Chapter13pp122-133

OCR (A) specifications: 5.2.2e,k,l

Chapter 13 Practical circuits Worksheet Worked examples Practical 1: The e.m.f. and internal resistance of a lemon cell Practical 2: Calibrating an electrical thermometer Practical 3: Investigating turbidity End-of-chapter test Marking scheme: Worksheet Marking scheme: End-of-chapter test

Worksheet Intermediate level 1 2

Suggest why a chemical cell has internal resistance. Use the terms below to write a word equation for the e.m.f. of a power supply. • • •

3

4

5

[1]

terminal p.d. e.m.f. of power supply p.d. across internal resistance

[1]

A d.c. power supply of e.m.f. 12 V has an internal resistance of 2.3 Ω. It is accidentally shorted-out across its terminals by a short length of wire of negligible resistance. a

Calculate the current drawn from the supply.

[2]

b

Suggest why it may be dangerous to have a supply shorted-out in this way.

[1]

A cell of e.m.f. 1.5 V is connected across a length of wire of resistance 2.6 Ω. A highresistance voltmeter placed across the terminals of the cell measures 0.85 V. Calculate: a

the potential difference across the internal resistance;

[2]

b

the internal resistance of the cell.

[2]

The diagram shows a potential divider circuit. The battery has negligible internal resistance. Calculate the potential difference across the 6.0 Ω resistor.

5.0 V

[3]

Higher level 6

6.0 Ω

18.0 Ω

A length of wire of resistance 7.3 Ω is connected across the terminals of a cell of e.m.f. 1.4 V. A high-resistance voltmeter measures a p.d. of 0.81 V across the terminals of the cell. Calculate: a

the ‘lost volts’ (the p.d. across the internal resistance of the cell);

[2]

b

the internal resistance of the cell;

[2]

c

the ratio: power dissipated by the 7.3 Ω resistor power delivered by the cell

7

8

[3]

Two chemical cells are connected in series. Each cell has e.m.f. 1.4 V and internal resistance 0.38 Ω. The combination of the cells is connected across an electronic circuit of resistance 1.8 Ω. Calculate: a

the potential difference across the electronic circuit;

[4]

b

the potential difference across the terminals of each cell.

[2]

The diagram shows a potential divider circuit. The voltmeter has infinite resistance and the battery has negligible internal resistance. a

b

The variable resistor is set on its maximum resistance of 200 Ω. Calculate the voltmeter reading.

R

[3]

The resistance R of the variable resistor is gradually altered from its maximum resistance value of 200 Ω to zero. Use a sketch graph to describe how the voltmeter reading changes with R. [3]

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V

6.0 V

180 Ω

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9 The diagram shows a simple electrical thermometer based on a negative temperature coefficient (NTC) thermistor. At 30 °C the thermistor has a resistance of 2.4 kΩ and this decreases to 430 Ω at 100 °C. The battery has negligible internal resistance. Calculate the maximum input voltage into the datalogger.

5.0 V

[4] 3.6 kΩ

to datalogger

Extension 10 A cell has e.m.f. 1.5 V and internal resistance 0.50 Ω. It is connected across a variable resistor of resistance R. Copy and complete the table (I = current drawn from the cell; V = terminal p.d.; P = power dissipated by external resistor). With the aid of a sketch graph, describe how the power dissipated by the external resistor is affected by its resistance. [5]

R (Ω)

I (A)

V (V)

P (W)

0 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 Total: ––– Score: 40

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Worked examples Example 1 For the circuit shown below, calculate the reading shown by the high-resistance voltmeter. 4.5 V

battery 0.8 Ω

V

2.0 Ω

e.m.f. current = total resistance Therefore: I=

E R+r

I=

4.5 = 1.61 A ≈ 1.6 A 2.0 + 0.8

The current in a series circuit is the same at all points. Hence, the p.d. across the external resistor is given by: V = IR = 1.61 × 2.0

You must use 2.0 Ω here. Substituting 0.8 Ω will give the ‘lost volts’.

V ≈ 3.2 V

Tip There is another method for getting the correct answer. You can also use the potential divider equation: V=

Rb × Vin Ra + Rb

V=

2.0 × 4.5 0.8 + 2.0

V ≈ 3.2 V

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Example 2

9.0 V

For the circuit shown, calculate the reading on the high-resistance voltmeter connected across the thermistor, which has a resistance of 1.2 kΩ. You may assume that the supply has negligible internal resistance. The p.d. across the thermistor is given by the potential divider equation: V=

Rb × Vin Ra + Rb

V=

1.2 × 9.0 ≈ 1.8 V 4.7 + 1.2

4.7 kΩ

V

The resistance values are often confused. You are measuring the p.d. across the resistor of resistance Rb.

