CURRENT ELECTRICITY
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What is Current?
Current is the flow of electrons in a circuit. To use electricity, we need electrons to flow in the same direction around a circuit, we usually use copper cables to form the circuit because copper is a very good electrical conductor, which means the atoms that make copper have a loosely bound electron in that outermost or valence shell, which is free to move around inside the metal.
Contents to be covered are;
Meaning of current electricity and electric current.
Simple electric circuits.
Construction of simple electric circuits.
Introduction
Current electricity is a fundamental aspect of engineering science that is crucial for designing, analysing, and enhancing electric circuits, electronic devices, and power systems in various technological fields.
In this chapter, you will learn the concepts of current electricity, voltage, Ohm’s law, electrical components, electrical energy and power. The competencies developed will enable you to design and make simple electric circuits for various applications. It will also enable you to utilise electrical appliances efficiently in various contexts.
Think
How would life on Earth be without electricity?
Concept of current electricity
A conductor, such as an electric wire (Figure 2.1(a)), contains charges (negative or positive); when these charges are compelled to move, they constitute an electric current. To initiate this movement, we must establish a potential difference across the conductor ends. Various methods have been utilised to create a potential difference; one employs a battery as shown in Figure 2.1 (b), while another involves connecting a charged capacitor. To ensure a steady flow of current, we must use a battery, as the capacitor discharges immediately once the current flows.
Figure 2.1: Flow of charges
Electric current ( I ) is defined as the rate of flow of electric charge ( Q ) through a conductor over time ( t ). This relationship is expressed mathematically as
\[I=\frac{\operator name {charge} (Q)} {\operator name {time} (t)}\]
But, Q=ne,
where n is the number of electrons and e is the charge of an electron.
Therefore,
I=net
In simpler terms, current measures the amount of charge that passes a given point in a circuit per unit of time. This charge is carried by charged particles such as electrons or ions. A higher current indicates a greater flow of charge.
Example 2.1Given that the charge of an electron is 1.6×10−19C find the number of electrons that pass in one second through any cross-section of a conductor with a steady current of 1 ampere. Solution |
Electromotive force
The electromotive force (e.m.f) is the potential difference across the terminals of a source when no current is flowing. It provides the energy required to move electrons through a conductor, leading to an electric current. Note that the word “force” in this case does not mean the force of interaction between bodies. One may draw an analogy of e.m.f to water pressure. When the pressure is high, more water flows through a pipe.
Similarly, with higher e.m.f, more electrons flow through a conductor. Sources of e.m.f include electrochemical cells such as dry cells and car batteries, thermoelectric devices, solar cells and electric generators. Labels of potential difference values written on a battery or cell refer to its e.m.f. Figure 2.2(a) shows a dry cell whose e.m.f is 1.5 V , and (b) a car battery whose e.m.f is 12 V .
Potential difference (p.d)
When an electric device such as a bulb is connected to a cell, electric current flows through the device, and some of the electrical energy is converted into light and heat. The amount of energy converted per unit charge equals the potential difference (p.d) across the device. Alternatively, electric potential difference (p.d) is the work done in moving a unit charge in a circuit from one point to another. Potential difference (p.d) is also called voltage.
p.d(V)= work done (W) charge (Q)
Like the e.m.f., the SI unit for p.d is the volt (V). One volt is equivalent to one joule per coulomb. Figure 2.2 shows a circuit diagram with the dashed box indicating a region where the potential difference between two points of a resistor can be measured.
The difference between e.m.f. and p.d is that e.m.f is the potential difference across the cell terminals when no current flows or no load is connected. In contrast, p.d is measured when current flows through the circuit or when a load is connected. The load here refers to the resistance, R, of the electrical appliance connected to the source of e.m.f. Figure 2.3 shows a circuit with an e.m.f source and a resistor as a load.
Figure 2.3: Electromotive force (e.m.f) source and potential difference (p.d)
Measurement of the e.m.f of a cell and the p.d across a conductor
The e.m.f. of a cell and p.d are measured using a high resistance device called a voltmeter. A voltmeter measures the difference in potential between two points. It is always connected in parallel with the cell or the load because the e.m.f. of a cell is compared to the p.d across the voltmeter’s terminals. To measure e.m.f. and p.d, the positive terminal of the voltmeter is connected to the positive terminal of a cell, and the negative terminal of the voltmeter is connected to the negative terminal of the cell. Note: Measuring e.m.f by a voltmeter only provides an estimated value, since a small amount of current is drawn by the meter contrary to the requirement that no current is drawn for e.m.f measurement. Activity 2.1 will assist on understanding how one can measure of a cell and potential difference of a conductor.
