Electrical Resistance: Unit, Formula, Ohm’s Law & Resistivity

Ans. Ohm’s law states that if the physical conditions like temperature, mechanical strain, length of the conductor etc are kept unchanged then the current flowing across the conductor will be directly proportional to the applied potential difference. Mathematically it is represented as : 

V \(\propto\) I or V = RI 

V = Potential difference (voltage)

I = Current 

R= Resistance (constant of proportionality)

The following cases illustrate the failure of Ohm’s law:

1. Ohm’s law shows a linear relation between V and I for metallic conductors. However, in most cases, the graph follows a curved path. This is because of the heating effect of the current. So when the temperature rises, resistance in the conductor also increases. This can only be represented correctly by a curve.
2. For semiconductors and p-n junction diodes, the V-I relation is different and depends on the sign of applied potential difference.

Ans. Electric Current is the amount of charge ‘Q’ flowing through a particular area of a cross-section in unit time ‘t’, i.e.

Electric current, I = Q/t

The SI unit of electric current is Ampere. One ampere of current is that current which flows when one coulomb of electric charge flows through a particular area of cross-section of the conductor in one second, i.e.

1A = 1 Cs^-1.

If the electron flow is from B to A then the conventional current flows from A to B. This is because the conventional current is the flow of positive charge i.e movement of charge from higher potential to a lower potential. In metals free electrons carry current. So electron movement is opposite to the movement of conventional charge.

Ans. The resistance of conductor is given by the relation, 

R = V/I 

Given: 

V = 20V 

I = 0mA = 0 A 

Putting the values in the relation, 

R = V/I 

⇒ R = (20)/(0) 

The resistance is closest to infinity, which means the material is an insulator. 

Ans. Here,

V = 1.8 V ; I = 1.2 A ; 

L= 3m ; Area of cross section = 5.4 * 10^-6 m² .

The resistance of the wire is given as : R = V/I = 1.8/1.2 = 1.5 ohm

So conductance = (resistance)-1 = 1/R = 0.67 Siemen

We know for conductivity : 

 \(\sigma\) = length/(R*A) = 3 / (1.5 * 5.4 * 10^-6) = 3.7 * 105 Sm^-1

Ans. The resistance of a conductor is given by the relation, 

R = V/I 

Given: 

V = 20V 

I = 50mA = 0.05 A 

Plugging in the values inside the relation, 

R = V/I 

⇒ R = (20)/(0.05) 

⇒ R= 400 Ohms.

Ans. Denoted by \(\sigma\), conductivity is the reciprocal of resistivity. It can be expressed as 

 \(\sigma\) =\(\frac{j}{E}\)

The SI unit of conductivity is ( ohm^-1metre^-1 ) or (\(\Omega\)m)^-1 

Ans.

We know that the resistance of a semiconductor decreases with an increase in temperature. So to keep the reading of the ammeter unchanged or constant the total resistance in the circuit must remain unchanged.

Therefore to keep the ammeter reading constantly while the semiconductor is heated, the value of R which is the resistance in the resistor will have to be increased. 

Ans. Let \(\rho \) be the resistivity of the wire. Further, suppose that the initial length and area of the cross-section of the wire are l₁ and A₁ respectively.

On stretching the wire by 0.1 % the values of length and cross-section area become l₂ and A₂ respectively. 

Now if R₁ and R₂ are the resistances of the wires in the two cases, then,

R₁ = \(\rho \) ( l₁/ A₁) and R₂ =\(\rho \) (l₂ / A₂)

Since the volume of the wire must remain constant after the elongation, so we can write :

A₁l₁ = A₂l₂ 

A₂ = (A₁l₁) / l₂ 

Therefore R2 =\(\rho \) * (\(\frac{(1_2)^2}{A_1I_1}\))

As the wire has been stretched by 0.1 % 

l₂ = l₁ * \(\frac{100.1}{100}\)  ; So substituting the value of l₂ in the expression for R₂ we get R₂ = 1.002 R₁.

Therefore percentage % increase in the resistance of the wire = \(\frac{R_2 -R_1}{R_1}\) * 100

\(\frac{1.002R_1 -R_1}{R_1}\) * 100 = 0.2 %. 

Ans. The factors affecting the electrical resistance of a metallic conductor are discussed below:

Length of the conductor: Electrical resistance is directly proportional to the length of the

metallic conductor i.e. R \(\propto\)L .

Cross-sectional area: Electrical resistance is inversely proportional to the cross-sectional area of the metallic conductor. i.e R\(\propto\)  1A . A denotes the cross-sectional area of the metal conductor. On the basis of the above two relations, electrical resistance can be represented as

R =\(\frac{\rho}{LA} \),  \(\rho \) denotes resistivity.

