Resistance Formula

Last Updated : 02 Sep, 2022

If one wants to control the current flowing inside the wires, then one needs to know about resistance. In a circuit, it is the opposition to current flow. It is measured in ohm(Ω). It can be calculated using Ohm’s law. It is defined as the ratio of the applied voltage to the current. Therefore,

R = V/I

Where,

R = resistance

I = current

V = voltage

The Electrical Resistance formula can be calculated using the length and area of a wire. The formula states that the value of resistance through a wire which is directly proportional to the length and inversely proportional to the cross-sectional area.

Mathematically, this relationship is often written as,

Resistance of the conductor is directly proportional to it’s length,

R α L

Resistance of the conductor is inversely proportional to it’s area of cross-section,

$R α \frac{1}{A}$

Removing the proportionality and the constant is called as Resistivity,

$R = ρ\frac{l}{A}$

Where,

R = resistance

ρ = resistivity of the conductor

l = length of the conductor

A = area of the cross-section of the conductor

The Above Equation can also be written as,

$ρ = R\frac{A}{l}$

$l = R\frac{A}{ρ}$

$A = ρ\frac{l}{R}$

The resistance depends on the material it is made of. Objects made of electrical insulators like rubber tend to have very high resistance, while objects made of electrical conductors like metals tend to have very low resistance.

Water Pipe Analogy for Electrical Resistance

When the length of the pipe is long, the resistance to the flow of water will be high.
When the cross-section area of the pipe is high, the resistance to the flow of water is low.

Electrical Resistance relation with Power

Electric power (P) is the product of current times voltage. The SI unit of Power is Watt(W). The electric power is calculated by Ohm’s law and by the values of voltage, current, and resistance.

P = VI

Where,

P = Electric Power

V = Voltage

I = Current

From Ohm’s Law we know that,

V = IR

Replacing the value of V in the above equation,

P = I²/R

When the values for current and resistance are given, equation is,

P = V²/R

Effect of Temperature on Electrical Resistance

The resistance of the materials changes by the change in temperature. The amount of the change differ with the types of material.

Metals

The electrical resistance of pure metals increases when the temperature increases. Thus, metals have a positive temperature coefficient of resistance. e.g., copper, aluminum, silver, etc.

Alloys

The electrical resistance of alloys increases with an increase in temperature but alloys have a low value of positive temperature coefficient of resistance. e.g., nichrome, etc.

Semi-Conductors, Insulators & Electrolytes

The electrical resistance of semiconductors & insulators decreases with an increase in temperature. There is decrease in the value of resistance. Therefore, these materials have a negative temperature coefficient of resistance.

Factors Affecting Electrical Resistance

Length of the conductor

The greater the length of wire more will be the more resistance offered by the wire.

Area of a cross-section of the conductor

The resistance of the wire decreases, as the area of cross-section of wire, is increased.

Material of the conductor.

Different materials have different resistances. For Metals, the resistance offered is very low but for Insulators, the resistance offered is quite large.

Temperature of the Material

The electrical resistance of pure metals and alloys increases, when temperature increases but for insulators, the electrical resistance decreases with an increase in temperature.

Resistivity

Electrical resistivity is a property of a material which is fundamental in nature and it measures how strongly the property resists electric current. The SI unit of electrical resistivity is ohms meter and the symbol is row(ρ).

For Ideal Cases, where the cross-section and physical composition of the material is uniform across the sample. The resistivity can be written as:

$ρ = R\frac{A}{l}$

For less ideal cases, the current and electric field varies in different parts of the material. We use a General Expression,

$ρ= \frac{E}{J}$

Where,

ρ is that the electrical resistance of metal Ω.m

E is that the electric field in V.m^-1

J is that the current density in A.m-2

Sample Problems

Question 1: What is Ohm’s law?

Answer:

Ohm’s law states that the current through a conductor between two points is directly proportional to the voltage across the two points.

Question 2: Two Wires of Length 50m and 40m respectively have the Same Area of Cross Section and are Made Up of the Same Material. Which Wire has Higher Resistance?

Answer:

We know that R α L, wire of length 50m will offer higher resistance. The resistance of a small wire is low while the resistance of a long wire is high. t

Question 3: Two Wires L1 and L2 have lengths L and 2L respectively. The area of cross-section is 2A and A respectively. Both the wires are made up of the same material. Find the ratio of resistance in Wire L₁ and L₂?