Activity 2.1Aim: To measure the e.m.f. of a cell and the potential difference of a conductor Materials: dry cell ( 1.5 V ), voltmeter, bulb, two switches, connecting wires Procedure 1. Connect the circuit as shown in Figure 2.4.
Figure 2.4 2. Close switch S1, then note the voltmeter reading. Questions(a) Compare the voltmeter reading when only S1 is closed and when both S1 and S2 are closed. Explain your results. (b) What is the e.m.f. of the cell? (c) What is the potential difference across the bulb? The voltmeter reading when the current flows through the bulb (when both switches are closed) is less than when no current flows through the bulb (only switch S1 is closed). When no current flows out of the cell or battery, the voltmeter reading is the cell’s e.m.f. When both switches are closed and the current flows through the circuit, the voltmeter reading is the potential difference across the bulb. |
Electric circuits and symbols
Circuit diagrams are visual representations of electrical circuits, using circuit symbols to represent electrical components and their connections. You can create these diagrams by placing components and connecting them with lines. These diagrams are crucial for understanding and building electrical circuits. Various IT tools, such as virtual labs and Smart Draw, may assist in developing the skills to construct an electric circuit.
Task 2.1Draw a circuit with components A , B, C, and D as shown in Figure 2.5 using online interactive simulation software such as Phet simulation.
Figure 2.5 |
Principles for constructing an electric circuit
When drawing a circuit diagram, follow the following steps.
Voltmeter connection: A voltmeter must be connected in parallel with the component across which the voltage is being measured.
Ammeter connection: An ammeter, which measures current, is always connected in series with the circuit component(s) to ensure that the current flowing through the component(s) also flows through, and is measured by, the ammeter.
Components: For beginners, connect a single component, e.g., a resistor, bulb, in series with a source of e.m.f. However, depending on the requirement, if multiple components are used, they may be connected in parallel and/or series configurations, and then the system is connected in series with the source of e.m.f.
Other rules
1. Draw the circuit symbols first.
2. Using a ruler, draw all the connecting wires.
3. Make all the connecting wires and leads straight lines with corner angles of 90 degrees as shown in figure 2.6.
4. Do not cross any of the lines representing conducting wires.
Figure 2.6: Simple electric circuit
Example 2.2Draw an open and a closed electric circuit using a bulb, battery, switch and wire using the circuit symbols. Solution
Figure 2.7 |
Resistance to an electric current
The flow of charge through a wire in a circuit can be compared to the flow of water through a pipe connecting two water reservoirs. Water flows through the pipe if there is a pressure difference between the two reservoirs. As it flows through a pipe, it loses energy due to friction between the moving water molecules and the pipe’s inner walls.
Similarly, electric charges will flow between two points of a conductor if one point has a higher electric potential than the other; that is, there is a potential difference. As charges flow in a material, they encounter numerous collisions with atoms, resulting in a resistance to the flow of charges. This is known as electrical resistance, denoted by the letter R. In other words, electrical resistance is the opposition to the flow of electrical current through a material.
Any material that offers low resistance to the current flow is termed a conductor. On the other hand, a material is said to be an insulator if it does not allow the flow of electric current through it. Some materials, known as semiconductors, have resistance between that of conductors and insulators. The SI unit of electrical resistance is the ohm, denoted by the symbol Ω. It is worth noting that electrical current loses its energy as it flows through a conductor.
Types of resistors
A resistor is an electrical component with two terminals used to restrict the flow of electric current in a circuit. That is, resistors control the magnitude of the current in a circuit in accordance with Ohm’s law. There are two basic types of resistors depending on their applications and characteristics. These are fixed resistors and variable resistors.
Fixed resistors
The most commonly used type of resistor is the fixed resistor. A fixed resistor has a constant value of resistance. Applications of fixed resistors include protecting electrical components from excess current, dividing voltage in a circuit, and controlling the working of some circuits. In a circuit, the fixed resistor is represented by the symbols shown in Figure 2.8.