Nature of the material: The property of electrical resistance is dependent on the intrinsic 

qualities of the conducting material. Conductors like Aluminium, Gold have very low resistance. Insulators like wax, rubber, glass, etc have very high resistance to the flow of charge.

Temperature of the conductor: Electrical resistance depends on the temperature of the conducting material. As the conductor gets heated, the thermal energy of the ions increases and they start moving vigorously. This results in more impedance in the current’s path.

Temperature dependence of resistance can be given as : 

R₂ = R₁ (1 + \(\alpha \)(T₂ - T₁) )

R1 and R2 are resistances at T₁ and T₂ temperatures respectively. \(\alpha\)denotes the temperature coefficient.

Ans.

Let I be the total current in the circuit. If R’ is the effective resistance of the 4 resistors connected in parallel of 12 \(\Omega\) each, then 

1/R’ = (1/12) +(1/12) +(1/12) +(1/12)

Or R’ = 3 \(\Omega\) 

Therefore the effective resistance of the entire circuit network R = R’ + R’ + R’ = 3R’ = 9\(\Omega\)

The current of the circuit , I = E/R = 9/9 = 1 Ampere .

Since all the four resistors are of the same resistance, the same current will pass through each resistor. Therefore the current through each resistor. Therefore the current through each resistor 

I’ = ¼ I = 0.25 Ampere

Ans. The resistors 3\(\Omega\), 4\(\Omega\) and 6\(\Omega\) are connected in parallel across the emf of 2 V.

Let Rp  the effective resistance of the parallel combination of 3\(\Omega\), 4\(\Omega\) and 6\(\Omega\).

Then , 1/R_p = 1/3 +1/4 +1/6 = 9/12 Therefore R_p = 4/3 \(\Omega\)

Total resistance of the circuit , R = R_P + r = 4/3 + \(\frac{2}{3}\) = 2\(\Omega\)

Therefore the current in the circuit is :

I = E/R = 2/2 = 1 Ampere.

Now the potential difference across the parallel combination = I * RP = 1 * 4/3 = 4/3 V

As the potential difference across all the 3 resistors is the same = 4/3 V.

Therefore the current through the 3\(\Omega\) resistor = (4/3) / 3 = 4/9 Ampere.

Ans. Let l be the length, A be the area of cross-section of the wire and \(\rho \) be the resistivity of the material 

Then R =\(\frac{\rho}{LA} \)

Suppose when the length of the wire gets doubled it becomes 2l = l’

Its area of cross-section becomes A’

As the volume of the wire must remain constant after elongation 

Therefore Al = A’l’

A’ = (Al) / (2l) = A/2

Let R’ be the new resistance of the wire. Then, 

R’ =   \(\frac{\rho}{L'A'}\)' = 4 \(\frac{\rho}{LA} \)

Comparing the equations for R and R’ we get R’ = 4R = 4*5 = 20 \(\Omega\)

Ans. The resistance of insulators decreases as the temperature increases. This is due to the electron movement from the conduction band to the valence band increases as the energy gap between these two bands is large. Therefore, conductance increases and the resistance decreases.

Ans. As the temperature increases, the resistance of pure metals also increase. This is due to the increase in the number of electrons in the conduction band which reduces the mobility thereby increasing the resistance.

Ans. We are given:

• The value of the resistance of the resistor is R=2.55 kΩ=2.55×10³ Ω.
• The value of the capacitance of the capacitor is C=2.15 μ F=2.15×10⁻⁶ F.
• The value of the frequency of the generator is f=41.5 Hz.

The formula for the capacitive reactance is:

X_C=1/2πfC

Substitute the required values in the above formula and solve it.

Using the formula of impedance and the required values, we have:

Ans. Resistance, current, and voltage are related together by Ohm's law as V=IR. Thus, the voltage of the device is obtained as

V​=IR

=15×20

=300V​

Ans. Using Ohm's law, V=I R, we have

V​=IR

=(0.2A)(1.4×1000 Ω)

=280V​

Ans. Ohmic materials are the ones that have constant resistance over a wide range of applied voltages. In other words, in an ohmic conductor, the ratio of the voltage across it to the current through it, which is defined as resistance, is always a constant. 

Thus, ohmic materials have a linear current-voltage relationship, and its curve passes through the origin. In contrast, the materials having a resistance that changes with a potential drop or current are called non-ohmic. The curve of a non-ohmic material is not linear. Examples of non-ohmic materials that violate Ohm's law are diodes and transistors. 

With these explanations, as the curve of (A) is linear, and passes through the origin, it is an ohmic conductor whose slope gives the reciprocal of the resistance. As in the previous problem, its resistance is calculated as RA​=5Ω. 

The resistor (B) has a nonlinear relationship between the voltage across it to the current, so it is a non-ohmic conductor with variable resistance.