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Resistance Variation Due To Temperature

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.

Variation of Conductivity with Temperature

• 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.

Relation between Conductivity and Resistivity

$$Conductivity\propto \dfrac{1}{Resistivity}$$

Thus, resistivity is increased when conductivity decreases.

Conclusion

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.

Illustration Questions

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

×

At room temperature, resistance is $$R_0$$.

As filament warms up

$$R_t>R_0$$

Thus, resistance increases with increase in temperature.

Hence, option (B) is correct. 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

Option B is Correct

Variation of Resistance with Small Temperature

• 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 T0 respectively. $$\alpha$$is a constant at T0 and is known as temperature coefficient of resistivity.

As $$R\propto\rho$$

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

Conductor

• Temperature coefficient of resistivity $$(\alpha)$$

$$\alpha>0$$

• As temperature increases, resistance increases.  Semiconductors

• Temperature coefficient of resistivity $$(\alpha)$$

$$\alpha<0$$

Superconductors

• 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 metallic wire has a resistance R1 $$\Omega$$ at $$t_1\;^{\circ}C$$ temperature and resistance changes at $$t_2\;^{\circ}C$$ temperature. Calculate change in resistance if temperature coefficient of wire is $$\alpha \;^{\circ}C^{-1}$$.

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

×

Change in resistance due to temperature is given as

$$R(t_2)=R(t_1)[1+\alpha\,(t_2-t_1)]$$

or,  $$R(t_2)=R(t_1)+R(t_1)\alpha\,(t_2-t_1)$$

or,  $$R(t_2)-R(t_1)=R(t_1)\alpha\,(t_2-t_1)$$

or,  $$R_2-R_1=R_1\,\alpha\,(t_2-t_1)$$

$$\Delta R=\alpha R_1 (t_2-t_1)$$

A metallic wire has a resistance R1 $$\Omega$$ at $$t_1\;^{\circ}C$$ temperature and resistance changes at $$t_2\;^{\circ}C$$ temperature. Calculate change in resistance if temperature coefficient of wire is $$\alpha \;^{\circ}C^{-1}$$.

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

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

Option A is Correct

Condition for the Total Resistance to be Independent of Temperature

• 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 wire of resistance R1 and temperature coefficient $$\alpha_1$$ is connected end to end to a second wire of resistance R2 of same cross-sectional area and temperature coefficient $$\alpha_2$$. Which one of the following conditions must be satisfied if total resistance is independent of temperature for small temperature changes ?

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 $$\dfrac{\alpha_1}{R_2}=\dfrac{\alpha_2}{R_1}$$

×

Here two resistors are connected in series

$$R_{eq}=R_1+R_2$$ Taking derivative

$$dR_{eq}=dR_1+dR_2$$

We know

$$\Delta R=R_0\,\alpha \,\Delta T$$

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

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

$$\Rightarrow \alpha _{eq}\,R_{eq}\Delta T = (\alpha_1\,R_1+\alpha_2R_2)\Delta T$$

$$\Rightarrow \alpha_{eq}=\dfrac{\alpha_1\,R_1+\alpha_2R_2}{R_{eq}}$$

For total resistance to be independent of temperature;

The term

$$R_1\alpha_1+R_2\alpha_2=0$$

A wire of resistance R1 and temperature coefficient $$\alpha_1$$ is connected end to end to a second wire of resistance R2 of same cross-sectional area and temperature coefficient $$\alpha_2$$. Which one of the following conditions must be satisfied if total resistance is independent of temperature for small temperature changes ?

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

$$\dfrac{\alpha_1}{R_2}=\dfrac{\alpha_2}{R_1}$$

Option A is Correct

Calculate the lengths $$\ell_1$$ and $$\ell_2$$ of wires made of copper and nichrome respectively which are used to achieve overall resistance of $$R=10\,\Omega$$, if the resistance per unit length of wire of copper and nichrome is $$x_1=5\,\Omega/m$$ and $$x_2=10\,\Omega/m$$ respectively. Assume total resistance is independent for small temperature changes. Given  $$\alpha_{cu}=-0.5\times10^{-3}\,{^\circ{C^{-1}}}$$,      $$\alpha_{nichrome}=0.4\times10^{-3}\,{^\circ{C}^{-1}}$$

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 $$\ell_1=.1\,m,\,\ell_2=3\,m$$

×

Length of wire 1 used = $$\ell_1$$

Length of wire 2 used =  $$\ell_2$$

Overall resistance $$R=10\,\Omega$$

Total resistance of copper R1= Total length × Resistance per unit length

$$R_1=\ell_1\times x_1$$

$$R_1=5\ell_1\,\Omega$$

Total resistance of nichrome R2 = Total length × Resistance per unit length

$$R_2=\ell_2\times x_2$$

$$R_2=10\ell_2\,\Omega$$

Total resistance  $$R=R_1+R_2$$

or,   $$5\ell_1+10\ell_2=10$$ ...(1)