Figure 2.8: Circuit symbols for fixed resistors
Diverse resistor materials are used to fabricate fixed resistors. These materials influence the resistor’s properties, such as tolerance and noise. Fixed resistors can be made of carbon, wound wire, metal oxide and metal films.
Task 2.2
Variable resistors
These are resistors whose electric resistance can be adjusted (from zero to its maximum value) to suit the requirements of a circuit. Examples of variable resistors include rheostats, voltage dividers, thermistors, photoresistors and magneto-resistors. Some types of variable resistors and their circuit symbols are shown in Table 2.5.
Table 2.5:
Equivalent resistors
Depending on the purpose, fixed resistors can be connected in either a parallel or a series configuration. The resultant resistance in the circuit is called total resistance, effective resistance, or equivalent resistance.
Resistors in series
When two or more resistors are connected end to end consecutively, and the same current flows through each resistor, such a connection is said to be a series connection. A single-loop circuit with series combination of resistors and a battery, is shown in Figure 2.9.
Figure 2.9: A circuit with resistors connected in series
The current flowing in the circuit is the same at all points. The sum of the potential difference across all the resistors equals the potential difference across the battery.
Therefore,
V=V1+V2+V3
From Ohm’s law, the current, I, in the circuit is given by,
I=VRT
where RT is the total resistance, thus,
V=IRTIRT=IR1+IR2+IR3IRT=I(R1+R2+R3)RT=R1+R2+R3
When resistors are connected in series, the total resistance in the circuit is the sum of the individual resistances. Therefore, for n resistors connected in series, the total resistance is given by,
RT=R1+R2+R3+…+Rn
Note that when resistors are identical, the total resistance is given by RT=nR, where R is the resistance of an individual resistor and n is the number of identical resistors.
Resistors in parallel
Resistors are said to be in parallel connection when two or more resistors are placed side by side with their corresponding ends joined together, such that the same p.d is applied to each resistor. Figure 2.10 shows a circuit with parallel connection of resistors and a battery.
Figure 2.10: A circuit with resistors connected in parallel
When resistors are connected in parallel, the p.d. across each resistor is the same as the total voltage in the circuit. Though the voltage is the same for all branches, the currents flowing in each branch are different. Thus, Ohm’s law can be applied in each branch. To find the different currents, one considers the current through each junction. The sum of the currents in each resistor in Figure 2.10 equals the total current in the circuit. That is,
I=I1+I2+I3
From Ohm’s law, the current, I, in the circuit is given by,
I=VRT
Where, RT is the equivalent resistance.
Applying Ohm’s law to each resistor gives:
VRT=VR1+VR2+VR3VRT=V(1R1+1R2+1R3)
Therefore,
1RT=1R1+1R2+1R3
When resistors are connected in parallel, the reciprocal of the total resistance is equal to the sum of the reciprocals of the individual resistances. Generally, when n resistors are connected in parallel, the reciprocal of the equivalent resistance is given by,
1RT=1R1+1R2+1R3+…+1Rn
For identical n resistors connected in parallel, R1=R2=R3=…=Rn=R. The reciprocal of total resistance is given by 1RT=nR. Therefore, the total resistance is given by RT=Rn.
It should be noted that when resistors are connected in series, their total resistance is higher than that of individual resistors. However, for resistors connected in parallel, the total resistance is lower than that of individual resistors. Activity 2.2 will help in practical learning on effect of arrangement of resistors.
Activity 2.2
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Aim: To investigate the effective resistance for resistors in series and in parallel
Materials: two dry cells of 1.5 V each, connecting wires, bulbs Procedure 1. Connect two cells, each of 1.5 V in series and then connect them to one of the bulbs, A, as shown in Figure 2.11
Figure 2.11 Observe the brightness of the bulb. 2. Connect another bulb, B , in series with bulb A as shown in Figure 2.12.
Figure 2.12 Observe the brightness of the bulbs. 3. Connect another bulb, C , in series with bulbs, A and B as shown in Figure 2.13. Observe how the brightness of bulbs A and B changes.
Figure 2.13 4. Connect bulb B in parallel with bulb A as shown in Figure 2.14. Observe the brightness of the bulbs compared to the initial brightness of bulb, A.
