Current Electricity

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Home Concept Boosters CBSE Class 12 Physics Current Electricity

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How to Use This Page

Read each concept carefully, then check the formula, common mistake, and exam tip before moving to the next. This page covers Current Electricity completely for CBSE Class 12 Physics.

Key Concepts

Class 12 · Physics · Current Electricity

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Core concepts you must know

Class 12 · Ch 3

Drift Velocity Formula

The average velocity with which free electrons get drifted towards the positive end of a conductor under the influence of an external electric field.

$$\vec{v}_d = \frac{-e \vec{E}}{m_e} \tau$$

Relation: Current & Drift Velocity Formula

Electric current $I$ is directly proportional to the drift velocity $v_d$, the electron density $n$, and the cross-sectional area $A$ of the conductor.

$$I = ne A v_d$$

Mobility & Electric Current Formula

Mobility ($\mu$) is the magnitude of drift velocity per unit electric field. Current can be expressed in terms of mobility.

$$\mu = \frac{v_d}{E} \quad \text{and} \quad I = n A \mu E e$$

Current Density & Vector Ohm’s Law Formula

Current density ($\vec{j}$) is the electric current per unit cross-sectional area. It is a vector quantity related to the electric field by conductivity ($\sigma$).

$$\vec{j} = \frac{I}{A} = nq\vec{v}_d = \sigma \vec{E}$$

Resistance & Resistivity Formula

Resistance ($R$) opposes current flow. Resistivity ($\rho$) is the inherent property of the material, depending on electron density ($n$) and relaxation time ($\tau$).

$$R = \frac{V}{I} = \rho \frac{l}{A} \quad \text{where} \quad \rho = \frac{m}{ne^2 \tau}$$

Combination of Resistors Formula

In series, resistances add directly to maximize opposition. In parallel, their reciprocals add up, minimizing the total resistance.

$$\text{Series: } R_{eq} = R_1 + R_2 + R_3 \quad | \quad \text{Parallel: } \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$

Temperature Dependence of Resistivity Formula

For conductors, resistivity increases linearly with temperature ($\alpha$ is positive). For semiconductors, it decreases exponentially due to the energy gap ($E_g$).

$$\text{Conductors: } \rho = \rho_0[1 + \alpha(T – T_0)] \quad | \quad \text{Semiconductors: } \rho = \rho_0 e^{\frac{E_g}{kT}}$$

Resistance Thermometer Formula

Using the linear expansion of resistance to measure unknown temperatures. The temperature coefficient ($\alpha$) dictates the rate of change.

$$R_t = R_0(1 + \alpha t) \quad \text{and} \quad t = \frac{R_t – R_0}{R_{100} – R_0} \times 100$$

EMF, Terminal Voltage & Internal Resistance Formula

EMF ($\varepsilon$) is the maximum potential of a cell. When current is drawn, terminal voltage ($V$) drops due to internal resistance ($r$).

$$\varepsilon = I(R + r) \quad \text{and} \quad r = \left(\frac{\varepsilon – V}{V}\right)R$$

Grouping of Cells Formula

Cells can be grouped to maximize current. Series grouping is effective when external resistance is high; parallel when it is low. Mixed grouping balances both.

$$\text{Series: } I = \frac{n\varepsilon}{R + nr} \quad | \quad \text{Parallel: } I = \frac{m\varepsilon}{mR + r} \quad | \quad \text{Mixed: } I = \frac{n\varepsilon}{R + \frac{nr}{m}}$$

Slide Wire Bridge (Meter Bridge) Formula

A practical application of the Wheatstone bridge used to find an unknown resistance ($X$) using a balancing length ($l$) out of $100$ cm.

$$X = \frac{100 – l}{l} R$$

Potentiometer Applications Formula

Used to accurately compare EMFs of two cells without drawing current, and to find the internal resistance of a primary cell.

$$\frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1}{l_2} \quad \text{and} \quad r = \left[\frac{l_1}{l_2} – 1\right]R$$

Power, Energy & Heating Effect Formula

Electrical power ($P$) is the rate of energy consumption. Joule’s heating effect ($Q$) is the heat dissipated when current passes through a resistor.

$$P = I^2R = \frac{V^2}{R} \quad \text{and} \quad Q = \frac{I^2Rt}{J}$$

Concept Deep Dive

Drift Velocity vs. Thermal Velocity

Why do lights turn on instantly?

