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Current Electricity - Ohm’s Law and Drift Velocity

Grade 12CBSEPhysics

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

🔑Concepts

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Electric Current (II): The rate of flow of electric charge through any cross-section of a conductor, I=dqdtI = \frac{dq}{dt}. The SI unit is Ampere (AA).

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Drift Velocity (vdv_d): The average velocity with which free electrons get drifted towards the positive terminal of the conductor under the influence of an applied electric field. It is of the order of 10−4 m/s10^{-4} \text{ m/s}.

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Relaxation Time (τ\tau): The average time interval between two successive collisions of an electron with the positive ions in the conductor.

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Ohm's Law: At constant temperature and physical conditions, the current flowing through a conductor is directly proportional to the potential difference across its ends (V∝IV \propto I), giving V=IRV = IR.

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Resistance (RR): The opposition offered by a conductor to the flow of current. R=ρlAR = \rho \frac{l}{A}, where ρ\rho is resistivity, ll is length, and AA is the area of cross-section.

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Current Density (JJ): The amount of current flowing per unit area of cross-section held perpendicular to the direction of flow, J=IAJ = \frac{I}{A}.

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Mobility (μ\mu): The magnitude of the drift velocity per unit electric field, μ=vdE\mu = \frac{v_d}{E}. Its SI unit is m2V−1s−1m^2 V^{-1} s^{-1}.

📐Formulae

I=qt=nAevdI = \frac{q}{t} = nAe v_d

vd=eEτm=eVτmlv_d = \frac{e E \tau}{m} = \frac{e V \tau}{m l}

R=VI=mlne2τAR = \frac{V}{I} = \frac{m l}{n e^2 \tau A}

ρ=mne2τ\rho = \frac{m}{n e^2 \tau}

J⃗=σE⃗\vec{J} = \sigma \vec{E}

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

μ=vdE=eτm\mu = \frac{v_d}{E} = \frac{e \tau}{m}

Rt=R0(1+αΔT)R_t = R_0(1 + \alpha \Delta T)

💡Examples

Problem 1:

A potential difference of 3 V3 \text{ V} is applied across a conductor of length 0.1 m0.1 \text{ m}. If the drift velocity of electrons is 7.5×10−4 m/s7.5 \times 10^{-4} \text{ m/s}, calculate the mobility of the electrons.

Solution:

Given: V=3 VV = 3 \text{ V}, l=0.1 ml = 0.1 \text{ m}, vd=7.5×10−4 m/sv_d = 7.5 \times 10^{-4} \text{ m/s}. \nFirst, find the Electric Field: E=Vl=30.1=30 V/mE = \frac{V}{l} = \frac{3}{0.1} = 30 \text{ V/m}. \nMobility μ=vdE=7.5×10−430=2.5×10−5 m2V−1s−1\mu = \frac{v_d}{E} = \frac{7.5 \times 10^{-4}}{30} = 2.5 \times 10^{-5} \text{ m}^2 \text{V}^{-1} \text{s}^{-1}.

Explanation:

Mobility is defined as the drift velocity acquired per unit electric field applied. We first derive the field from the potential and length.

Problem 2:

A wire of resistance RR is stretched to triple its original length. What will be its new resistance, assuming density and resistivity remain constant?

Solution:

Let initial length be ll and area be AA. Volume V=AlV = Al is constant. \nWhen l′=3ll' = 3l, the new area A′A' must satisfy A′l′=Al  ⟹  A′(3l)=Al  ⟹  A′=A3A'l' = Al \implies A'(3l) = Al \implies A' = \frac{A}{3}. \nInitial Resistance R=ρlAR = \rho \frac{l}{A}. \nNew Resistance R′=ρl′A′=ρ3lA/3=9ρlA=9RR' = \rho \frac{l'}{A'} = \rho \frac{3l}{A/3} = 9 \rho \frac{l}{A} = 9R.

Explanation:

Resistance depends on the geometry of the conductor. When a wire is stretched, its length increases and its cross-sectional area decreases such that the total volume remains the same. The resistance increases by the square of the stretching factor.

Ohm’s Law and Drift Velocity - Revision Notes & Key Formulas | CBSE Class 12 Physics