 Mathematics OHM’S LAW AND RESISTANCE OF A SYSTEM OF RESISTORS

### Topic Covered

color{red} ♦ OHM’S LAW
color{red} ♦FACTORS ON WHICH THE RESISTANCE OF A CONDUCTOR DEPENDS
color{red} ♦ RESISTANCE OF A SYSTEM OF RESISTORS

### OHM’S LAW

Is there a relationship between the potential difference across a conductor and the current through it?

Let us explore with an Activity.

ul"Activity 12.1"

♦ Set up a circuit as shown in Fig. 12.2, consisting of a nichrome wire XY of length, say 0.5 m, an ammeter, a voltmeter and four cells of 1.5 V each. (Nichrome is an alloy of nickel, chromium, manganese, and iron metals.) ♦ First use only one cell as the source in the circuit. Note the reading in the ammeter I, for the current and reading of the voltmeter V for the potential difference across the nichrome wire XY in the circuit. Tabulate them in the Table given.

♦Next connect two cells in the circuit and note the respective readings of the ammeter and voltmeter for the values of current through the nichrome wire and potential difference across the nichrome wire.

♦ Repeat the above steps using three cells and then four cells in the circuit separately.

♦ Calculate the ratio of V to I for each pair of potential difference V and current I. ♦ Graph between V and I, and observe the nature of the graph.

In this Activity, you will find that approximately the same value for V//I is obtained in each case. Thus the V–I graph is a straight line that passes through the origin of the graph, as shown in Fig. 12.3. Thus, V//I is a constant ratio. In 1827, a German physicist Georg Simon Ohm (1787–1854) found out the relationship between the current I, flowing in a metallic wire and the potential difference across its terminals.

He stated that the electric current flowing through a metallic wire is directly proportional to the potential difference V, across its ends provided its temperature remains the same. This is called Ohm’s law. In other words –

V \ \ ∝ \ \ I ........(12.4)
or  V//I =  constant
= R
or V = IR .........(12.5)

In Eq. (12.4), R is a constant for the given metallic wire at a given temperature and is called its resistance. It is the property of a conductor to resist the flow of charges through it.

Its SI unit is ohm, represented by the Greek letter Ω. According to Ohm’s law,
R = V//I .........(12.6)

If the potential difference across the two ends of a conductor is 1 V and the current through it is 1 A, then the resistance R, of the conductor is 1 Ω. That is, 1 ohm  = text(1 volt)/text( 1 ampere)

Also from Eq. (12.5) we get  I = V//R ..........(12.7)

It is obvious from Eq. (12.7) that the current through a resistor is inversely proportional to its resistance.

If the resistance is doubled the current gets halved. In many practical cases it is necessary to increase or decrease the current in an electric circuit.

A component used to regulate current without changing the voltage source is called variable resistance. In an electric circuit, a device called rheostat is often used to change the resistance in the circuit.

We will now study about electrical resistance of a conductor with the help of following Activity.

ul"Activity 12.2"

♦ Take a nichrome wire, a torch bulb, a 10 W bulb and an ammeter (0 – 5 A range), a plug key and some connecting wires.
♦ Set up the circuit by connecting four dry cells of 1.5 V each in series with the ammeter leaving a gap XY in the circuit, as shown in Fig. 12.4. ♦ Complete the circuit by connecting the nichrome wire in the gap XY. Plug the key. Note down the ammeter reading. Take out the key from the plug. [Note: Always take out the key from the plug after measuring the current through the circuit.]
♦ Replace the nichrome wire with the torch bulb in the circuit and find the current through it by measuring the reading of the ammeter.
♦ Now repeat the above step with the 10 W bulb in the gap XY.
♦ Are the ammeter readings differ for different components connected in the gap XY? What do the above observations indicate?
♦ You may repeat this Activity by keeping any material component in the gap. Observe the ammeter readings in each case. Analyse the observations.

In this Activity we observe that the current is different for different components. Why do they differ? Certain components offer an easy path for the flow of electric current while the others resist the flow.

