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From the study of subatomic particles to the laws of motion, Physics Topics offer insights into the workings of the world around us.
What is Resistance? What is the formula for specific resistance or resistivity?
The resistance of a conductor (R) under constant temperature depends on the length (l), cross sectional area (A) and material of the conductor.
1. Rule of length: The resistance of a conductor is direcdy proportional to its length if the area of cross section remains same, i.e.,
R ∝ l, when A is constant
2. Rule of cross section: The resistance of a conductor is inversely proportional to its area of cross section if the length of the conductor remains unchanged, i.e.,
R ∝ \(\frac{1}{A}\) when l is constant A
∴ R ∝ \(\frac{l}{A}\) or, R = \(\rho \cdot \frac{l}{A}\) …. (i)
Here, ρ is the constant of proportionality, called resistivity or specific resistance of the material of the conductor. Its value depends on the material of the conductor.
If l = 1 and A = 1 then from equation (1) it is obtained R = ρ.
i.e., resistance of the conductor becomes numerically equal to its resistivity when both Z and A are in unity. This leads to the definition of resistivity.
Definition: Electrical resistance of a conductor of unit cross sectional area and unit length is defined as the resistivity of the material of the conductor.
Unit of resistivity: As ρ = \(\frac{R A}{l}\)
unit of ρ = \(\frac{\mathrm{ohm} \cdot \mathrm{cm}^2}{\mathrm{~cm}}\) = ohm ᐧ cm (Ω ᐧ cm)
If length is expressed in metre, the SI unit is ohm-metre (Ω ᐧ m).
1 Ω ᐧ m = 1Ω ᐧ 100 cm = 100 Ω ᐧ cm
For example, resistivity of copper at 20 °C is 1.76 × 10-6 Ω ᐧ cm means that resistance across a copper conductor of cross sectional area 1 cm2 and length 1 cm at 20 °C is 1.76 × 10-6 Ω.
Resistivity of different substances:
Resistivity of a few conductors and insulators at 20°C is given below.
From the above table it appears that resistivity of silver is the lowest. This means, silver is the best conducting substance. However, since silver is costly, good conductors are mostly made of copper.
Resistivity and conductivity: If an amount of charge Q passes through a conductor in time t, then current I = \(\frac{Q}{t}\). If the length of the conductor is l, cross sectional area is A and potential difference between its two ends is VA – VB, then
VA – VB = IR = \(\frac{Q}{t} \cdot \frac{\rho l}{A}\) or, Q = \(\frac{1}{\rho} \cdot \frac{A\left(V_A-V_B\right) t}{l}\)
This relation is identical with that of conduction of heat,
Q = \(\frac{K A\left(T_2-T_1\right) t}{l}\)
So, if \(\frac{1}{\rho}\) is taken as σ, then σ may be analogically called electrical conductivity. It is obvious that the two physical quantities, conductivity and resistivity are the reverse of each other. Hence another way of saying that silver is the best conductor is to say that silver has the least resistivity.
An analogous term conductance has been introduced as a counter to resistance. The concepts obviously bear reciprocal relation. The SI unit of conductance is siemens (S) which was formerly called reciprocal ohm or mho (ʊ). If Y be the conduc-tance, then Y = \(\frac{1}{R}\).
As conductance is opposite to resistance, the conductance of a body decreases for the factors for which its resistance increases. Like resistance, the conductance of a body depends on its tem-perature and other physical conditions.
An important comment regarding magnitude of resistance: Consider a copper wire of length 1 m, cross sectional area 1 mm2 or 0.01 cm2. The magnitude of its resistance is
R = ρ\(\frac{l}{A}\) = 1.76 × 10-6 × \(\frac{100}{0.01}\) = 0.0176 Ω
So the resistance of the wire is very negligible. If a wire of about 57 m in length having the same cross section is taken, its resis-tance will be only 1 Ω. So this type of wire cannot be regarded as a resistor. Such wires are used for connecting purposes in an electrical circuit. So it can be taken for granted that the wires which connect different electrical instruments in an electrical circuit have almost no resistance.
Short circuit: The circuit obtained by connecting the two electrodes of a source of electricity directly with the help of a wire having almost no resistance is called short circuit. For example, if the positive and the negative terminals of a battery are connected by a copper wire, the wire becomes very hot. As the resistance of such a wire is very small, current passing through it is hign. So, there is a chance of damage of both the connecting wire and the source of electricity. To avoid short circuit, fuse wire is used in domestic electrical lines.
Numerical Examples
Example 1.
The length, radius and resistivity of two wires are each in the ratio 1:3. The resistance of the comparatively thin wire is 20 ft. Determine the resistance of the other wire.
Solution:
Example 2.
If the length of a copper wire is increased by 0.1%, show that the resistance of the copper wire will increase by 0.2%.
Solution:
If the initial length be , then the final length will be
l2 = l1 × (100.1%) = l1 × \(\frac{100.1}{100}\) = 1.001 l1
If the temperature of the wire remains constant, its volume will remain constant.
