Home / CIE AS & A Level Physics 9702: Topic 9: Electricity- Unit : 9.3 Resistance and resistivity Study Notes

CIE AS & A Level Physics 9702: Topic 9: Electricity- Unit : 9.3 Resistance and resistivity Study Notes

Resistance and Ohm

Ohm’s Law states that the current through the conductor is directly proportional to the potential difference between its ends provided its temperature and other physical conditions remain constant.
Mathematically
$
I \alpha V \Rightarrow V=R I \Rightarrow R=\frac{V}{l}
$
The proportionality constant $R$ in the equation is the electrical resistance of the device. It is constant for a metallic conductor under steady physical conditions. Materials which obey Ohm’s law are called ohmic conductors.

Resistance of a conductor is defined as the ratio of the potential difference across it to the current flowing through it.
From $R=\frac{V}{l}, 1 \Omega=1 \vee A^{-1}$ defines the ohm
The ohm is the resistance of a conductor if a current of one ampere flows through when there is a potential difference of one volt across it.

Solve problems using
$
P=V I, P=I^2 R, V=I R
$

Example 5

A $12 \mathrm{~V} 24 \mathrm{~W}$ bulb is connected in series with a variable resistor and a 18 battery of negligible internal resistance. The variable resistor is adjusted until the bulb operates at its normal rating.

Determine
(i) the current in the bulb
(ii) the resistance of the bulb
(iii) the p.d. across the variable resistor;
(iv) the power dissipation in the variable resistor.

Answer/Explanation

Solution:
(i) $P=V I$
$24=(12) !$
$I=2.0 \mathrm{~A}$
(ii) $\quad V=I R$
$12=(2.0) R$
$R=6.0 \Omega$
(iii) p.d. across variable resistor $=18-12=6.0 \mathrm{~V}$
(iv) $P=V I=(6.0)(2.0)=12 \mathrm{~W}$

Resistance  & Resistivity

The resistance $R$ of a sample is directly proportional to its length $l$ and inversely proportional to its cross-sectional area $A$.
$
R \propto \frac{I}{A}
$
The relationship could be expressed as an algebraic equation by introducing a constant of proportionality as follows:
$
R=\frac{\rho l}{A}
$

The constant $\rho$ is now recognised as a property of the material and is called its resistivity. Hence
$
\rho=\frac{R A}{I}
$
where $\rho$ is the resistivity of the material, in $\Omega \mathrm{m}$
$R$ is the resistance of the sample, in ohms $(\Omega)$
$A$ is the cross-section area of the sample, in $\mathrm{m}^2$
$I$ is the length of the sample, in metres $(\mathrm{m})$
Resistivity is useful when comparing various materials on their ability to conduct electricity. A high resistivity means a sample of the material is a poor conductor. A low resistivity means a sample of the material is a good conductor.

Resistivity

• Resistivity is defined as the electrical property of a material that determines the resistance of a piece of given dimensions.

  •  It is equal to $\rho=\frac{R A}{l}$ where $R$ is the resistance, $A$ the cross-sectional area, and / the length, and is the reciprocal of conductivity. It is measured in ohm metres. It is denoted by the symbol $\boldsymbol{\rho}$.

Solve problems using $R=\frac{\rho L}{A}$

Example 6
The resistivity of a material is $3.1 \times 10^{-5} \Omega \mathrm{m}$. Determine the resistance of a sample of the material given that its length is $20 \mathrm{~cm}$ and its cross-section area is $2.0 \mathrm{~mm}^2$.

Answer/Explanation

Solution:
$
R=\frac{\rho l}{A}=\frac{\left(3.1 \times 10^{-5}\right)(0.20)}{(2.0)(0.001)^2}=3.1 \Omega
$

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