Solution:

$R = ρ\frac{l}{A}$

So,

$R_1 = ρ\frac{L}{2A}$ ⇢ (1)

$R_2 = ρ\frac{2L}{A}$ ⇢ (2)

Dividing equation 1 by equation 2 above two equations,

$\frac{R_1}{R_2} = \frac{ρ\frac{L}{2A}}{ρ\frac{2L}{A}}$

$\frac{R_1}{R_2} = \frac{1}{4}$

Question 4: Calculate the resistance of a copper wire of length 5m and area of cross-section 2 × 10^-6 m². The resistivity of copper is 1.7 × 10^-8 Ωm.

Solution:

Length of copper wire = 5m

Area of cross-section of copper wire = 2 $\times$ 10^-6 m²

Resistivity of copper wire = 1.7 × 10^-8Ωm

We know that,

$R = ρ\frac{l}{A}$

So putting the values of ρ, l and A in the above equation,

$R = \frac{(1.7 \times 10^{-6})\times (5)}{2 \times 10^{-6}}$

R = 4.25 × 10^-2Ω

Question 5: What is Resistivity?

Answer:

Electrical resistivity is a fundamental property of a material that measures how strongly it resists electric current. The SI unit of electrical resistivity is ohms meter and the symbol is row (ρ).

Question 6: Explain the relation of Power with Resistance?

Answer:

Power is Voltage times Current,

P = VI ⇢ (1)

From Ohm’s Law,

V = RI ⇢ (2)

Replacing value of equation (2) in equation (1),

P = I²R

It can also be rewritten as,

$P = \frac{V^2}{R}$

Question 7: What current will be taken by a 1500W appliance if the supply voltage is 220V?

Solution:

Power of the appliance = 1500W

Voltage supply to the appliance = 220V

From Power- current relation we know that,

P = VI

The above equation can be rewritten as,

$I = \frac{P}{V}$

$I = \frac{1500}{220}$

I = 6.81A

Question 8: Calculate the current and resistance of a 50W,100V electric bulb.

Solution:

Power of the appliance = 100W

Voltage = 100V

From Power-current relation we know that,

P = VI

Also,

$I = \frac{P}{V}$

$I = \frac{50}{100}$

I = 0.5A

So, the current = 0.5A

From Ohm’s Law,

$R = \frac{V}{I}$

$R = \frac{100}{0.5}$

R = 200Ω

The Resistance offered by the Electric Bulb = 200Ω

Question 9: What happens to the resistance of insulators with a rise in temperature?

Answer:

With an increase in temperature, the resistance of insulators decreases. Thus, such material has a negative temperature coefficient of resistance.

Question 10: A metal wire of resistivity 6 × 10^-6 Ωm and length 20m has a resistance of 10Ω. Calculate its radius.

Answer:

Resistivity of the wire = 6 × 10^-6Ωm

Length of the wire = 20m

Resistance of the wire = 10Ω

$R = ρ\frac{l}{A}$

The above equation can be rewritten as,

$A = ρ\frac{l}{R}$ ⇢ (1)

Replacing the values in equation (1),

$A = \frac{(6\times 10^{-6})\times (20)}{10}$

A = 12 × 10^-6m²⇢ (2)

Generally the area-cross section of a wire is a circle. So the area of a circle is,

A = πr² ⇢ (3)

r = radius of the cross-section of the wire

Putting the value of (2) in (3),

πr² = 12 × 10^-6

r = 1.954 × 10^-3m

Question 11: Initially, the length of the wire is L. The length of the wire is changed from L to 2L keeping the area of cross-section the same. Find the initial to a final ratio of resistance?

Solution:

Initially,

The Length of wire = Lm

The area of cross-section of wire = Am²

Resistance of wire = R2Ω

Finally,

The Length of wire = 2Lm

The area of cross-section of wire = Am²

Resistance of wire = R2Ω

The ratio of initial to final is,

$\frac{R_1}{R_2} = \frac{\frac{ρ\times L}{A}}{\frac{ρ\times 2L}{A}}$

$\frac{R_1}{R_2} = \frac{1}{2}$