We know that motion of electrons in an electric circuit constitutes an electric current. The electrons, however, are not completely free to move within a conductor.

They are restrained by the attraction of the atoms among which they move. Thus, motion of electrons through a conductor is retarded by its resistance. A component of a given size that offers a low resistance is a good conductor.

A conductor having some appreciable resistance is called a resistor. A component of identical size that offers a higher resistance is a poor conductor. An insulator of the same size offers even higher resistance.

### FACTORS ON WHICH THE RESISTANCE OF A CONDUCTOR DEPENDS

ul"Activity 12.3"

♦ Complete an electric circuit consisting of a cell, an ammeter, a nichrome wire of length l [say, marked (1)] and a plug key, as shown in Fig. 12.5. ♦ Now, plug the key. Note the current in the ammeter.
♦ Replace the nichrome wire by another nichrome wire of same thickness but twice the length, that is 2 l [marked (2) in the Fig. 12.5].
♦ Note the ammeter reading.
♦ Now replace the wire by a thicker nichrome wire, of the same length l [marked (3)]. A thicker wire has a larger cross-sectional area. Again note down the current through the circuit.
♦ Instead of taking a nichrome wire, connect a copper wire [marked (4) in Fig. 12.5] in the circuit. Let the wire be of the same length and same area of cross-section as that of the first nichrome wire [marked (1)]. Note the value of the current.
♦ Notice the difference in the current in all cases.
♦ Does the current depend on the length of the conductor?
♦ Does the current depend on the area of cross-section of the wire used?

It is observed that the ammeter reading decreases to one-half when the length of the wire is doubled. The ammeter reading is increased when a thicker wire of the same material and of the same length is used in the circuit.

A change in ammeter reading is observed when a wire of different material of the same length and the same area of cross-section is used. On applying Ohm’s law [Eqs. (12.5) – (12.7)], we observe that the resistance of the conductor depends (i) on its length, (ii) on its area of cross-section, and (iii) on the nature of its material.

Precise measurements have shown that resistance of a uniform metallic conductor is directly proportional to its length (l) and inversely proportional to the area of cross-section (A). That is,

R \ \ ∝ \ \ l .........(12.8)

and R \ \ ∝ \ \ 1//A ............(12.9)
Combining Eqs. (12.8) and (12.9) we get

 R \ \ alpha \ \ l/A
or  R = rho l/A  ............(12.10)

where ρ (rho) is a constant of proportionality and is called the electrical resistivity of the material of the conductor.

The SI unit of resistivity is Ω  m. It is a characteristic property of the material. The metals and alloys have very low resistivity in the range of 10^(–8) Ω m to 10^(–6) Ω m.

They are good conductors of electricity. Insulators like rubber and glass have resistivity of the order of 10^(12) to 10^(17) Ω m. Both the resistance and resistivity of a material vary with temperature.

Table 12.2 reveals that the resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidise (burn) readily at high temperatures.

For this reason, they are commonly used in electrical heating devices, like electric iron, toasters etc. Tungsten is used almost exclusively for filaments of electric bulbs, whereas copper and aluminium are generally used for electrical transmission lines. Q 3214591450 (a) How much current will an electric bulb draw from a 220 V source, if the resistance of the bulb filament is 1200 Ω?
(b) How much current will an electric heater coil draw from a 220 V source, if the resistance of the heater coil is 100 Ω?

Class 10 Chapter 12 Example 3 Solution:

(a) We are given V = 220 V; R = 1200 Ω.
From Eq. (12.6), we have the current I = 220 V//1200 Ω = 0.18 A.

(b) We are given, V = 220 V, R = 100 Ω.
From Eq. (12.6), we have the current I = 220 V//100 Ω = 2.2 A.
Note the difference of current drawn by an electric bulb and electric heater from the same 220 V source !
Q 3224591451 The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 120 V?
Class 10 Chapter 12 Example 4 Solution:

We are given, potential difference V = 60 V, current I = 4 A.

According to Ohm’s law,  R = V/I = (60 V)/(4A) = 15 Omega

When the potential difference is increased to 120 V the current is given by

current  = V/R = (120 V)/(15) = 8 A.