Figure 2.14 5. Connect bulb C in parallel with bulbs A and B as in the circuit shown in Figure 2.15. Observe how the brightness of the bulb’s changes. Disconnect the circuit.
Figure 2.15 Questions (a) What happened to the brightness of the bulbs as more bulbs were added in series? (b) What happened to the brightness of the bulbs as more bulbs were added in parallel? The brightness of the bulbs was reduced as more bulbs were added in series. This is because the total resistance increases when more bulbs (resistors) are added in series and consequently, the potential difference across individual bulbs is reduced. When bulbs are connected in parallel, the brightness of individual bulbs remains the same with an increase in the number of bulbs in the circuit. This is because the potential difference across each bulb remains the same when the number of bulbs is increased in a parallel connection. One might wrongly perceive that bulbs connected in a parallel connection are brighter only because of the combined brightness of all the bulbs. |
Example 2.3Figure 2.16 shows an electric circuit where, VB=9 V,R1=4 Ω,R2=5 Ω, and R3=6 Ω.
Figure 2.16 (a) What is the total resistance of the circuit? (b) What is the value of the current flows in the circuit? (c) What is the potential drop across each resistor? (d) What is the electric potential at point A? Solution (a) RT=R1+R2+R3=4 Ω+5 Ω+6 Ω =15 Ω (b) I=VRT =9 V15 Ω=0.6 A (c) p.d across R1 VR1=IR1=0.6 A×4 Ω=2.4 V p.d across R2 VR2=IR2=0.6 A×4 Ω=3.0 V p.d across R3 VR3=IR3=0.6 A×5 Ω=3.60 V (d) To find the electric potential at point A, take note that, the potential is measured with reference to the negative terminal of the source. This becomes our reference, 0 V . Therefore, we start at the negative terminal of the battery, where the potential is 0 V . Travelling around the circuit in the direction of electron flow, we pass through R3 where there is a voltage drop of 3.6 V , then pass R2 where the p.d is 3.0 V up to point A. Therefore, the potential at point A is, VA=VR1+VR2=3.6 V+3.0 V=6.6 V |
Example 2.4In the circuit shown in Figure 2.17 the battery has a voltage V=10 V, R1=4 Ω,R2=5 Ω, R3=6 Ω.
Figure 2.17 From the given circuit, find: Solution Given V=10 V,R1=4 Ω, R2=5 Ω, R3=6 Ω; Required to find the total resistance RT. Using: (a) 1RT=1R1+1R2+1R3 1RT=14 Ω+15 Ω+16 Ω=3760 ΩRT=1.62 Ω Total current, I, is given by I=VRT=10 V1.62 Ω=6.17 A The total current flowing in the circuit is 6.17 A . (b) The current flowing through each resistor is given by; (c) The total current flowing through the circuit is, |
Example 2.5Determine the current reading on the ammeter in the circuit shown in Figure 2.18.
Figure 2.18 Solution For three resistors in parallel, 1RT=1R1+1R2+1R3 For the first set of three resistors in parallel, 1RT=124 Ω+18 Ω+112 Ω=14 ΩRT=4 Ω For the second set of resistors in parallel, 1RT=1R4+1R5+1R61RT=16 Ω+13 Ω+16 Ω=23 ΩRT=1.5 Ω The effective resistance of the resistors in parallel, i.e., 4 Ω and 1.5 Ω, is now in series with resistors of 15 Ω and 20 Ω. Therefore, the total resistance is given by, =4 Ω+1.5 Ω+20 Ω+15 Ω=40.5 Ω The current is obtained through, I=VRT=9 V40.5 Ω=0.22 A Therefore, the current reading on the ammeter in the circuit is 0.22 A . |
Internal resistance of a cell
A cell always resists the passage of current in a circuit. This resistance is called internal resistance. The internal resistance of a cell acts as if it is a resistor connected in series with the cell, so when current flows through the cell, there is a potential drop across this resistor. Consider a cell of e.m.f, E, and internal resistance, r connected across an external resistor, R, as shown in Figure 2.19 (a).
