(2) 7.4 Emf and Internal Resistance 09

The electromotive force, E (e.m.f.) is defined as the work done by a source in driving one coulomb of charge around a complete circuit.

Total energy needed to move one coulomb of charge around a complete circuit (souce of electrical energy)

Unit of e.m.f. is the volt, V = J/C or J C-1

Sek Men Sains Muzaffar Syah

When the switch is is closed, the circuit is complete. The bulb lights up showing that a current flows in the circuit. The flow of current is the flow of charge. What causes the charge to flow in the circuit? The cell is the source of

energy and the light bulb is the energy consuming device.

The light bulb converts electrical energy into heat and light energy. The electric charges that flow round the circuit transfer energy from the

source to the device. In a cell, chemical reaction converts chemical energy into electrical

energy. This energy pushes the free electrons to move them from the negative terminal to the positive terminal of the cell.

Work is done by the source in driving the charges around a complete circuit

The voltage label on a battery or cell indicates its e.m.fThe label 1.5 V on a dry cell indicates the e.m.f. of the cell is 1.5 V.A cell has an e.m.f. of 1.5 V if a flow of 1 C of charge produces 1.5 J of electrical energy to the whole circuit.

Open circuit

Open circuit

Closed circuit

Closed circuit

7.4

89

Idea of Electro-motive force

Define electro-motive force (e.m.f), E

What does the label 1.5 V on the battery mean?


No current flows through the circuit

The voltmeter reading is 1.5 V. The e.m.f. = the reading of the

voltmeter which is connected directly across the terminals of the cells.

e.m.f. = 1.5 V

Current flows through the circuit The reading of the voltmeter will

drop a little if a lamp is connected in series to the cell.

If the voltmeter reading is 1.2 V, then the potential difference across the lamp = 1.2 V.

Electromotive Force, E Potential Difference, V

Indicates the electrical energy given to 1 C of charge flowing through the cell or source.

Indicates the electrical energy that is transformed to other forms of energy when 1 C of charge passes through a component in a closed circuit.

Used in reference to source of electrical energy

Used in reference to electrical component in a circuit.

Represented by the voltmeter reading in an open circuit (when switch is opened)

Represented by the voltmeter reading in a closed circuit (when switch is closed)

Measured in JC-1 or Volts,V Measured in JC-1 or Volts,V

Activity 1: To distinguish between e.m.f. and potential difference

No current flow

R

Voltmeter reading,e.m.f.

Voltmeter reading,potential difference, V < e.m.f., E

E , r

Current flowing

90

Compare e.m.f. and potential difference

Open circuit & Closed circuit


Switch S is let in the open position. What happens to the bulb is observed. The readings of the ammeter and the voltmeter are recorded.

Switch S is closed and what happens to the bulb is observed. The readings of the ammeter and the voltmeter are recorded.

Position of switch

Open Closed

State of the bulbAmmeter reading/AVoltmeter reading/V

Conclusion1. What is the voltmeter reading

when switch S is closed?2. What is the e.m.f. of the cell?

Discussion1. Why does the ammeter

reading is zero when switch S is open and have a reading when switch S is closed.

The reading of the ammeter when switch S is open is zero

because there is no flow of _______ in an open circuit,

Current flow is ___________

When switch S is closed, there is an ammeter reading due to

the flow of _________ in the closed circuit produce flow of

___________

2. Compare the difference in the two voltmeter readings measured.

The reading of the voltmeter when switch S is open is

___________ than when switch S is closed.

3. What is the work done in moving 1 C of charge around the complete circuit?

The reading of the voltmeter when the circuit is open is the

________ value of the cell.

The voltmeter reading = ____________

So _____ J of electrical energy is required to move 1 C of

charge around the complete circuit.

4. What is the work done in The reading of the voltmeter when the circuit is closed is the

91


moving 1 C of charge across the light bulb?

__________ _________across the bulb.

The voltmeter reading = __________

Therefore, ________J of electrical energy is dissipated by 1 C

of charge across the light bulb. The electrical energy is

transformed to _________ and __________ energy.

5. Why there is a difference in the two voltmeter readings measured?

There is a _________ in potential difference across the cell.

Drop in potential = e.m.f - potential Difference difference (bulb)

This is caused by the ________ __________ of the cell.

In an open circuit when there is no current flow, the potential difference, V across the cell is the electromotive force, E of the cell.

In a closed circuit when there is a current flow, the potential difference, V across the cell is smaller than the e.m.f. of the cell.

This drop in potential difference across the cell is caused by the internal resistance of the cell.

e.m.f. = 1.5 V means the cell gives 1.5 J of electrical energy to each coulomb that passes through it. The energy dissipated in the resistor or bulb is less than 1.5 J per coulomb. The current that flows through the circuit also flows through the battery.

Internal resistance in the battery causes this loss of energy or drop in potential difference across the battery.

Some of the energy per charge the battery provides will be used to overcome the internal resistance of the battery and change to heat energy.

In practice, the potential difference a battery provides to an external circuit will always be less than its e.m.f.

The internal resistance, r of a source or battery is the resistance against the moving charge due to the electrolyte in the source or cell.Work is needed to drive a charge against the internal resistance.This causes a drop in potential difference across the cell as the charge flows through it and loss of heat energy in the cell.

E.m.f of the cell = EP.d to the external circuit = VDrop in P.d. inside the cell = Ir Ir = E – V

Or E = V + Ir

Or E = IR + Ir

92

Why there is drop in potential difference?

Explain Internal Resistance, rr

Explain equation relates, E, V, I, and r


Exercise 1: Solve problems involving e.m.f and internal resistance

1. A cell with e.m.f. 2 V and internal resistance 1 Ω is connected to a resistor of 4 Ω. What is the reading on the voltmeter when it is connected across the 4 Ω resistor?

