply

Series + Parallel Circuit

AV

18.5 Current In Series Circuit

1. In a series circuit, the current through every component is the SAME.

A

I1 I4

I3I2

I1 = I2 = I3 = I4

Example 1

12 V

3 A

Q1: What is the name given to the device X?

X

A1: Lamp

Q2: The current passing through X is 3A, what is the current passing through the resistor and the cell?

A2: 3A

Recall: All the resistors are connected in series, the current passing through each resistor is the same.

R1 R2 R3 R4

V1 V2 V3 V4

I I

18.5 P.D In Series Circuit

18.5 P.D In Series Circuit

2. Let V be the potential difference across the combination of the 4 resistors in series.

V = V1 + V2 + V3 + V4

R1 R2 R3 R4

V1 V2 V3 V4

I I

V

18.5 P.D In Series Circuit

Points to note:

3. V1 = V2 = V3 = V4 if only the resistors are identical i.e. having the same resistance.

4. In a series circuit, the component that has the LARGEST resistance will have the LARGEST potential difference across it.

Example 2

12 V

Q1: What is the total p.d across the resistor and the lamp? A1: P.d across resistor and lamp

= e.m.f of cell, E = 12 V

Q2: If the p.d across the resistor is 9 V, what is the p.d across the lamp?A2: P.d across lamp, V = 12 V - 9 V = 3 V

18.5 Resistance In Series Circuit

Let R be the combined/total/effective resistance of the 4 resistors.

R1 R2 R3 R4

V1 V2 V3 V4

I I

V

18.5 Resistance In Series Circuit

How is R related to R1, R2, R3 and R4?

RI I

V

The arrangement can then be simplified from this …

R1 R2 R3 R4

V1 V2 V3 V4

I I

V

to…

18.5 Resistance In Series Circuit

Using equation V = IR, we have…

V1 = IR1 V2 = IR2 V3 = IR3 V4 = IR4

and… V = IR

R1 R2 R3 R4

V1 V2 V3 V4

I I

V

RI I

V

18.5 Resistance In Series Circuit

5. V = V1 + V2 + V3 + V4

IR = IR1 + IR2 + IR3 + IR4

R = R1 + R2 + R3 + R4

R1 R2 R3 R4

V1 V2 V3 V4

I I

V

RI I

V

18.5 Resistance In Series Circuit

R = R1 + R2 + R3 + R46. For resistors in series, the

combined /total/effective resistance is the SUM of the individual resistances.

7. The effective resistance is GREATER than any of the individual resistance.

Example 3

12 V

1) Calculate the effective resistance of the circuit.

2) What is the current passing through the resistor?

0.5 3 2.5

R = 0.5 + 3 + 2.5 = 6

I = V/R = 12/6 = 2 A

Examples on Series CircuitExample 4:

Find the combined resistance of the three resistors in series given that R1 = 1 , R2 unknown and R3 = 2 The current I recorded in the ammeter is 1 A and the voltmeter reading V across R2 = 3 V.

I

R2

A

R3 R1

V

R2 = V / I

= 3 / 1 = 3

Combined resistance = R1 + R2 + R3

= 1 + 3 + 2 = 6

I = 1 A

R2

A

R3 (2) R1 (1)

V3 V

Examples on Series CircuitExample 5:

A cell of e.m.f. 1.5 V was connected in series with two resistors, as shown below

1.5 V

6 4

Calculate

i) the effective resistance of the circuit,

ii) the current flowing in the circuit,

iii) the potential difference across the 4 resistor.

1.5 V

6 4

i) Effective resistance, R = R1 + R2 = 4 + 6 = 10

ii) Current flowing in circuit, I = V / R = 1.5 / 10

= 0.15 A

1.5 V

6 4

iii) P.d across 4 resistor,

V4 = IR

= 0.15 4

= 0.6 V

1.5 V

6 4 0.15 A

WAIT! That is not all….

What is the p.d across the 6 resistor?

18.6 Current In Parallel Circuit

In a parallel circuit, the current from the source is shared by 2 or more branches.

Lamp 1 Current from source is shared by lamp 1 and 2

Lamp 2 A B

Current In Parallel Circuit

I1 = I2 + I3 = I4

Lamp 1

Lamp 2

I1

I2

I3

I4

Current In Parallel Circuit

1. The SUM of the currents in the separate branches of a parallel circuit is EQUAL to the current from the source.

I1 = I2 + I3 = I4

Current In Parallel CircuitPoints to note:

2. In a parallel circuit, the component with the SMALLEST resistance will allow the LARGEST current to pass.3. I1 = I2 if only the lamps are identical i.e. having the same resistance.

Example 8

Which ammeter shows a faulty reading?

