7.5.1 Electrical Energy

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  1. From the definition of potential difference, the electrical work done is given by the equation W = QV, where
    W = work
    Q = charge
    V = potential difference
  1. Since the work done must be equal to the energy to do the work, therefore we can also say that, the electrical energy ( E ) is also given by the formula

Example
Given that the potential difference across a bulb is 240V and the current that flow through the bulb is 0.25A. Find the energy dissipated in the bulb in 30s.

Answer:
Formula of current,
E = QV
hence
Q = It

Energy dissipated,

E = QV
E = (It)Q
E = (0.25)(30)(240)
E = 1800V

 

7.4.2 Internal Resistance

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The internal resistance of a source (cell or generator) is the resistance against the moving charge in the source.

Load Resistance

The load resistance in a circuit is the effective resistance against the moving charge outside the source of electric.

Terminal Potential Difference

Terminal potential difference or terminal voltage is the potential difference across the two terminal (the positive terminal and the negative terminal) of an electric source (cell or generator).

Internal Resistance and Potential Difference Drop

  1. If the internal resistance is ignored, the terminal potential difference is equal to the e.m.f.
  2. If the internal resistance is present, the terminal potential difference will be lower than the e.m.f.
  3. The relationship between e.m.f. and the terminal potential difference is given by the following equation.
    Equation
    E = V + Ir
    or
    E = IR + Ir

    E = e.m.f.
    V = terminal potential difference
    I = current flows in the circuit
    r = internal resistance
    R = the load resistance

Example 1:
A cell has internal resistance 0.5Ω and the potential difference across the cell is 4V when a 2A current flows through it. Find the e.m.f. of the cell.

Answer:
r = 0.5Ω
V = 4V
I = 2A
E = ?

E = V + Ir
E = (4) + (2)(0.5)
E = 5V

Example 2:

A cell with e.m.f. 3V and internal resistance, 1Ω is connected to a 5Ω resistor, and a voltmeter is connected across the resistor as shown in the diagram on the left. Find the reading of the voltmeter.

Answer:
E = 3V
r = 1Ω
R = 5Ω
V = ?

E = I(R + r)
(3) = I(5 + 1)
3 = 6I
I = 3/6 = 0.5A

V = IR
V = (0.5)(5)
V = 2.5V

Measuring e.m.f. and Internal Resistance



Three methods can be used to measure the e.m.f. and internal resistance.
  1. Open circuit-Close circuit
  2. Simultaneous equation
  3. Linear Graph
Open Circuit
In open circuit ( when the switch is off), the voltmeter shows the reading of the e.m.f.

Close Circuit
In close circuit ( when the switch is on), the voltmeter shows the reading of the potential difference across the cell.

With the presence of internal resistance, the potential difference across the cell is always less than the e.m.f..

Example 1:


The diagram above shows a simple circuit that connect some baterries to a resistor. The voltmeter shows a reading of 5.0V when the switch is off and 4.5V when the switch is on. What are the e.m.f. and the internal resistance of the cell?

Answer:
When the switch is off, the reading of the voltmeter shows the e.m.f. of the batteries. Therefore.
e.m.f. = 5.0V

When the switch in turned on, the reading of the voltmeter shows the potential difference of the resistor. Therefore,
V = 4.5V

The current that pass through the resistor,

E = V + Ir
(5.0) = (4.5) + I(0.5)
0.5I = 5.0 - 4.5
I = 0.5/0.5
I = 1A


Example 2:
Diagram (a)
Diagram (b)


A cell is connected to a circuit as shown in diagram (a). The graph in diagram (b) shows the change of the reading of the voltmeter, V against time, t. If t is the time where the switch is close, find
(a) the e.m.f. of the cell
(b) the internal resistance of the cell.

Answer:
(a) Before the switch turned on, the reading of the ammeter shows the e.m.f. of the cells.

From the graph, the e.m.f. = 3.0V

(b)
e.m.f., E = 3.0V
Potential difference across the resistor, V = 2.5V

Current that pass through the resistor,


 

7.4.1 Electromotive Force

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  1. In a circuit, electromotive force is the energy per unit charge converted from the other forms of energy into electrical energy to move the charge across the whole circuit.
    In equation,

  2. where
    E = e.m.f.,
    W = energy converted from non-electrical forms to electrical form
    Q = positive charge.
  1. The unit of e.m.f. is JC-1 or V (Volt)
  2. The unit of e.m.f. is JC-1 or V (Volt). Electromotive force of 1 Volt means that 1 Joule of electrical energy is supplied to the circuit to move 1 Coulomd of charge across the whole circuit.