Figure 2.19: A simple circuit, (a) switch open and (b) switch closed
Suppose the switch is closed as shown in Figure 2.19 (b), current, I, flows through the circuit. From Ohm’s law, the p.d, V, across the resistor, R, is given by,
V=IR
The potential drop (V1) caused by the cell’s internal resistance is given by,
V1=Ir
The total voltage in the circuit is the sum of the potential drop due to the cell’s internal resistance and that of the resistor, R. That is,
E=V+Ir
When no current is drawn in the circuit, e.m.f is equal to p.d across the cell E=V. Generally, the total e.m.f of the cell (s) is given by,
E=V+V1=IR+IrE=I(R+r)
The equation E=I(R+r) can be used to determine the internal resistance of a cell and the e.m.f of the cell experimentally. From Figure 2.19, different R(Ω) values can be used to give different values of current, I (A), and the two sets of values are plotted as shown in Figure 20.20 based on the equation:
R=E(1I)−r
With R in the vertical axis and the reciprocal of current in the horizontal axis.
Figure 2.20: The graph of resistance, R, against the reciprocal of current
The numerical value of e.m.f and r can be determined from the graph. The slope of the graph represents the value of e.m.f, E, and the vertical intercept represents the value of internal resistance, r, of the cell. Similarly, when the axes of the variables are interchanged, the equation becomes:
1I=1ER+rE
Therefore, the slope of this graph (Figure 2.21) represents the reciprocal of the cell’s e.m.f, and the vertical intercept represents the product of the slope and the cell’s internal resistance.
Figure 2.21: The graph of the reciprocal of current against resistance
The following activity 2.3 will aid on practical understanding in measurement of e.m.f and internal resistance.
Activity 2.3Aim: To determine the electromotive force, E, and internal resistance, r , of a cell Materials: A cell, a set of standard resistors (2−10 Ω), an ammeter, a switch Procedure
Figure 2.22 2. Set R at 10 Ω, close switch S and read the ammeter and record current, I. 3. Repeat step 2 for R=8 Ω,6 Ω,4 Ω and 2 Ω. 4. Using a spreadsheet or otherwise, record results as shown in Table 2.6.
Questions (a) Use ICT software or otherwise to plot a graph of R against 1I. (b) Determine the vertical intercept. (c) What is the value of r ? (d) Determine the slope of the graph. (e) What is the value of E ? |
Example 2.6An electric cell with minimal voltage, Em, has a resistance of 3 Ω connected across it. If the voltage falls to 0.6E, calculate the internal resistance of this cell. Solution Given: From E=I(R+r)E=0.6E+IrIr=E−0.6EIr=0.4E From Ohm’s law V=IR3I=0.6EI=0.6E3 Substituting Equation (2) into (1) and solving for r, you get r=2 Ω. |
Exercise 2.11. A student measures the voltage of a cell in two scenarios. In scenario 1, he connects the voltmeter directly to the cell and records a value of 1.5 V . In scenario 2, he adds an unknown resistor to the circuit and records a voltage of 1.32 V . (a) What conclusions can be drawn from these results? (b) Explain the terms associated with the values recorded in scenarios 1 and 2 . 2. The following results were obtained in an experiment to determine the value of resistance.
From the experimental results, using ICT tools or otherwise: (a) Plot a graph of V against I (b) Determine the value of resistance R 3. Study Figure 2.23 and the given instructions to answer the questions that follow.
Figure 2.23
(a) Why does the reading on the voltmeter drop when the switch is closed? (b) Calculate the unknown external load resistance R. (c) Calculate the internal resistance 4. A student sets up the schematic diagram as in Figure 2.24 to investigate the internal resistance of a cell where R=2 Ω is a fixed resistor, r is the internal resistance of the cell, A is an ammeter and V is a voltmeter across the cell. When the switch is closed, the ammeter reads 0.4 A , the voltmeter reads 1.2 V . If the emf, E of the cell is known to be 1.6 V : (a) Draw the circuit diagram clearly (b) Calculate the effective resistance of the entire circuit. (c) Use the readings to calculate the internal resistance r of the cell.
Figure 2.24 |
Potential difference across a conductor, current and resistance
Performing Activity 2.4 can determine the relationship between p.d across a conductor, the current flowing through the conductor, and the conductor’s resistance and performing activity 2.5 one can determine the unknown resistance in circuit by ohms law.