2. A bulb M is connected to a battery by means of a switch. A voltmeter is also connected across the battery. When the switch is open, the voltmeter reads 6.0 V. When the switch is closed, the voltmeter reads 4.8 V.

(a) What is the e.m.f. of the battery?(b) If the resistance of the bulb M is 8 Ω,

what is the current passing through M when the switch is closed?

(c) Find the value of the internal resistance, r, of the battery.

3. When switch S is opened, the voltmeter reading is 1.5 V. When the switch is closed, the voltmeter reading is 1.35 V and the ammeter reading is 0.3 A.

Calculate: (a) e.m.f(b) internal resistance (c) resistance of R

Activity 2: Determine e.m.f. and internal resistance, r

93


Apparatus Dry cell, rheostat, ammeter, voltmeter, switch, connecting wires

Diagram

Procedure Turn on the switch, adjust the rheostat until the ammeter reading is I = 0.2 A

Record the voltmeter reading as V Repeat the experiment with current, I = 0.3 A, 0.4 A, 0.5 A and 0.6 A.

ResultsI /A

V / V

94


Plot graph of V against I on a graph paper.

Sketch the shape of the graph.

1. From the graph of V against I plotted, state the relationship between the potential difference across the cell and the current flow.

2. By using the equation E = V + Ir,(a) write down V in terms of E, I and r.

95


(b) explain how you can determine the values of E and r from the graph plotted in this experiment. (compare eq, in (a) with y = mx + c)

(c) Determine the value of E and r from the graph.

3. Why the potential difference across the cell decreases when the current increases?

When the switch is opened, the voltmeter = ______ , current, I = ______

When the switch is closed, ________ flows through the circuit and there is a drop in

__________ ________ due to the ___________ resistance, r in the cell (V = Ir)

The readings of the voltmeter decreases as V = ____________

The bigger the current, the ___________ the drop in potential difference (Ir) and the

___________ the reading of the voltmeter, V.

Determine:(a) the e.m.f of the cell

(b) the internal resistance of the cell

Comparison between total e.m.f and total internal resistance in a series and parallel circuit . each cell has E = 1.5 V and r = 0.5 Ω

Total e.m.f = 2 x 1.5 = 3.0 V Total e.m.f = 1.5 V Total r = 2 x 0.5 = 1.0 Ω Total r = ½ x 0.5 = 0.25 Ω

96


TUTORIAL 7.4

1. Which of the following describe the e.m.f. of a cell?A. The difference in the energy needed

to drive a unit charge through the load resistors and the energy needed to drive a unit charge through the cell.

B. The difference in energy to drive a unit charge across each resistor.

C. The energy supplied to a unit charge within the cell.

D. The total energy used to drive a unit charge round the complete circuit.

2. Voltmeter V has infinite resistance and ammeter A has negligible resistance. Which circuit can be used to measure the electromotive force (e.m.f) of a battery?(2005)

3. Which of the following graphs relates the reading of ammeter and the voltmeter in figure above?

4. Three resistors with the resistance shown are connected to a battery. The electromotive force and the internal resistance of the battery are E and r respectively.

Which of the following statements is true?A. E = Reading of voltmeters V1 + V2

B. Readings A1 = A2

C. Readings V1 > V2

D. Effective resistance of the circuit = 3R

5. The electromotive force and internal resistance of the dry cell in the circuit below is 1.5 V and 0.5 Ω respectively.

What are the reading of the voltmeter when the switch is closed? (2004)A. 0 VB. 1.5 VC. Less than 1.5 VD. More than 1.5 V

6. Diagram below shows an electric circuit.

Calculate the internal resistance of the battery in this circuit. (2007)A. 0.06 ΩB. 0.07 ΩC. 0.25 ΩD. 3.75 Ω

97


7. The graph V against I below shows the results of the experiment to determine the electromotive force ,E, and the internal resistance, r , of a dry cell.

(a) Based on the graph in the figure above,

(i) what happen when I increases?

____________________________

(ii) determine the value of the potential difference , V, when the current , I = 0.0 A.Show on the graph , how you determine V.

____________________________

(iii) name the physical quantity that represents the value in (a)(ii).

____________________________

(b) The internal resistance, r, of the dry cell is given by

r = -mwhere m is the gradient of the graph. Calculate the value r.

(c)(i) From the graph , state the value of V, when I = 0.80 A.

Show on the graph how you determine the value V.

_____________________________

(ii)The resistance R is given by R = V

I Calculate R when I = 0.80 A

(d) State one precaution that should be taken during this experiment.

____________________________

98


Figure 10.1 (a) and Figure 10.2 (a) show two circuits used to investigate the relationship between potential difference and electric current. Figure 10.1(b) and Figure 10.2(b) show the potential difference against electric current graph respectively for Figure 10.1(a) and Figure 10.2(b).

(a) What is meant by electric current? [ 1 mark ]

(b) Compare the placement of the voltmeter in Figure 10.1(a) and Figure 10.2(a). Then compare the relationship between potential difference and current for both graphs. Explain the results shown in both graphs. Hence state the related physics law and concepts applicable in each situation.

[ 5 markah ]

99

I

V

00

V

VFigure 10.1(a)[Rajah 10.1(a)]

I

V

Figure 10.1(b)Rajah 10.1 (b)

00

Figure 10.2(a)[Rajah 10.2(a)]

Figure 10.2(b)[Rajah 10.2(b)]

(2) 7.4 Emf and Internal Resistance 09

Documents

Transcript of (2) 7.4 Emf and Internal Resistance 09