Electromotive Force
Potential Difference
Similarities:
Have same unit (Volt)
Can be measured by Voltmeter
Definition
The electromotive force (e.m.f.) is defined as the energy per unit charge that is converted from chemical, mechanical, or other forms of energy into electrical energy in a battery or dynamo.		 Definition
The potential difference (p.d.) between two points is defined as the energy converted from electrical to other forms when one coulomb of positive charge passes between the two points.
Symbol:
Denote by the symbol, E.		 Symbol:
Denote by the symbol, V

Example 1
When a 1Ω resistor is connected to the terminal of a cell, the current that flow through it is 8A. When the resistor is replaced by another resistor with resistance 4Ω, the current becomes 2⅔A. Find the
a. internal resistance of the cell
b. e.m.f. of the cell

Answer:
Experiment 1
R1 = 1Ω
I1 = 8A


Experiment 2
R2 = 4Ω
I2 = 2⅔A


Solve the simultaneous equation
E = 12V, r = 0.5Ω


Example 2
The diagram on the left shows that the terminal potential difference of a batteries is 1.2V when a 4 Ω resistor is connected to it. The terminal potential become 1.45V when the resistor is replaced by another resistor of resistance 29Ω
Find the
a. internal resistance, r
b. e.m.f. of the batteries.
Answer:
Experiment 1

V1 = 1.2V
R1 = 4Ω

I = V/R
I = (1.2)/(4)
I = 0.3A


E = V + Ir
E = (1.2) + (0.3)r
E - 0.3r = 1.2 ------------(eq1)

Experiment 2
V2 = 1.45V
R2 = 29

I = V/R
I = (1.45)/(29)
I = 0.05A

E = V + Ir
E = (1.5) + (0.05)r
E - 0.05r = 1.45 -----------------(eq2)

Solve the simultaneous equation eq1 and eq2

E = 1.5V, r = 1Ω

The Linear Graph


From the equation,

E = V + Ir
Therefore
V = -rI + E

Y axis = Potential difference (V)
X axis = Current (I)
Gradient od the grapf, m = - internal resistance (r)
Y intercept of the graph, c = e.m.f.

Example:

The graph shows the variation of potential difference with current of a battery.
What is the internal resistance and e.m.f. of the battery?

Answer:
e.m.f. = y-intercept = 3V

internal resistance,
r = -gradient of the graph
r = - (-3)/(6) = 0.5Ω

 

7.3.3 Current and Potential Difference

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Series Circuit

The current flow into a resistor = the current flow inside the resistor = the current flows out from the resistor

IA = IB = IC


In a series circuit, the current at any points of the circuit is the same.

Parallel Circuit

The current flow into a parallel circuit is equal to the sum of the current in each branches of the circuit.
I = I1 + I2
Example

If the resistance of the 2 resistors is the same, current will be divided equally to both of the resistor.


Series Circuit


The sum of the potential difference across individual resistor in between 2 points in a series circuit is equal to the potential difference across the two points.

V = V1 + V2

Example

Parallel Circuit


The potential difference across all resistors in a parallel circuit is the same.

V = V1 = V2

Example

Finding Current in a Series Circuit

  1. In a series circuit, the current flow through each of the resistor is equal to the current flows through the whole circuit.
  2. Potential difference (V) across the whole circuit is equal to the e.m.f. (E) if the internal resistance is ignored.
  3. Effective resistance for the whole circuit = R1 + R2
  4. By Ohm’s Law


Example 1

For the circuit in the diagram above,
  1. find the reading of the ammeter.
  2. find the current flows through the resistor.

Answer:
a.
Potential difference across the 2Ω resistor, V = 12V
Resistance, R =2Ω

V = IR
(12) = I(2)
I = 12/2 = 6A
b.
The current flows through the resistor
= the reading of the ammeter
= 6Ω


Example 2

For the circuit above,
  1. find the reading of the ammeter.
  2. find the current flows through each of the resistors.


Answer:
a.
Potential difference across the 2 resistors, V = 12V
The equivalent resistance of the 2 resistors, R =6Ω

V = IR
(12) = I(6)
I = 12/6 = 2A

b.
The current flows through the resistors
= the reading of the ammeter
= 2Ω

Finding Current in a Parallel Circuit


In parallel circuit, the potential difference (V) across each of the resistors is equal to the e.m.f. (E) if the internal resistance of the cell is ignored.

By Ohm’s Law



Example 1:


For the diagram above,
a. find the reading of the ammeter.
b. find the current in each of the resistors.

Answer:
a.
Effective resistance of the 2 resistors in parallel

Potential difference across the 2 resistance, V = 3V
V = IR
(3) = I(2)
I = 3/2 = 1.5A

Reading of the ammeter = 1.5A
b.
Current in the 3Ω resistor,
V = IR
(3) = I(3)
I = 3/3 = 1A

Current in the 6Ω resistor,
V = IR
(3) = I(6)
I = 3/6 = 0.5A


Example 2


Figure above shows 3 identical resistors connected in parallel in a circuit. Given that the resistance of the resistors are 6Ω each. Find the reading of all the ammeters in the figure above.
Answer:
The potential diffrence across the resistors V = 6V
For ammeter A,
The equivalent resistance = 6/3 = 2Ω.
V = IR
(6) = I(2)
I = 6/2 = 3A

For ammeter A1,
V = IR
(6) = I(6)
I = 6/6 = 1A

For ammeter A2,

V = IR
(6) = I(6)
I = 6/6 = 1A

Finding Potential Difference in a Series Circuit


In the circuit above, if the current in the circuit is I, then the potential difference across each of the resistor is

Example


Find the potential difference across each of the resistors in the diagram above.