The current through the heater becomes 8 A.
Q 3234591452 Resistance of a metal wire of length 1 m is 26 Ω at 20°C. If the diameter of the wire is 0.3 m m, what will be the resistivity of the metal at that temperature? Using Table 12.2, predict the material of the wire.
Class 10 Chapter 12 Example 5 Solution:

We are given the resistance R of the wire = 26 Ω, the diameter
d = 0.3 m m = 3 xx 10^(-4) m, and the length l of the wire = 1 m.
Therefore, from Eq. (12.10), the resistivity of the given metallic wire is
ρ = (RA//l) = (R π d^2 // 4 l )
Substitution of values in this gives
ρ = 1.84 xx 10^(–8) Ω m
The resistivity of the metal at 20°C is 1.84 xx 10^(–8) Ω m. From
Table 12.2, we see that this is the resistivity of manganese.
Q 3244591453 A 4 Ω resistance wire is doubled on it. Calculate the new resistance of the wire.

Class 10 Chapter 12 Example 6 Solution:

We are given, R = 4 Ω.
When a wire is doubled on it, its length would become half and
area of cross-section would double. That is, a wire of length l and
area of cross-section A becomes of length l//2 and area of crosssection
2A. From Eq. (12.10), we have

 R = rho l/A

R_1 = rho ( l//2)/(2A)

where R_1 is the new resistance.
Therefore,  R_1/R = rho (l//2)/(2 A) // rho l/A = 1/4

or ,  R_1 = R/4 = (4 Omega)/4 = 1 Omega.

The new resistance of the wire is 1 Ω.

### RESISTANCE OF A SYSTEM OF RESISTORS

In preceding sections, we learnt about some simple electric circuits. We have noticed how the current through a conductor depends upon its resistance and the potential difference across its ends.

In various electrical gadgets, we often use resistors in various combinations. We now therefore intend to see how Ohm’s law can be applied to combinations of resistors. There are two methods of joining the resistors together.

Figure 12.6 shows an electric circuit in which three resistors having resistances R_1, R_2 and R_3, respectively, are joined end to end. Here the resistors are said to be connected in series. Figure 12.7 shows a combination of resistors in which three resistors are connected together between points X and Y.
Here, the resistors are said to be connected in parallel. ### Resistors in Series

What happens to the value of current when a number of resistors are connected in series in a circuit? What would be their equivalent resistance? Let us try to understand these with the help of the following activities.

ul"Activity 12.4"

♦ Join three resistors of different values in series. Connect them with a battery, an ammeter and a plug key, as shown in Fig. 12.6. You may use the resistors of values like 1 Ω, 2 Ω, 3 Ω etc., and a battery of  6 V for performing this Activity.
♦ Plug the key. Note the ammeter reading.
♦ Change the position of ammeter to anywhere in between the resistors. Note the ammeter reading each time.
♦ Do you find any change in the value of current through the ammeter?

You will observe that the value of the current in the ammeter is the same, independent of its position in the electric circuit

It means that in a series combination of resistors the current is the same in every part of the circuit or the same current through each resistor.

ul"Activity 12.5"

♦ In Activity 12.4, insert a voltmeter across the ends X and Y of the series combination of three resistors, as shown in Fig. 12.8.
♦ Plug the key in the circuit and note the voltmeter reading. It gives the potential difference across the series combination of resistors. Let it be V. Now measure the potential difference across the two terminals of the battery. Compare the two values.
♦ Take out the plug key and disconnect the voltmeter. Now insert the voltmeter across the ends X and P of the first resistor, as shown in Fig. 12.8. ♦ Plug the key and measure the potential difference across the first resistor. Let it be V_1.
♦ Similarly, measure the potential difference across the other two resistors, separately. Let these values be V_2 and V_3, respectively.
♦ Deduce a relationship between V, V_1 , V_2 and V_3.

You will observe that the potential difference V is equal to the sum of potential differences V_1, V_2, and V_3. That is the total potential difference across a combination of resistors in series is equal to the sum of potential difference across the individual resistors.