Activity 2.4Aim: To determine the relationship between potential difference, current and resistance Materials: a d.c. power supply or 1.5 V cell, ammeter, bulb, switch, voltmeter, rheostat, connecting wires Procedure
Figure 2.25 2. With the switch open, measure and record the potential difference across the bulb and the current through the circuit by using the voltmeter and ammeter, respectively.
Questions |
Current flows when a cell is connected across the ends of a conductor. For a conductor such as copper wire, the current flowing through it is directly proportional to the potential difference across the ends of the conductor. This complies with Ohm’s law, which states that “The current passing through a conductor at constant temperature is proportional to the potential difference between its ends.”
That is,
I∝V or I=kV
where k is the constant of proportionality called conductance of a conductor, denoted by G. Hence,
G=IV; 1G=VI
Nevertheless, the reciprocal of conductance is the resistance of the conductor. That is,
1G=R.
Similarly, Ohm’s law can be expressed as:
V∝I it implies V=IR
where R is the constant of proportionality called resistance of a conductor. Hence,
R=VI
The ohm can be defined as the resistance of a conductor such that, when a potential difference of 1 volt is applied to two points of a conductor, a current of 1 ampere flows through it. Therefore,
ohm=voltampere
By keeping temperature and other physical properties constant, the resistance (R) of a conductor remains constant. This relationship enables us to determine currents and voltages in electric circuits.
Aim: To determine the value of the unknown resistance by using Ohm’s law Procedure
Figure 2.26 2. Close switch S and adjust the rheostat, Rh, so that a current of 0.1 A passes through the unknown resistor, R.
Questions |
A cell develops a potential difference of 2 V across a resistor of 4 ohms. Calculate the current flowing through the resistor and the conductance of the resistor. Solution Using Ohm’s law, V=IR, I=VRI=2 V4 Ω=0.5 A Conductance G is the reciprocal of resistance G=1R=14 Ω=0.25 Ω−1 Therefore, the current flowing in the resistor is 0.5 A and the conductance of the resistor is 0.25 Ω−1. |
Factors that determine the resistance of a conductor
The resistance of a conductor is determined by its temperature, length, cross-sectional area and type of material.
Temperature
As temperature rises, the atoms in the conductor vibrate more, leading to increased collisions between electrons and atoms, which raises resistance. In some materials, the resistance varies almost linearly with temperature. For instance, the resistance of copper ( Cu ) varies approximately linearly with temperature. For some other materials, resistance does not vary linearly with temperature. However, the effect of temperature on the resistance of some alloys, such as constantan and manganin, is minimal.
Length of a conductor
When the length of a conductor is increased, while other factors are kept constant, the resistance of the conductor also increases. This is because electrons and atoms collide more in a long conductor than in a short conductor. This means that the resistance (R) of the wire is proportional to the length (l) of the wire. That is,
R∝l
Cross-sectional area
A conductor with a larger cross-sectional area has more charge carriers to carry the electrical current than of a smaller cross area. This means the resistance R of a conductor is inversely proportional to the cross-sectional area A of the conductor, that is,
R∝1A
Nature of material
The resistance of a conductor also depends on the material used to make the conductor. For example, a conductor made from steel will have higher resistance than one made of copper of identical dimensions at the same temperature. For example, steel has a higher resistivity than copper. A material’s property that resists current flow is known as the material’s resistivity, denoted by ρ. Since, R∝l and R∝1A; then,
R∝lA
Hence,
R=lA
where r is the constant of proportionality. This constant of proportionality is the resistivity of the material, given as:
ρ=RAl
Resistivity is, therefore, the measure of the ability of a material to oppose the flow of an electric current. The SI unit of resistivity is the ohm-metre (Ωm). Activity 2.6 will aid in developing practical knowledge on determination of resistivity of wires.
Aim: To determine the resistivity of constantan wire Materials: battery, ammeter, switch, voltmeter, rheostat, connecting wires, micrometer screw gauge, a constantan wire of length 20 cm Procedure
Figure 2.27 2. Close the switch S and adjust the rheostat so that the ammeter reads 0.2 A . Table 2.3
6. Measure the diameter, D, of the constantan wire and calculate its cross-sectional area, A. Questions Different materials have different values of resistivity. The resistivities of some materials are given in Table 2.4.
