Temperature Dependence of Resistivity MCQ Quiz - Objective Question with Answer for Temperature Dependence of Resistivity - Download Free PDF

Concept:

The resistance of the conductor changes when the temperature of that conductor changes.

New resistance is given by

\({{R}_{t}}={{R}_{0}}\left( 1+\alpha \text{ }\!\!\Delta\!\!\text{ }T \right)\)

Where Rt = the resistance of conductor after temperature changes

R0 = the resistance of conductor before temperature changes

α = temperature coefficient

ΔT = final temperature – initial temperature

Calculation:

Given;

T₁ = 27°C, R₁ = 65 Ω

\({R_2} = \frac{{Supply\;voltage}}{{Steady\;current}} = \frac{{220}}{{2.8}} = 78.6\;{\rm{\Omega }}\) 

Now, using the relation

R₂ = R₁ [1 + α (T₂ – T₁)]

\({T_2} - {T_1} = \frac{{{R_2} - {R_1}}}{{{R_1}}} \times \frac{1}{\alpha }\)

\( = \frac{{78.6 - 65}}{{65}} \times \frac{1}{{1.7\; \times\; {{10}^{ - 4}}}}\)

T₂ – T₁ = 1231

T₂ = 1231 + T₁

= 1231 + 27 = 1258°C

Ans.(4)

Sol.

Slope of V - I graph

Ohm's Law states that V = IR

From the graph

Slope of V-I graph will be resistance

So, \(\frac{V_{1}}{I_{1}}=\tan θ=\mathrm{R}_{1}\)

& \(\frac{V_{2}}{l_{2}}=\tan (90-θ)=\cot θ=R_{2}\)

Calculation of T₂ - T₁

Temperature dependence of resistance is given by:

R₁ = R₀(1 + αT₁)

R₂ = R₀(1 + αT₂)

Subtracting above equations, we get

R₂ - R₁ = α(T₂ - T₁)

∴ T₂ - T₁ is proportional to R₂ - R₁

⇒ T₂ - T₁ ∝ (cot - tan θ) = \(\frac{\cos θ}{\sin θ}-\frac{\sin θ}{\cos θ}=\frac{\cos ^{2} θ-\sin ^{2} θ}{\sin θ \cos θ}\)

T₂ - T₁ ∝ \(\frac{2 \cos 2 θ}{\sin 2 θ}=2 \cot 2 θ\)

Hence, T₂ - T₁ ∝ cot 2θ

The correct answer is: The graph that shows an exponential decrease in resistivity with increase in temperature.

Explanation:

In semiconductors, as temperature (T) increases, the number density of free electrons (n) increases significantly due to thermal excitation.

This results in a higher availability of charge carriers that can contribute to electrical conduction.

Although the relaxation time (τ) or the average time between collisions decreases with temperature, the impact of increasing carrier density dominates.

Hence, overall conductivity increases, and resistivity (ρ = 1/σ) decreases.

Explanation:

Resistivity is independent of temperature for wire-bound resistors 

Concept:

Temperature Dependence of Resistance:

• The resistance of a material changes with temperature according to the formula:
• R_t = R₀ (1 + α(T - T₀))
• Where: R_t = resistance at temperature T, R₀ = resistance at reference temperature T₀, α = temperature coefficient of resistance.
• The temperature coefficient of resistance (α) is given as 1.35 × 10^-4 °C^-1.

Calculation:

Given,

R₀ = 100 Ω (at T₀ = 27°C)

R_t = 137 Ω (at T)

α = 1.35 × 10^-4 °C^-1

Using the formula:

137 = 100 (1 + 1.35 × 10^-4 (T - 27))

⇒ 137/100 = 1 + 1.35 × 10^-4 (T - 27)

⇒ 1.37 = 1 + 1.35 × 10^-4 (T - 27)

⇒ 0.37 = 1.35 × 10^-4 (T - 27)

⇒ (T - 27) = 0.37 / (1.35 × 10^-4)

⇒ T - 27 = 2740.74

⇒ T = 2740.74 + 27 = 2767.74°C

∴ The temperature is approximately 2767°C.
Hence, the correct option is 1)

CONCEPT:

Relaxation time: In a conductor, the time gap between two successive collisions of electrons, when current flows is called the relaxation time.

Relaxation time is calculated by:

\(τ = \frac{m}{ne^2ρ}\)

where m is the mass of the electron, n is the number of free electrons per unit volume, e is the charge on one electron, and ρ is the resistivity.

Application:

The relaxation time is inversely proportional to the resistivity of the material

Given,

T_A = 2.7 × 10-4 s

T_B = 1.35 × 10-4 s

\(\frac{ρ_B}{ρ_A}=\frac{T_A}{T_B} \)

\(\frac{ρ_B}{ρ_A}=\frac{2.7}{1.35}=2 \)

CONCEPT:

Tungsten: 

• The metal tungsten is used for the filaments in incandescent bulbs.
• It has a high melting point and retains its strength when heated.
• Filaments of the light bulbs are made up of the Tungsten element.
• Its symbol is ‘W’ because of its scientific name ‘Wolfram’ and its and the atomic number 74. 