That is, V = V_1 + V_2 + V_3 ........(12.11)

In the electric circuit shown in Fig. 12.8, let I be the current through the circuit. The current through each resistor is also I.

It is possible to replace the three resistors joined in series by an equivalent single resistor of resistance R, such that the potential difference V across it, and the current I through the circuit remains the same.

Applying the Ohm’s law to the entire circuit, we have

V = I R ............(12.12)

On applying Ohm’s law to the three resistors separately, we further have

 V_1 = I R_1 ..........[12.13(a)]

V_2 = I R_2 ...........[12.13(b)]

and V_3 = I R_3 ...........[12.13(c)]

From Eq. (12.11),

I R = I R_1 + I R_2 + I R_3
or

R_s = R_1 +R_2 + R_3 ..........(12.14)

We can conclude that when several resistors are joined in series, the resistance of the combination R_s equals the sum of their individual resistances, R_1, R_2, R_3, and is thus greater than any individual resistance.

Q 3254591454 An electric lamp, whose resistance is 20 Ω, and a conductor of 4 Ω resistance are connected to a 6 V battery (Fig. 12.9). Calculate (a) the total resistance of the circuit, (b) the current through the circuit, and (c) the potential difference across the electric lamp and conductor.
Class 10 Chapter 12 Example 7 Solution:

The resistance of electric lamp, R_1 = 20 Ω,
The resistance of the conductor connected in series, R_2 = 4 Ω.
Then the total resistance in the circuit
R = R_1 + R_2
R_s = 20 Ω + 4 Ω = 24 Ω.
The total potential difference across the two terminals of the battery
V = 6 V.
Now by Ohm’s law, the current through the circuit is given by
I = V//R_s
= 6 V//24 Ω
= 0.25 A.

Applying Ohm’s law to the electric lamp and conductor separately,
we get potential difference across the electric lamp,
V_1 = 20 Ω xx 0.25 A
= 5 V;
and,
that across the conductor, V_2 = 4 Ω xx 0.25 A
= 1 V.
Suppose that we like to replace the series combination of electric lamp and conductor by a single and equivalent resistor. Its resistance must be such that a potential difference of 6 V across the battery terminals will cause a current of 0.25 A in the circuit. The resistance R of this equivalent resistor would be
R = V//I
= 6 V// 0.25 A
= 24 Ω.
This is the total resistance of the series circuit; it is equal to the sum of the two resistances.

### Resistors in Parallel

Now, let us consider the arrangement of three resistors joined in parallel with a combination of cells (or a battery), as shown in Fig.12.7.

ul"Activity 12.6"

♦ Make a parallel combination, XY, of three resistors having resistances R_1, R_2, and R_3, respectively. Connect it with a battery, a plug key and an ammeter, as shown in Fig. 12.10. Also connect a voltmeter in parallel with the combination of resistors. ♦ Plug the key and note the ammeter reading. Let the current be I. Also take the voltmeter reading. It gives the potential difference V, across the combination. The potential difference across each resistor is also V. This can be checked by connecting the voltmeter across each individual resistor (see Fig. 12.11).

♦ Take out the plug from the key. Remove the ammeter and voltmeter from the circuit. Insert the ammeter in series with the resistor R_1, as shown in Fig. 12.11. Note the ammeter reading, I_1. ♦ Similarly, measure the currents through R_2 and R_3. Let these be I_2 and I_3, respectively. What is the relationship between I, I_1, I_2 and I_3?

It is observed that the total current I, is equal to the sum of the separate currents through each branch of the combination.

I = I_1 + I_2 + I_3 ........(12.15)

Let R_p be the equivalent resistance of the parallel combination of resistors.
By applying Ohm’s law to the parallel combination of resistors,

we have

I = V//R_p .........(12.16)

On applying Ohm’s law to each resistor, we have

I_1 = V //R_1 ; I_2 = V // R_2 ; and I_3 = V // R_3 ........(12.17)

From Eqs. (12.15) to (12.17), we have

V//R_p = V//R_1 + V//R_2 + V//R_3

or

1//R_p = 1//R_1 + 1//R_2 + 1//R_3 .........(12.18)

Thus, we may conclude that the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances.