Answer:
The potential difference across the 2 resistors, V = 12V
The equivalent resistance of the 2 resistors, R = 2 + 4 = 6Ω
Current pass through the 2 resistor,


For resistor R1,
V = IR
V = (2)(2) = 4V

For resistor R2
V = IR
V = (2)(4) = 8V


 

7.3.2 Resistance in Series and Parallel Circuit

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  1. In a series circuit, the effective resistance is equal to the sum of the individual resistance, as shown in the following equation. 

  2. In a parallel circuit, the effective resistance of the resistors can be calculated fro the following equation. 


Example:


Find the equivalent resistance of the connection shown in the diagram above.

Answer:
(a)
R = 2 + 3 + 6 = 11Ω

(b)

(c)

(d)

  1. In a series circuit, the more resistors with equal resistance in the circuit, the higher the effective resistance of the circuit.
  2. In a parallel circuit, the more resistors with equal resistance in the circuit, the lower the effective resistance of the circuit.

Finding the Resistance of an Individual Resistor

Example


Given that the equivalent resistance of the connection in the figure above is 0.8Ω, find the resistance of the resistor R.

Answer:


Example


Given that the equivalent resistance of the connection in the figure above is 4Ω, find the resistance of the resistor R.

Answer:

 

7.2.4 Factors Affecting the Resistance in a Conductor

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The resistance R of a given conductor depends on:
  1. its length l,
  2. its cross-sectional area A
  3. its temperature and
  4. the type of material.

Length


Resistance is directly proportional to the length of the conductor.

Cross Sectional Area


Resistance is inversely proportional to the cross sectional area of the conductor.

Temperature


A conductor with higher temperature has higher resistance.

Material

Difference materials have difference resistivity. The resistance of copper wire is lower than iron wire.

Since resistance is directly proportional to the length and inversely proportional to the cross sectional area of the conductor. If two resistors of same material have same temperature, we can relate the resistance of the two resistors by the following equation.



 

7.2.3 Resistance

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  1. The resistance R of a material is defined as the ratio V : I, where V is the potential difference across the material and I is the current flowing in it.

  2. The SI unit of resistance is the ohm (W). One ohm is the resistance of a material through which a current of one ampere flows when a potential difference of one volt is maintained.

Finding Resistance from the Potential Difference - Current Graph

In the graph potential difference against current, the gradient of the graph is equal to the resistance of the resistor.

Resistance, R = Gradient of the Graph


Example:


Figure above shows the graph of potential difference across a wire against its current. Find the resistance of the wire.

Answer:
Resistance

Ohmic Conductor

  1. Conductors that obey Ohm’s law are said to be ohmic conductor.
  2. Examples of Ohmic conductor: Metal, Copper sulphate solution with copper electrodes

Non-Ohmic Conductor

  1. Conductors which do not obey Ohm’s law are called non-ohmic conductor.
  2. Example: Semiconductor Diode, Vacuum tube diode

(Examples of characteristic of non-Ohmic conductor)

 

7.2.2 Relationship Between Current and Potential Difference

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Ohm’s Law

  1. The relationship between the current passing through 2 points in a conductor and the potential difference between the 2 points is given by Ohm's law.
  2. Ohm’s Law states that the current flowing in the metallic conductor is directly proportional to the potential difference applied across it’s ends, provided that the physical conditions ( such as temperature ) are constant.

    where k is a constant

Example:
What is the current flow through an 800Ω toaster when it is operating on 240V?

Answer:
Resistance, R = 800Ω
Potential difference, V = 240V
Current, I = ?

 

7.2.1 Potential Difference

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Potential and Potential Difference

  1. The electric potential V at a point in an electric field is the work done to bring a unit ( 1 Coulomb) positive charge from infinity to the point.
  2. The potential difference (p.d.) between two points is defined as the work done in moving 1 Coulomb of positive charge from 1 point in an electric field to another point.
  3. In mathematics

    or

  1. Example, in the diagram above, if the work done to move a charge of 2C from point A to point  is 10J, the potential difference between A and B,

Example:
During an occasion of lightning, 200C of charge was transferred from the cloud to the surface of the earth and 1.25×1010J of energy was produced. Find the potential difference between the cloud and the surface of the earth.

Answer:
Work done, W = 1.25×1010J
Charge transferred, Q = 200C
Potential difference, V = ?

Arrangement of Ammeter


To use the ammeter in the measurement of an electric current, the ammeter must be connected in series to the circuit.

Arrangement of Voltmeter


To use the voltmeter in the measurement of potential difference across an object, the voltmeter must be connected in parallel to the circuit.