Learn resistance variation due to temperature, practice to calculate temperature coefficient of resistance, equivalent resistance & resistors in series, effect of temperature on conductivity and resistivity.

- As the temperature of a conductor is increased, the thermal agitation increases and the collision becomes more frequent. Due to this, conductivity decreased because the drift speed also decreased.

\(Conductivity\propto \dfrac{1}{Resistivity}\)

Thus, resistivity is increased when conductivity decreases.

Resistivity of conductor increased with increase in temperature, hence resistance also increases.

**Important Points**

The resistance of conductor depends on its length and area of cross-section as these quantities also vary with temperature. But the changes \(\left[\dfrac{Length}{Area}\right]\)are very small so they are considered to be constant.

A Immediately after it is turned on and glow of metal filament is increasing

B After it has been on for a few milliseconds and the glow is steady

C Always remains the same

D None of these

- For small temperature variations

\(\rho(T)=\rho (T_0)\,[1+\alpha(T-T_0)]\)

where \(\rho(T)\) and \(\rho(T_0)\) are resistivity at temperature T and T_{0} respectively. \(\alpha\)is a constant at T_{0} and is known as temperature coefficient of resistivity.

As \(R\propto\rho\)

\(R(T)=R(T_0)\,[1+\alpha(T-T_0)]\)

- Temperature coefficient of resistivity \((\alpha)\)

\(\alpha>0\)

- As temperature increases, resistance increases.

- Temperature coefficient of resistivity \((\alpha)\)

\(\alpha<0\)

- Those materials whose resistivity becomes zero below a certain temperature called critical temperature, are superconductors.
- Above critical temperature, superconductor behaves as conductor and resistance increases as temperature increases.

A \(\alpha R_1(t_2-t_1)\)

B \(\alpha R_1(t_1-t_2)\)

C \(\alpha R_2(t_2-t_1)\)

D \(\alpha R_2(t_1-t_2)\)

- The resistance at temperature T
_{0}is R(T_{0}) and at temperature T is R(T). - The resistance at temperature T is given by

\(R(T)=R(T_0)\,[1+\alpha (T-T_0)]\)

where \(\alpha\) is the temperature coefficient of resistivity.

A \(22.4\,\Omega\)

B \(30\,\Omega\)

C \(40\,\Omega\)

D \(50\,\Omega\)

- Consider two resistance are connected in series.

\(R_{eq}=R_1+R_2\)

Taking derivative

\(dR_{eq}=dR_1+dR_2\)

We know,

\(\Delta R=R_0\,\alpha \,\Delta T\)

\(\Rightarrow dR_{eq}=\alpha _1R_1\Delta T+\alpha_2R_2\,\Delta T\)

\(\Rightarrow dR_{eq}=(\alpha _1R_1+\alpha_2R_2)\,\Delta T\)

A \(.4\,\Omega\)

B \(.2\,\Omega\)

C \(.16\,\Omega\)

D \(.14\,\Omega\)

- Consider two resistance are connected in series.

\(R_{eq}=R_1+R_2\)

Taking derivative

\(dR_{eq}=dR_1+dR_2\)

We know,

\(\Delta R=R_0\,\alpha \,\,\Delta T\)

\(\Rightarrow dR_{eq}=\alpha _1R_1\Delta T+\alpha_2R_2\,\Delta T\)

\(\Rightarrow dR_{eq}=(\alpha _1R_1+\alpha_2R_2)\,\Delta T\)

\(\Rightarrow \alpha _{eq} \,R_{eq} \,\Delta T=(\alpha_1R_1+\alpha_2R_2)\Delta T\)

\(\Rightarrow \alpha_{eq}=\dfrac{\alpha_1R_1+\alpha_2R_2}{R_{eq}}\)

- For total resistance to be independent of temperature the term

\(R_1\alpha_1+R_2\alpha_2=0\)

or, \(R_1\alpha_1=-R_2\alpha_2\)

\(\alpha_1=\dfrac{-R_2\alpha_2}{R_1}\)

and \(\alpha_2=\dfrac{-R_1\alpha_1}{R_2}\)

A \(R_1\alpha_1+R_2\alpha_2=0\)

B \(R_1\alpha_1=R_2\alpha_2\)

C \(\dfrac{R_1}{\alpha_1}=\dfrac{R_2}{\alpha_2}\)

D \(\dfrac{\alpha_1}{R_2}=\dfrac{\alpha_2}{R_1}\)

A \(\ell_1=0.88\,m,\,\ell_2=0.55\,m\)

B \(\ell_1=3\,m,\,\ell_2=2\,m\)

C \(\ell_1=4\,m,\,\ell_2=5\,m\)

D \(\ell_1=.1\,m,\,\ell_2=3\,m\)