EXPLANATION:

• As the resistance is less, heat energy is produced is very low which is not sufficient for an electric bulb to glow so the resistance is kept high.
• Tungsten is very resistant to corrosion and has the highest melting point (melting point = 3695 K) and the highest tensile strength of any element.
• Tungsten is used for making bulb filaments of incandescent lamps because it has the highest melting point and does not melt even while it is glowing for long hours. Therefore option 4 is correct.

NOTE:

• Light bulb filaments aren't resistive because of the tungsten.
• They're resistive because of their very long length, and very thin wire.​

CONCEPT:

• The resistance of the conductor changes when the temperature of that conductor changes.
• New resistance is given by

⇒ Rt = R0(1 + αΔT)

Where Rt = the resistance of conductor after temperature changes

R0 = resistance of conductor before temperature changes

α = temperature coefficient

ΔT = final temperature – initial temperature

CALCULATION:

Given that,

Co-efficient of resistance (α) = 0.005 per °C

Let resistance of bulb filament be R_o at 0°C

Resistance of the bulb filament at 100° C (R) = 100Ω

• The change in temperature in the first case will be

⇒ ΔT = 100° C - 0° C = 100° C

• resistance is given by

⇒ Rt = R0(1 + αΔT)

⇒ 100 = R0(1 + 0.005 × 100)

⇒ 100 = 1.5 R0    -------- (1)

The temperature of the bulb filament at R = 200 Ω is T'

• Hence the resistance of bulb’s filament is given by

⇒ R’ = R (1 + α ΔT)

⇒ 200 = R0(1 + 0.005 × T')   ------ (2)

On dividing equation 1 and 2, we get

\(⇒ \frac{100}{200} = \frac{1.5 R_0}{ R_0(1 + 0.005T')}\)

On solving above equation, we get

CONCEPT

• Fuse: A fuse is an electrical safety device that operates to provide overcurrent protection of an electrical circuit.
• Properties of fuse: 
  • High resistance: The resistance of the fuse must have high so that it can carry a short circuit current when a fault occurs.
  • Low melting point: The meting point of fuse wire should be low so that it can melt immediately when excess current flows through it.
  • It is used in a series of equipment as a safety device.
  • The material used for making a fuse wire is an alloy of lead and tin.

​Working of fuse:

EXPLANATION

• The conductivity of the fuse must be high so that it can carry current easily. So, statement 1 is correct.

A fuse contains a strip of metal or a metal wire having a small cross-section area.

It was discussed that fuse is used for the protection of a circuit from overcurrent. Whenever there is a passage of a high amount of current, the fuse breaks the circuit by melting itself.
For this to happen, the melting point of wire is kept low.

As a high amount of current is passing, resistances of the fuse wire have to be large due to which the heat produced will be high and it is broken down immediately.

Hence, fuse wire is a wire of high resistance and low melting point.

• The meting point of fuse wire should low so that it can melt immediately when excess current flows through it. statement 2 is wrong.
• It is used as safety equipment for sensitive and costly devices as it can break the circuit when an excess current passes through it. So statement 3 is correct.
• Generally, the fuse wire is made up of an alloy of lead and tin because it has a low melting point, good conductivity, and does not get oxidized easily. statement 4 is correct.

⇒ Option 2 is correct.

The Correct Answer is Option 2 i.e decreases.

CONCEPT:

• Conductor: The material that allows the electric current to flow through them is called the conductor.
  • The resistance of a conductor increases by increasing the temperature.
• Semiconductor: It is a substance or material that is neither a good conductor of electricity nor a good insulator, but has properties of electrical conductivity somewhere between the two.
  • The two most frequently used semiconductors are (i) germanium (Ge) and (ii) silicon (Si).

​EXPLANATION:

• In a semiconductor, as the temperature increases, the electrons get excited and jump from the valance band into the conduction band and thereby increases conductance resulting in the decrease of resistance. 
• Hence on increasing temperature, the resistance of semiconductors decreases. So option 2 is correct.

CONCEPT:

Temperature coefficient of resistance (α) :

• The change in electrical resistance of a substance with respect to a per degree change in temperature.
• α depends on the type of material.

Variation of resistance with temperature:

Let R0  and RT represent the resistance at a temperature of T0 and T respectively. Therefore

⇒ RT = R0 [1 + α (T - T0)]

Where α is the temperature coefficient of resistance.

EXPLANATION:

α for conductors:  

• In metal, if the temperature increases, the resistance of the metal increases with the rise in temperature.
• So, we consider the temperature coefficient of resistance as positive for metal.

α for semiconductor:

• If temperature increases, the number of electrons come to the conduction bands from valence bands by crossing the forbidden energy gap.
• As the number of free electrons increases, the resistance of this type of non-metallic substance decreases with an increase in temperature.