Tip You can use your knowledge of series circuits to get the answer. current =

e.m.f. total resistance

total resistance = 4700 + 1200 = 5900 Ω I=

9.0 ≈ 1.53 × 10–3 A 5900

The potential difference across the thermistor is given by: V = IR V = 1.53 × 10–3 × 1200 ≈ 1.8 V

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Practical 1 The e.m.f. and internal resistance of a lemon cell Safety There are not likely to be any major hazards in carrying out this experiment. However, teachers and technicians should always refer to the departmental risk assessment before carrying out any practical work.

Apparatus • fresh lemon • variable resistor (100 kΩ) • clean zinc and copper electrodes

• digital voltmeter • digital ammeter (100 µA fsd) • connecting leads

Introduction You will be familiar with cells such as chemical cells, solar cells, etc. In this experiment, you have the opportunity to make your own cell and investigate its properties in terms of its e.m.f. and its internal resistance. The cell is a little unusual – it is made from a lemon!

Procedure The diagram below shows how to use a lemon to make a chemical cell. zinc electrode

copper electrode E

lemon

r

I A

equivalent to E

r

100 kΩ

V

1 2 3 4 5

Connect up the circuit. By altering the resistance of the variable resistor, measure the terminal potential difference V across the lemon cell for a range of values of current I. Record your results in a table. Plot a graph of V against I and draw a straight line of best fit. The magnitude of the gradient of the straight-line graph is equal to the internal resistance r of the lemon cell. The intercept on the V axis is equal to the e.m.f. of the lemon cell. Use your graph to determine the e.m.f. and internal resistance of your lemon cell.

Guidance for teachers To achieve consistent results, it is advisable to remove any oxides from the electrodes using an emery cloth. In trials, the following typical results were obtained for a fresh lemon: e.m.f. ≈ 1.0 V

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internal resistance ≈ 120 kΩ

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Practical 2 Calibrating an electrical thermometer Safety Take care when pouring the boiling water into the beaker. Teachers and technicians should follow their school and departmental safety policies and should ensure that the employer’s risk assessment has been carried out before undertaking any practical work.

Apparatus • • • •

low-voltage d.c. supply 1 kΩ resistor (or equivalent) digital voltmeter electric kettle

• • • •

100 ml beaker thermometer NTC thermistor connecting leads

Introduction In this experiment you will calibrate a thermistor-based potential divider circuit and use the results to estimate your body temperature.

Procedure

1

Connect up a potential divider circuit incorporating a thermistor as shown in figure 13.10 on page 111 of Physics 1.

2 3 4

Put the thermistor in a waterproof plastic bag and place into the beaker.

5 6 7 8 9

128

Pour boiling hot water into the beaker. Measure and record the temperature of the water and the potential difference across the thermistor. Take several readings of temperature and potential difference as the water cools. Plot a graph of potential difference against temperature. Remove the thermistor from the beaker and the plastic bag and hold it tightly between the palms of your hands. Record the constant value of the potential difference across the thermistor. Use this reading and the calibration graph to determine your body temperature. How reliable is your estimate for your body temperature? How does this value compare with a reading using the glass thermometer?

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13 Practical circuits

Practical 3 Investigating turbidity Safety There are not likely to be any major hazards in carrying out this experiment. However, teachers and technicians should always refer to the departmental risk assessment before carrying out any practical work.

Apparatus • • • •

low-voltage d.c. supply 300 Ω resistor or equivalent digital voltmeter pipette with small quantity of milk

• • • •

100 ml beaker light-dependent resistor (LDR) light source (table lamp or ray box) connecting leads

Introduction Potential divider circuits are very useful as sensing circuits, especially when used with thermistors and light-dependent resistors (LDRs). In this experiment you will investigate the effect of varying light intensity on a potential divider circuit based on an LDR.

Procedure

1

Connect up the potential divider circuit incorporating an LDR as shown in figure 13.12 on page 112 of Physics 1.

2

Set up the light source, beaker, LDR and pipette as shown in the diagram.

3

Put 100 ml of clean water in the beaker.

4

Measure the potential difference across the LDR.

5

Add one drop of milk and measure the potential difference.

6

7 8

Repeat for more drops of milk until the water is very turbid (cloudy).

pipette of milk

to potential divider light water LDR

Plot a graph of potential difference against the number of drops of milk. The intensity of light falling on the LDR decreases as the number of drops of milk increases. The graph will show that the potential difference across the LDR increases as more drops of milk are added to the water. Can you explain these observations in terms of the potential divider equation? V=

Rb × Vin Ra + Rb

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End-of-chapter test Answer all questions.

1

Two identical cells are connected in series. Each cell has internal resistance of 0.4 Ω and e.m.f. of 1.5 V. a

For these two cells, state:

b

2

i

the total e.m.f.;

[1]

ii

the total internal resistance.

[1]

The two cells in series are now connected to an electrical device of constant resistance 2.5 Ω. Calculate: i

the p.d. across the device;

[3]

ii

the power dissipated by the internal resistance of each cell.

[3]

The graph shows how the terminal p.d. V depends on the current I drawn for a particular cell.

V (V) 4.5 V

Use the graph to determine: a b

the internal resistance of the cell;

[2]

the e.m.f. of the cell.