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What is the resistance of a copper wire of length 20 m and a diameter of 0.080 cm ? (Resistivity of copper, ρcu=1.68×10−8Ω m.) Solution Given that But A=πr2 Hence, Therefore, the resistance of the wire is 0.67 Ω. |
A nichrome wire has a cross-sectional area of 4×10−8 m2 and a resistivity of 1.0×10−6 Ω m. If a resistor of resistance 11 Ω is to be made from this wire, calculate the length of the required wire. Solution Given that, Therefore, the length of the wire is 0.44 m. |
A constantan wire has a length of 45 cm, a diameter of 0.37 mm and a resistivity of 4.9×10−7 Ω m. Solution (a) ρ=4.9×10−7 Ω m and l=0.45 m But, R=ρlA and A=πr2; where, Hence, Resistance of the wire is 2.05 Ω. Therefore, the current flowing in the wire will be 0.73 A. |
Measurement of resistance
Resistance measurement is essential in electrical circuits. Two accurate methods used are the Wheatstone bridge, which balances known and unknown resistances, and the potentiometer, which compares voltage drops to determine resistance. These methods are clearly explained in the sections below, providing step-by-step procedures, principles of operation, and practical applications to enhance understanding.
Wheatstone bridge
From Activity 2.5, Ohm’s law was used to determine the value of the unknown resistance of a conductor. A similar approach cannot be applied to accurately measure a very low value of resistance in the milli-Ohms (mΩ) range. However, when resistors are connected in the series-parallel arrangement as shown in Figure 2.28, the value of the unknown resistance of a conductor, even in mΩ, can be measured. This diamond-like arrangement is called the Wheatstone bridge circuit. It consists mainly of four resistors known as the bridge arms and a galvanometer in between them.
Figure 2.28: The Wheatstone bridge
Working principle of a Wheatstone bridge
To understand how the Wheatstone bridge works, let us consider a bridge circuit in Figure 2.29.
Figure 2.29: The Wheatstone circuit
If the resistors P=5 Ω,Q=1 Ω, R=10 Ω,X=2 Ω and J=6 Ω, what is the current flowing through the circuit?
The circuit in Figure 2.29 does not have any resistors arranged in series or in parallel. Thus, one cannot reduce the circuit to an equivalent resistance using parallel and series formulae. However, if you observe the circuit carefully, you realize that the resistors on one side of the joint resistor, J, are in the same ratio as the resistor on the other side of J. That is, PQ=5 Ω1 ΩandRX=10 Ω2 Ω=5
Therefore, PQ=RX. A bridge circuit that has this property is said to be a balanced Wheatstone bridge. Now, suppose the resistor J is removed from the circuit so that resistor P is in series with resistor R, and resistor Q is in series with resistor X. Moreover, the resistors P and R are in parallel with resistors Q and X. Therefore, the voltages at points E and F are the same. Since PQ=RX, the voltage drop at P is the same as the voltage drop at Q, meaning that voltage at point B(VB) is the same as voltage at point C(VC). This means the potential difference between points B and C is zero. Thus, when the resistor J is connected, current through it is zero. That is, when the bridge circuit is balanced, IJ=0 A. Suppose a resistor J is replaced with a galvanometer G as shown in Figure 2.29. The variable resistor RV can be varied to a point where the current through the galvanometer IG=0 A.
At this point, the bridge circuit is balanced, and therefore, PQ=RVX which means, X=QRVP
Performing Activity 2.7 gives more understanding on Wheatstone bridge.
Aim: To determine the resistance of an unknown resistor X, using a Wheatstone bridge. Procedure Questions |
Slide wire bridge (Metre bridge)
A slide wire bridge (Metre bridge) is a modified Wheatstone bridge. It is made of one known resistor R, an unknown resistor RX, and a wire of uniform cross section area, as shown in Figure 2.30.
Figure 2.30: Meter bridge
The ratio RBRA is adjusted by sliding a jockey along the length of the wire. Because the wire has a uniform cross-sectional area, the ratio RBRA is equivalent to respective lengths LB and LA. That is, RBRA=ρLBAρLAA=LBLA
But RBRA=RXR⇒LBLA=RXR
This expression can therefore be written as, RX=LBLA×R This expression can be used to determine the value of the unknown resistor.