Hence, the temperature coefficient of resistance is negative for non-metallic substances and semiconductors.

• One other reason for this opposite behaviour is that temperature coefficient of semiconductors is negative and for metals its positive.
• Out of given options, except Germanium all are metals, so its temperature coefficient will be negative.

CONCEPT:

• Resistance: The obstruction offered to the flow of current is known as resistance.
• The formula of resistance is:

\(\Rightarrow R = \frac{{ml}}{{n{e^2}\tau A}}\)

Where m = mass of an electron, n = Number of electron per unit volume of the conductor, τ = relaxation time, l = length of the conductor, and A = cross-section area of conductor

• The relation between resistance and temperature is given by:

\(\Rightarrow {R_t} = {R_o}\left[ {1 + \alpha \left( {T - {T_o}} \right)} \right]\)

Where Rt = resistance at temperature t°, Ro = resistance at temperature 0°, and α = temperature co-efficient of resistance.

EXPLANATION:

Metal:

• For metals, the number density n of free electrons is almost independent of temperature.
• As temperature increases, the thermal speed of free electrons increases, and also the amplitude of vibration of the metal ions increases.
• Consequently, the free electrons collide more frequently with the metal ions.
• The mean collision time τ decreases.
• Hence the resistance of the metal (R ∝ 1/τ) increases with the increase in temperature. Therefore option 2 is correct.

Additional Information

Non-metal:

• In the case of non-metals, the relaxation time τ does not change with temperature but the number density of free electrons increases exponentially with the increase in temperature.
• Consequentially, the conductivity increases or resistivity decreases exponentially with the increase in temperature.

Concept:

The resistance of conductor changes when the temperature of that conductor changes.

New resistance is given by

Rt = R0(1 + αΔT)

Where R_t = the resistance of conductor after temperature changes

R₀ = resistance of conductor before temperature changes

α = temperature coefficient

ΔT = final temperature – initial temperature

Calculation:

Let resistance at 0° C is R₀

At 25° C given R_t = 55 Ω

R_t = R₀(1 + αΔT)

55 = R₀(1 + α × 25)       …(i)

At 75° C given resistance R_t = 57.2 Ω

57.2 = R₀(1 + α × 75)       ….(ii)

Now solving for (ii)/(i)

\(\frac{57.2}{55}=\frac{{{R}_{0}}\left( 1+\alpha \times 25 \right)}{{{R}_{0}}\left( 1+\alpha \times 75 \right)}\)

\(1.04=\frac{1+75\alpha }{1+25\alpha }\)

1.04 + 26 α = 1 + 75α

CONCEPT:

Fuse: 

• An electrical safety device that is used to protect the electrical circuit from overload is called a fuse.
• The material used for fuse elements must be of:
• low melting point, low ohmic loss,
• high resistivity,
• low cost and free from detraction.
• The material used for making fuse element has a low melting point such as tin, lead, or zinc.
• A low melting point is, however, available with a high specific resistance metal.
• The material mainly used for fuse element are tin, lead, silver, copper, zinc, aluminum,  and an alloy of lead and tin
• Working of fuse:

EXPLANATION:

• When there is a sudden increase of current in the circuit then a huge amount of heat is generated at the fuse. Therefore option 3 is correct.
• Due to this heat, a spark is generated.
• Due to the huge amount of heat, the fuse melts and the circuit opened up.

CONCEPT:

Resistivity (ρ): The property of a conductor that opposes the flow of electric current through them and independent of the shape and size of the materials but depends on the nature and temperature of the materials is called resistivity.

• The unit for resistivity is the ohm-meter (Ω-m).
• The resistivity of a material depends on its nature and the temperature of the conductor.
• The resistivity of a material doesn't depend on its shape and size (length and area).

The relation between resistivity and temperature (T) is given by (It is experimentally observed):

ρ = ρ₀ [1 + α (T - T₀)

Where ρ is resistivity at any temperature, α is the thermal coefficient of resistivity, and T is the temperature, ρ₀ is resistivity at a reference temperature (generally, reference temperature T₀ = 0 °C).

EXPLANATION:

Since the resistivity of the conductor/metal is given by:

ρ = ρ0 [1 + α (T - T0)

Here ρ₀, α and T₀ are constant terms.

• So the resistivity is linearly proportional to temperature. But in the case of metal, there are free electrons.
• When the temperature of the metals increases then the randomness of the free electrons will increase and the resistivity will vary quite randomly (not a straight line).
• Resistance will be increasing with the temperature but the graph will be quite parabolic (It is seen experimentally -No mathematical relation is there).

• Since there is a constant addition term with T in the right side that's why the graph will not start from the origin. So option 1 is correct.

Important Points

For semiconductor:

In a semiconductor, as the temperature increases, the electrons get excited and jump from the valance band into the conduction band and thereby increases conductance resulting in the decrease of resistivity. 

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