Now, volume = length × cross sectional area
So if the length increases from l1 to 1.001 l1 cross sectional area
A1 decreases to A2, where A2= \(\frac{A_1}{1.001}\).
Now, R1 = \(\rho \frac{l_1}{A_1}\) and R2 = \(\rho \cdot \frac{l_2}{A_2}\)
∴ \(\frac{R_2}{R_1}\) = \(\frac{l_2}{l_1} \times \frac{A_1}{A_2}\) = 1.001 × 1.001 = 1.002 = 100.2%
i.e., increase of resistance = 0.2% .
Alternative Method:
R = \(\rho \frac{l}{A}\)
By logarithmic differentiation, Now, V = lA = constant
By logarithmic differentiation, \(\frac{d l}{l}\) = –\(-\frac{d A}{A}\)
Hence \(\frac{d R}{R}\) = \(\frac{2 d l}{l}\) = 2 × 0.1% = 0.2%
Example 3.
A lump of copper of mass 10 g and of density 9 g ᐧ cm-3 is given. What should be the length and cross section of the wire made from it so that its resistance is 2 ohm. (Given, specific resistance of copper 1.8 × 10-6 Ω ᐧ cm ). [HS (XI) ’07]
Solution:
\(\frac{\text { mass }}{\text { density }}\) = volume = length × cross sectional area
or, \(\frac{10}{9}\) = lA or lA = \(\frac{10}{9}\) …… (1)
Again, resistance, R = \(\rho \frac{l}{A}\) or, 2 = 1.8 × 10-6 × \(\frac{l}{A}\)
or, \(\frac{l}{A}\) = \(\frac{2}{1.8} \times 10^6\) = \(\frac{10}{9} \times 10^6\) ……….. (2)
Now, by multiplying (1) and (2) we get, A2 = 10-6
Again dividing (1) by (2) we get, A2 = 10-6
or, A = 10-3 cm2 = 10-1 mm2 = 0.1 mm2
Example 4.
A wire of resistance 5 Ω is stretched 20% . If the vol-ume remains constant, find the new resistance.
Solution:
Volume of the wire = V = constant . Let the initial length of the wire = l
Initial cross sectional area, A1 = \(\frac{V}{l}\)
If the length of the wire is increased by 20%,
final length = l × \(\frac{120}{100}\) = 1.2l
∴ Final cross sectional area, A2 = \(\frac{V}{1.2 l}\)
Now, R1 = \(\rho \cdot \frac{l_1}{A_1}\) and R2 = \(\rho \cdot \frac{l_2}{A_2}\)
∴ \(\frac{R_2}{R_1}\) = \(\frac{l_2}{l_1} \times \frac{A_1}{A_2}\)
Example 5.
A lump of copper is stretched into a wire 5 mm in diameter. Another wire of 1 cm diameter is made from another lump of copper of the same mass. Find the ratio of the resistances of the two wires.
Solution:
5 mm = 0.5 cm
Here, ρ = resistivity, l = length, d = diameter,
A = \(\frac{\pi d^2}{4}\) = cross sectional area, V = lA = volume, m = mass, D = \(\frac{m}{V}\) = density
Both the lumps have same D and p and according to the question, mass m is equal.
∴ \(\frac{R_1}{R_2}\) = \(\left(\frac{d_2}{d_1}\right)^4\) = \(\left(\frac{1}{0.5}\right)^4\) = (2)4 = 16 or, R1 = R2 = 16 : 1
Example 6.
The length of a wire of cylindrical cross section is Increased by 100%. Find out the percentage change in the resistance, taking into account the consequent decrease in the diameter of the wire.
Solution:
Initial length = l; final length, l’ = 1 + l × \(\frac{100}{100}\) = 2l
If V be the volume, area of cross section, A = \(\frac{V}{l}\)
∴ Final area of cross section, A’ = \(\frac{V}{2 l}\)
From the formula R = \(\rho \frac{l}{A}\) we get,
\(\frac{R}{R^{\prime}}\) = \(\frac{l}{l^{\prime}} \cdot \frac{A^{\prime}}{A}\) = \(\frac{l}{2 l} \cdot \frac{V / 2 l}{V / l}\) = \(\frac{1}{4}\)
∴ R’ = 4R = R + 3R = R + \(\frac{300}{100} R\)
i.e., percentage change in resistance = 300%
Example 7.
What will be the resistance of a semicircle [Fig.] between points A and B? Given that radiai thickness = 3 cm, axial thickness 4 cm, inner radius = 6 cm and resistivity = 4 × 10-6 Ω ᐧ cm.
Solution:
Cross sectional area, A = 4 cm × 3 cm = 12 cm2
Linear length, l = πr = π(6 + \(\frac{3}{2}\)) = 7.5π cm
∴ Resistance = ρ\(\frac{l}{A}\) (ρ = resistivity = 4 × 10-6 Ω ᐧ m.)
= \(\frac{4 \times 10^{-6} \times \pi \times 7.5}{12} \Omega\) = 7.85 × 10-6Ω