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Current Electricity - Resistivity and Conductivity

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Resistivity (ρ\rho) is an intrinsic property of a material that measures its opposition to the flow of electric current, defined as R=ρlAR = \rho \frac{l}{A}.

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Conductivity (σ\sigma) is the reciprocal of resistivity, representing how easily a material allows current to flow: σ=1ρ\sigma = \frac{1}{\rho}.

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The microscopic form of Ohm's Law relates current density (JJ), conductivity (σ\sigma), and electric field (EE) as J=σEJ = \sigma E.

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Resistivity depends on the nature of the material and temperature, but is independent of the dimensions of the conductor.

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For metallic conductors, resistivity increases with temperature because the relaxation time (τ\tau) decreases due to more frequent collisions of electrons: ρ=mne2τ\rho = \frac{m}{ne^2\tau}.

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The temperature dependence of resistivity is given by ρT=ρ0[1+α(T−T0)]\rho_T = \rho_0 [1 + \alpha(T - T_0)], where α\alpha is the temperature coefficient of resistivity.

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SI units: Resistivity is measured in Ω⋅m\Omega \cdot m and Conductivity is measured in Ω−1m−1\Omega^{-1} m^{-1} or Siemens per meter (S⋅m−1S \cdot m^{-1}).

📐Formulae

R=ρlAR = \rho \frac{l}{A}

ρ=mne2τ\rho = \frac{m}{ne^2\tau}

σ=1ρ=ne2τm\sigma = \frac{1}{\rho} = \frac{ne^2\tau}{m}

J⃗=σE⃗\vec{J} = \sigma \vec{E}

ρT=ρ0[1+α(T−T0)]\rho_T = \rho_0 [1 + \alpha(T - T_0)]

α=RT−R0R0(T−T0)\alpha = \frac{R_T - R_0}{R_0(T - T_0)}

💡Examples

Problem 1:

A wire of resistance RR is stretched to triple its original length. Find the new resistance and the new resistivity of the wire.

Solution:

When a wire is stretched, its volume V=A×lV = A \times l remains constant. Let the original length be ll and area be AA. New length l′=3ll' = 3l. Since Al=A′l′Al = A'l', we have A′=Al3l=A3A' = \frac{Al}{3l} = \frac{A}{3}. New resistance R′=ρl′A′=ρ3lA/3=9(ρlA)=9RR' = \rho \frac{l'}{A'} = \rho \frac{3l}{A/3} = 9 \left( \rho \frac{l}{A} \right) = 9R. The resistivity ρ\rho remains unchanged.

Explanation:

Resistance depends on the dimensions (ll and AA), so it increases by the square of the stretching factor. However, resistivity is a material property and does not change with physical dimensions.

Problem 2:

The resistance of a platinum wire at the ice point is 5Ω5 \Omega and at steam point is 5.39Ω5.39 \Omega. Find the temperature at which the resistance is 5.79Ω5.79 \Omega.

Solution:

Given R0=5ΩR_0 = 5 \Omega at T0=0∘CT_0 = 0^\circ C and R100=5.39ΩR_{100} = 5.39 \Omega at T=100∘CT = 100^\circ C. Using α=R100−R0R0(100−0)\alpha = \frac{R_{100} - R_0}{R_0(100 - 0)}, α=0.39500=0.00078∘C−1\alpha = \frac{0.39}{500} = 0.00078 ^\circ C^{-1}. To find TT for RT=5.79ΩR_T = 5.79 \Omega: 5.79=5[1+0.00078(T−0)]5.79 = 5[1 + 0.00078(T - 0)]. 1.158=1+0.00078T  ⟹  0.158=0.00078T  ⟹  T≈202.56∘C1.158 = 1 + 0.00078T \implies 0.158 = 0.00078T \implies T \approx 202.56^\circ C.

Explanation:

The variation of resistance with temperature is linear for most metals over a moderate range, allowing us to use the temperature coefficient α\alpha to find unknown temperatures.

Resistivity and Conductivity - Revision Notes & Key Formulas | CBSE Class 12 Physics