Informative line

### 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