We have seen that in a series circuit the current is constant throughout the electric circuit. Thus it is obviously impracticable to connect an electric bulb and an electric heater in series, because they need currents of widely different values to operate properly.

Another major disadvantage of a series circuit is that when one component fails the circuit is broken and none of the components works.
If you have used ‘fairy lights’ to decorate buildings on festivals, on marriage celebrations etc., you might have seen the electrician spending lot of time in trouble-locating and replacing the ‘dead’ bulb each has to be tested to find which has fused or gone.

On the other hand, a parallel circuit divides the current through the electrical gadgets. The total resistance in a parallel circuit is decreased as per Eq. (12.18). This is helpful particularly when each gadget has different resistance and requires different current to operate properly.
Q 3264591455 In the circuit diagram given in Fig. 12.10, suppose the resistors R_1, R_2 and R_3 have the values 5 Ω, 10 Ω, 30 Ω, respectively, which have been connected to a battery of 12 V. Calculate (a) the current through each resistor, (b) the total current in the circuit, and (c) the total circuit resistance.
Class 10 Chapter 12 Example 8 Solution:

R_1 = 5 Ω, R_2 = 10 Ω, and R_3 = 30 Ω.

Potential difference across the battery, V = 12 V.
This is also the potential difference across each of the individual resistor; therefore, to calculate the current in the resistors, we use
Ohm’s law.
The current I_1, through R_1 = V// R_1
I_1 = 12 V//5 Ω = 2.4 A.

The current I_2, through R_2 = V// R_2
I_2 = 12 V//10 Ω = 1.2 A.
The current I_3, through R_3 = V//R_3
I_3 = 12 V//30 Ω = 0.4 A.
The total current in the circuit,
I = I_1 + I_2 + I_3
= (2.4 + 1.2 + 0.4) A
= 4 A
The total resistance R_p, is given by [Eq. (12.18)]

 1/R_p = 1/5 + 1/(10) + 1/(30) = 1/3

Thus, R_p = 3 Ω.
Q 3274591456 If in Fig. 12.12, R_1 = 10 Ω, R_2 = 40 Ω, R_3 = 30 Ω, R_4 = 20 Ω, R_5 = 60 Ω, and a 12 V battery is connected to the arrangement. Calculate (a) the total resistance in the circuit, and (b) the total current flowing in the circuit.
Class 10 Chapter 12 Example 9 Solution:

Suppose we replace the parallel resistors R_1 and R_2 by an
equivalent resistor of resistance, R′. Similarly we replace
the parallel resistors R_3, R_4 and R_5 by an equivalent single
resistor of resistance R″. Then using Eq. (12.18), we have
1// R′ = 1//10 + 1//40 = 5//40; that is R′ = 8 Ω.
Similarly, 1// R″ = 1//30 + 1//20 + 1//60 = 6//60;
that is, R″ = 10 Ω.
Thus, the total resistance, R = R′ + R″ = 18 Ω.
To calculate the current, we use Ohm’s law, and get
I = V//R = 12 V//18 Ω = 0.67 A.

### Note

We have seen that in a series circuit the current is constant throughout the electric circuit.

Thus it is obviously impracticable to connect an electric bulb and an electric heater in series, because they need currents of widely different values to operate properly (see Example 12.3).

Another major disadvantage of a series circuit is that when one component fails the circuit is broken and none of the components works.

If you have used ‘fairy lights’ to decorate buildings on festivals, on marriage celebrations etc., you might have seen the electrician spending lot of time in trouble-locating and replacing the ‘dead’ bulb – each has to be tested to find which has fused or gone. On the other hand, a parallel circuit divides the current through the electrical gadgets.

The total resistance in a parallel circuit is decreased as per Eq. (12.18). This is helpful particularly when each gadget has different resistance and requires different current to operate properly. 