If total resistance is independent of temperature

$$R_1\alpha_1+R_2\alpha_2=0$$

or,  $$5\ell_1(-0.5\times10^{-3})+(10\ell_2)(0.4\times10^{-3})=0$$

or, $$-2.5\ell+4\ell_1=0$$

$$4\ell_2=2.5\ell_1$$  ...(2)

From (1) and (2)

$$5\ell_1+10\ell_2=10$$

or,  $$8\ell_2+10\ell_2=10$$      $$[4\ell_2=2.5\ell_1]$$

or,  $$18\ell_2=10$$

$$\ell_2=\dfrac{10}{18}$$

$$\ell_2=0.55\,m$$

$$\ell_1=0.88\,m$$

Calculate the lengths $$\ell_1$$ and $$\ell_2$$ of wires made of copper and nichrome respectively which are used to achieve overall resistance of $$R=10\,\Omega$$, if the resistance per unit length of wire of copper and nichrome is $$x_1=5\,\Omega/m$$ and $$x_2=10\,\Omega/m$$ respectively. Assume total resistance is independent for small temperature changes. Given  $$\alpha_{cu}=-0.5\times10^{-3}\,{^\circ{C^{-1}}}$$,      $$\alpha_{nichrome}=0.4\times10^{-3}\,{^\circ{C}^{-1}}$$

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

$$\ell_1=.1\,m,\,\ell_2=3\,m$$

Option A is Correct

Calculation of Resistance at Certain Temperature

• The resistance at temperature T0 is R(T0) 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 copper coil has a resistance of $$R=20\,\Omega\, \,\,\,at\,\,\,T_1=0\;^\circ{C}$$   and temperature coefficient of resistivity is $$\alpha=4\times 10^{-3}\,{^\circ{C^{-1}}}$$. Find the value of resistance at $$T_2=30\;^\circ{C}$$.

A $$22.4\,\Omega$$

B $$30\,\Omega$$

C $$40\,\Omega$$

D $$50\,\Omega$$

×

The resistance at temperature T is given by

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

Given : $$R(T_0)=20\,\Omega,\,T_0=0\;^\circ{C},\,T=30\;^\circ{C},\,\alpha=4\times10^{-3}\,{^\circ{C^{-1}}}$$

$$R(T)=20[1+4\times10^{-3}\,(30-0)]$$

$$R(T)=20[1+.12]$$

$$R(T)=22.4\,\Omega$$

A copper coil has a resistance of $$R=20\,\Omega\, \,\,\,at\,\,\,T_1=0\;^\circ{C}$$   and temperature coefficient of resistivity is $$\alpha=4\times 10^{-3}\,{^\circ{C^{-1}}}$$. Find the value of resistance at $$T_2=30\;^\circ{C}$$.

A

$$22.4\,\Omega$$

.

B

$$30\,\Omega$$

C

$$40\,\Omega$$

D

$$50\,\Omega$$

Option A is Correct

Change in Equivalent Resistance when Connected in Series

• Consider two resistance are connected in series.  $$R_{eq}=R_1+R_2$$

Taking derivative

$$dR_{eq}=dR_1+dR_2$$

We know,

$$\Delta R=R_0\,\alpha \,\Delta T$$

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

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

Two resistances $$R_1=3\,\Omega$$ and $$R_2=4\,\Omega$$ made of copper with temperature coefficient of resistivity $$\alpha=4\times10^{-3}\,{^\circ{C^{-1}}}$$ are connected in series at $$T_1=25\;^\circ{C}$$. Find the total change  in equivalent resistance at $$T_2=30\;^\circ{C}$$.

A $$.4\,\Omega$$

B $$.2\,\Omega$$

C $$.16\,\Omega$$

D $$.14\,\Omega$$

×

Change in equivalent resistance is given as

$$d\,R_{eq}=(\alpha_1R_1+\alpha_2R_2)\Delta T$$

where  $$\Delta T=T_2-T_1$$

Given :  $$R_1=3\,\Omega,\,R_2=4\,\Omega,\,\alpha_1=\alpha_2=\alpha=4\times10^{-3}\,{^\circ{C}^{-1}},\,T_1=25\;^\circ{C},\,T_2=30\;^\circ{C}$$

Change in temperature

$$\Delta T=T_2-T_1=30-25=5\;^\circ{C}$$

$$dR_{eq}=(4\times10^{-3}\times3+4\times10^{-3}\times4)(5)$$

$$dR_{eq}=.14\,\Omega$$

Two resistances $$R_1=3\,\Omega$$ and $$R_2=4\,\Omega$$ made of copper with temperature coefficient of resistivity $$\alpha=4\times10^{-3}\,{^\circ{C^{-1}}}$$ are connected in series at $$T_1=25\;^\circ{C}$$. Find the total change  in equivalent resistance at $$T_2=30\;^\circ{C}$$.

A

$$.4\,\Omega$$

.

B

$$.2\,\Omega$$

C

$$.16\,\Omega$$

D

$$.14\,\Omega$$

Option D is Correct