4A

2A

2A

4A

2A

PS

R

Q

T

P.D In Parallel Circuit

V3

V2

V1

4. Each component joined in parallel have the same potential difference across it.

V1 = V2 = V3

Example 9

Fill in the blanks using = , or <.

V1 _____V2 I1 ______I2

I1

I2

10

5

V1

V2

Resistance In Parallel Circuit

I = I1 + I2

V = V1 = V2

Let R be the combined/total/ effective resistance of the 2 resistors.

R1

R2

V2

V1

V

I

I1

I2

Resistance In Parallel Circuit

The arrangement can then be simplified from this … to…

R

V

V

I

I R1

R2

V2

V1

V

I

I1

I2How is R related to R1 and R2?

Resistance In Parallel Circuit

4. I = I1 + I2

R1

R2

V2

V1

V

II1

I2

R

V

V

I

I

2

2

1

1

R

V

R

V

R

V

Resistance In Parallel Circuit

21

111

RRR

5. For resistors in parallel, the reciprocal of the combined /total/effective resistance is the SUM of the reciprocal of individual resistances.

6. The effective resistance is SMALLER than any of the individual resistance.

Example 10 What is the effective resistance of the three resistors?

Let effective resistance be R

1/R = 1/3 +1/6 + 1/6

= 2/3

R = 3/2 = 1.5

3

6

6

Examples on Parallel Circuit

3.0

6.0

4.0 V

The circuit diagram shows a 6.0 resistor and a 3.0 resistor in parallel and connected to a 4.0 V battery.

Example 11:

3.0

6.0

4.0 V

1) Calculate the effective resistance of the parallel resistors.

1/R = 1/3.0 + 1/6.0

= 1/2.0

R = 2.0

3.0

6.0

4.0 V

Effective resistance = 2.0

2) Calculate the current flowing through the battery.

Ibattery = V/R = 4.0/2.0

= 2.0 A

3.0

6.0

4.0 V

3) Calculate the current flowing through in each resistor.

P.d across the resistors = 4.0 V

I3.0 = V/R = 4.0/3.0 = 1.3 A

I6.0 = Ibattery - 1.3 = 2.0 – 1.3 = 0.7 A

3.0

6.0

4.0 V

4) A third resistor is connected in parallel with the original pair. Is the current through the battery larger, smaller or the same as before. Explain. Current will be larger because the effective resistance of the circuit is smaller than before.

Examples on Parallel Circuit

Example 12:

Textbook pg 328 Example 18.3

Series + Parallel Circuit

AR1 R2

R3

R2 is connected in parallel with R3.

R1 is connected in series with the ammeter and the resistor combination of R2 and R3.

Resistance In Series + Parallel Circuit

Example 13:

What is the effective resistance of the circuit below?

5

3

3

Resistance In Series + Parallel Circuit

First, add the resistors in parallel.

Let R1 be the effective resistance of the resistors in parallel.

1/R1 = 1/3 + 1/3

R1 = 1.5 5

3

3

Resistance In Series + Parallel Circuit

This is then reduced to two resistors in series.

5

3

3

5 R1 = 1.5

Resistance In Series + Parallel Circuit

Effective resistance of circuit,

R = 5 +1.5

= 6.5

5 R1 = 1.5

Examples on Series + Parallel Circuit

Example 14:

The diagram shows three resistors connected to a 6.0 V battery supply

A6.0

6.0

8.0 6.0 V

1) Calculate the combined resistance of the 8.0 and the 6.0 resistors in series. Let combined resistance be Rc

Rc = 8.0 + 6.0 = 14.0

A6.0

6.0

8.0 6.0 V

2) Calculate the effective resistance of the circuit.

Let effective resistance of circuit be R

1/R = 1/14.0 + 1/6.0 = 5/21

R = 21/5 = 4.2

A6.0

6.0

8.0 6.0 V

3) Calculate the current through the ammeter. Effective resistance of circuit = 4.2

Current through ammeter = Current in circuit = V/R = 6.0/4.2

= 1.4 A

A6.0

6.0

8.0 6.0 V

Examples on Series + Parallel Circuit

Example 15:

Textbook pg 329 Example 18.4

Examples on Series + Parallel Circuit

Example 16:

Textbook pg 334 Q3

Example 17:

Textbook pg 335 Q4

Short Circuit

Short Circuit occurs in a closed circuit when there is an alternative path for current to flow. This alternative path is of a much lower resistance than the original path.

Short Circuit

The lamp will not light up because, compared to the copper wire, it offers more resistance and thus current by-passed it.

ALamp

Ammeter

CellI I

II Copper wire

Short Circuit

During a short circuit, the current flowing is very large as the resistance is very small. This can be dangerous as a large current may cause heating resulting in a fire.

Example 18:

Textbook pg 334 Q2

Examples on Series + Parallel Circuit