Activity 2.8 aids on performing experiments regarding finding unknown resistor by meter bridge.
Aim: To determine the resistance of an unknown resistor using the metre bridge Procedure
Figure 2.31 2. Set R to 1 Ω, then close the switch, S, and slide the jockey over the metre bridge wire until the galvanometer reads zero. Record length l1 and l2. Table 2.7
Questions |
A metre bridge is set up as shown in Figure 2.32 using a standard 10 Ω resistor. The galvanometer shows zero deflection when the jockey contact is at 48 cm from end A. Determine the resistance of a resistor, X.
Figure 2.32 Solution From the metre bridge expression, X=LALB×R where R=10 Ω,LA=48 cm,LB=52 cm Hence, X=48 cm52 cm×10 Ω=9.23 Ω Therefore, the resistance X is 9.23 Ω. |
Potentiometer
A potentiometer is a reliable tool for measuring e.m.f. It consists of a length AB of uniform resistance wire, with a steady current flowing through it from an accumulator C (Figure 2.33). Any length from A is read from a metre rule AB, and the length of AB may be one metre.
Figure 2.33: A potentiometer
On closing the switch, S, there will be a potential drop across AB and hence between point A and any other point along AB. This potential drop will produce a current in the galvanometer in the direction AGJ so that, in whatever position of J on AB, the galvanometer will be deflected. Suppose a second cell E is introduced in series with the galvanometer, its current in the direction JGA will now be opposed by the current due to the cell D flowing in the direction AGJ. There will now be two opposing p.d between A and J. It should be possible to find a position J of the jockey on AB for which the p.d across AJ due to E balances the p.d across AJ due to D. Activity 2.9 illustrates how one can compare e.m.f. of two cells by using potentiometer.
Aim: To compare the e.m.f of two cells using a potentiometer Materials: A slide-wire potentiometer, galvanometer, 2 dry cells, jockey, connecting wires, switch, accumulator Procedure
Figure 2.34 2. Join the positive terminal of P to A, the same terminal to which the positive terminal of C is joined. Questions Theoretically, the e.m.f E1 of cell P balances the p.d V1 on the wire because no current flows in G at the balance point N. As a result, E1=V1. Similarly, if E2 is the e.m.f of cell Q and V2 is the p.d between A and M, then E2=V2. As a result, E1E2=V1V2. The p.d V across any length l is directly proportional to l if the wire carries a constant current. As a result, E1E2=l1l2. Thus, the e.m.f of the two cells can be compared. |
The balance length of a potentiometer wire for a cell of e.m.f E1=1.63 V is 85 cm. If the cell is replaced by another one of e.m.f E2=1.07 V, calculate the new balance length. Solution Given E1=1.63 V, l1=85 cm, E2=1.07 V. Therefore, l2=E2E1×l1=1.07 V1.63 V×85 cm=55.79 cm The new balance length is 55.79 cm. |
1. A wire of length 1.2 m and diameter 0.64 mm has a resistance of 2.4 Ω. Calculate the resistance of a wire of length 0.80 m and diameter 0.32 mm of the same material. 2. A wire of length 2 m and a cross sectional area of 0.5 mm2 has a resistance of 2.2 Ω. Calculate the resistivity of the material making up the wire. 3. A battery of e.m.f 12 V and internal resistance 1.5 Ω is connected to a 4 Ω resistor. Calculate the: 4. The p.d across the terminals of a cell is 1.1 V when a current of 0.20 A is being drawn from the cell. If the p.d across the cell is 1.3 V when a current of 0.10 A is being drawn, determine the: 5. Two batteries, V1 and V2, are connected in series with an 8 Ω resistor as shown in Figure 2.35.
Figure 2.35 Given that the internal resistances of the two batteries are 2 Ω and 1 Ω, respectively, determine the p.d across the 8 Ω. 6. Look at the three circuit diagrams in Figure 2.36. Rank the circuits from brightest bulb(s) to dimmest bulb(s). Explain your choices.
Figure 2.36 7. A sub-woofer needs a household voltage of 220 V to push a current of 5.5 A through its coil. What is the resistance of the coil? |
TOPIC 2: CURRENT ELECTRICITY – PHYSICS FORM TWO
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