[2]

3.0 A 0 0

3

I (A)

In the potential divider circuit shown, the voltmeter has an infinite resistance and the power supply has negligible internal resistance.

+

6.0 V

–

Calculate the voltmeter reading: a

when the switch S is open;

[3]

b

when the switch S is closed.

[3]

S

1.2 kΩ V

1.8 kΩ 1.2 kΩ

4

a

Describe how the resistance of a light-dependent resistor (LDR) is affected by the intensity of the incident light. [1]

b

The diagram shows a light-sensing circuit. The voltmeter has an infinite resistance and the power supply has negligible internal resistance. Describe how the voltmeter reading changes as a hand gradually obscures the intensity of light on the LDR.

+

–

[4]

V

Total: ––– Score: 23

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Marking scheme Worksheet 1 2 3

Due to the current flowing through the chemicals within the cell. [1] e.m.f. of power supply = terminal p.d. + p.d. across internal resistance [1] a

E = I (R + r) R = 0 (since supply is shorted-out)

4

so

I=

E [1]; r

I=

12 = 5.2 A [1] 2.3

b

Excessive heating of the power supply. [1]

a

E = V + Ir; p.d. across internal resistance = Ir = 1.5 – 0.85 [1] Vr = 0.65 V [1] V 0.85 0.65 =I= = 0.327 A [1]; r = ≈ 2.0 Ω [1] R 2.6 0.327

b

I=

5

V=

Rb × Vin [1]; Ra + Rb

6

a

E = V + Ir

V=

6.0 × 5.0 [1]; V = 1.25 V ≈ 1.3 V [1] 18 + 6.0

p.d. across internal resistance = Ir = 1.4 – 0.81 [1]; Vr = 0.59 V [1] V 0.81 = = 0.111 A [1]; R 7.3

b

I=

c

P = I2R [1] Ratio =

r=

0.59 = 5.3 Ω [1] 0.111

0.1112 × 7.3 [1] 0.1112 × (7.3 + 5.3)

Ratio = 0.58 (42% of the power is ‘lost’ internally in the cell) [1]

7

a

I=

E [1] R+r

total e.m.f. = 2 × 1.4 = 2.8 V I=

8

and

total internal resistance = 2 × 0.38 = 0.76 Ω [1]

2.8 = 1.094 A ≈ 1.1 A [1]; V = IR = 1.094 × 1.8 = 1.97 V ≈ 2.0 V [1] 0.76 + 1.8

b

V = E – Ir [1];

a

V=

V = 1.4 – (1.094 × 0.38) = 0.98 V [1]

Rb × Vin [1]; Ra + Rb

V=

200 × 6.0 [1] 180 + 200

V = 3.16 V ≈ 3.2 V [1] b

As the resistance decreases, the p.d. across the variable resistor decreases. [1] Voltmeter reading (V)

Correct values marked [1]

3.2 V

Correct shape curve [1] 200 Ω

R (Ω)

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9 Maximum p.d. across the 3.6 kΩ resistor is when the resistance of the thermistor is minimum. [1] V=

Rb × Vin [1]; Ra + Rb

V=

3600 × 5.0 [1]; V = 4.47 V ≈ 4.5 V [1] 430 + 3600

10 All columns correctly calculated (for R = 0.50 Ω, P = 1.13 W). [2] The completed table will show that maximum power is dissipated when the external resistor has resistance of 0.50 Ω. This resistance value is equal to the internal resistance. [1] If you want to deliver maximum power to an external load, then its resistance must ‘match’ the resistance of the supply (the cell in this case). [1] P (W)

maximum power when R = r

1.13 W

0.50 Ω

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R (Ω)

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Marking scheme End-of-chapter test 1

a

b

2

3

i

Total e.m.f. 2 × 1.5 = 3.0 V [1]

ii

Total internal resistance = 2 × 0.4 = 0.8 Ω [1]

i

I=

ii

P = I 2R [1];

V = IR = 0.91 × 2.5 [1]; V = 2.3 V [1]

P = 0.912 × 0.4 [1]; P = 0.33 W [1]

a

V = E – Ir

b

e.m.f. = intercept with the V axis [1];

a

V=

b

1 1 1 = + R 1.2 1.2 V=

4

E 3.0 = = 0.91 A [1]; R + r 2.5 + 0.8

so

gradient of line = internal resistance [1]; r =

Rb × Vin [1]; Ra + Rb so

1.8 × 6.0 [1]; 0.6 + 1.8

V=

4.5 = 1.5 Ω [1] 3.0

e.m.f. = 4.5 V [1]

1.8 × 6.0 [1]; V = 3.6 V [1] 1.2 + 1.8

total Ra = 0.6 kΩ [1] V = 4.5 V [1]

a

The resistance of a light-dependent resistor decreases as the light intensity falling on it increases. [1]

b

When the hand does not obscure the LDR its resistance is low [1] and the voltmeter reading is low. [1] As the hand obscures the LDR its resistance increases [1] and the voltmeter reading is high. [1]

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