DEPENDING FACTORS OF RESISTANCE OF A CONDUCTOR

DEPENDING FACTORS OF RESISTANCE OF A CONDUCTOR

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.

Figure 12.5: Electric circuit to study the factors on which the resistance of the conductor depends

*  Now, plug the key. Note the current in the ammeter.

*  Replace the nichrome wire by another nichrome wire of the same thickness but twice the length, that is 2l [marked (2) in 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)],

V  =  IR           ............. (12.5)

R = V/I            ............. (12.6)

I = V/R            ............. (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   \(\propto\)   l                     ............. (12.8)

and R   \(\propto\)   1/A          ............. (12.9)

Combining Eqs. (12.8) and (12.9) we get

R   \(\propto\)   \(\frac{l}{A}\)         

or  R  = \(\rho\frac{l}{A}\)              ............. (12.10)

where \(\rho \)(rho) is a constant of proportionality and is called the electrical resistivity of the material of the conductor. The SI unit of resistivity is \(\Omega \)m. It is a characteristic property of the material. The metals and alloys have very low resistivity in the range of 10^–8 \(\Omega \)m to 10^–6 \(\Omega \)m. They are good conductors of electricity. Insulators like rubber and glass have a resistivity of the order of 10¹² to 10¹⁷ \(\Omega \)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.

Table 12.2: Electrical resistivity* of some substances at 20°C. (* You need not memorise these values. You can use these values for solving numerical problems.)

Illustration 12.3:

(a) How much current will an electric bulb draw from a 220 V source, if the resistance of the bulb filament is 1200 \(\Omega\)?

(b) How much current will an electric heater coil draw from a 220 V source, if the resistance of the heater coil is 100 \(\Omega\)?

Sol:

We are given V = 220 V; R = 1200 \(\Omega\).

From Eq. (12.6), we have the current I = 220 V/1200 \(\Omega\)= 0.18 A.

We are given, V = 220 V, R = 100 \(\Omega\).

From Eq. (12.6), we have the current I = 220 V/100 \(\Omega\) = 2.2 A. Note the difference of current drawn by an electric bulb and electric heater from the same 220 V source!

Illustration 12.4:

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?

Sol:

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

According to Ohm’s law, \(R=\frac{V}{I}=\frac{60V}{4A}=15\Omega\)

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

current =\(I=\frac{V}{R}=\frac{120V}{15\Omega}=8A\)

The current through the heater becomes 8 A.

Illustration 12.5:

Resistance of a metal wire of length 1 m is 26 \(\Omega\) at 20°C. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature? Using Table 12.2, predict the material of the wire.

Sol:

We are given the resistance R of the wire = 26 \(\Omega\), the diameter

d = 0.3 mm = 3 \(\times\) 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

\(\rho\)= (RA/l ) = (R\(\pi\)d²/4l ) Substitution of values in this gives

\(\rho\)= 1.84 × 10^–6 \(\Omega\)m

The resistivity of the metal at 20°C is 1.84 × 10^–6 \(\Omega\)m. From Table 12.2, we see that this is the resistivity of manganese.

Illustration 12.6:

A wire of given material having length l and area of cross-section A has a resistance of 4\(\Omega\) What would be the resistance of another wire of the same material having length l/2 and area of cross-section 2A?

Sol:

For first wire

R₁    =   \(\rho\frac{l}{A}=4\Omega\)  

Now for the second wire

R₂    = \(\rho\frac{l/2}{2A}=\frac{1}{4}\rho\frac{l}{A}\)

R₂    =\(\frac{1}{4}R_1\) 

R₂    = \(1\Omega\)  

The resistance of the new wire is 1\(\Omega\)

Questions

1.  On what factors does the resistance of a conductor depend?

2.  Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

3.  Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

4.  Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

5.  Use the data in Table 12.2 to answer the following –

            (a)  Which among iron and mercury is a better conductor?

            (b)  Which material is the best conductor?

Source: This topic is taken from NCERT TEXTBOOK