Core Concept

Free electrons in a room-temperature wire move randomly at massive thermal velocities ($\approx 10^5 \text{ m/s}$). Because this movement is random, the net current is zero. When you apply a voltage, the electric field pushes all electrons in one direction with a very small drift velocity ($\approx 10^{-4} \text{ m/s}$, like a snail’s pace).

If drift velocity is so slow, why does a bulb light up instantly? Because the electric field sets up across the entire wire at the speed of light. Electrons everywhere in the wire start drifting at the exact same time.

$$I = neAv_d$$

Everyday Analogy

Imagine a long hose completely full of water. If you turn on the tap and push a drop of water in at one end, a different drop of water instantly falls out the other end. The water drops themselves move slowly (Drift Velocity), but the pressure wave pushes them all instantly (Electric Field).

Potentiometer vs. Voltmeter

Why use a bulky wire over a handy meter?

Lab Application

A regular voltmeter must draw a tiny bit of current from the circuit to deflect its needle. Because it draws current, there is a voltage drop ($Ir$) across the cell’s internal resistance, meaning the voltmeter measures the Terminal Voltage ($V$), not the true EMF ($\varepsilon$). A potentiometer operates on the null-deflection method. At the balance point, it draws exactly zero current from the cell. Therefore, $I = 0$, and the voltage measured is the true, pure EMF ($\varepsilon$).

$$V = \varepsilon – Ir \xrightarrow{\text{If } I=0} V = \varepsilon$$

Compare & Contrast

✗ Conductors (Metals)

High number density of free electrons ($n$).
When Temperature increases, relaxation time ($\tau$) decreases due to frequent collisions.
Resistivity ($\rho$) increases linearly.
Temperature coefficient ($\alpha$) is positive.

✓ Semiconductors

Very few free charge carriers at room temp.
When Temperature increases, covalent bonds break, and carrier density ($n$) increases exponentially.
Resistivity ($\rho$) decreases exponentially.
Temperature coefficient ($\alpha$) is negative.

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Remember

Alloys like Manganin and Constantan are used to make standard resistors because their $\alpha$ is practically zero, meaning their resistance doesn’t change with room temperature fluctuations!

Common Mistakes to Avoid

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Mistake 1

Misunderstanding “Stretching a Wire”: If a wire is stretched to double its length ($L’ = 2L$), students often assume resistance simply doubles. But stretching also halves the cross-sectional area to keep volume constant! The new resistance actually becomes 4 times the original ($R’ = n^2 R$).

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Mistake 2

EMF vs. Terminal Voltage Signs: During discharging (drawing current), Terminal Voltage $V = \varepsilon – Ir$ (so $V < \varepsilon$). But during charging a battery, current enters the positive terminal, making $V = \varepsilon + Ir$ (so $V > \varepsilon$). Do not blindly apply the minus sign.

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Mistake 3

Meter Bridge Balancing Length: The formula is $X/R = l/(100-l)$. If you interchange the positions of the unknown resistance $X$ and the known resistance box $R$ in the gaps, the balancing length also shifts from $l$ to $(100-l)$. Always match the left/right ratio physically to the formula.

Exam Tips

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Tip 1

For Power calculations, if components are in Series (current $I$ is constant), use $P = I^2R$ to see who dissipates more heat. If they are in Parallel (voltage $V$ is constant), use $P = V^2/R$.

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Tip 2

In Potentiometer numericals, the primary circuit (driver battery) is ONLY responsible for creating the potential gradient ($k$). Never mix the EMFs or resistances of the secondary circuit into the primary circuit’s current calculation.

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Did You Know

Superconductors are materials that lose all electrical resistance ($R=0$) when cooled below a critical temperature. Once a current is set up in a superconductor loop, it can flow continuously for years without any power source!

Expected Exam Questions

Board Pattern Questions

Class 12 · Current Electricity · CBSE Exam

Class 12 · Physics

Define mobility of an electron. Give its SI unit. [1 mark]

Answer Drift velocity per unit electric field; $\text{m}^2\text{V}^{-1}\text{s}^{-1}$ 📝

Explanation

Mobility ($\mu$) is defined as the magnitude of drift velocity acquired by charge carriers per unit applied electric field. The formula is $\mu = v_d / E$. Substituting their SI units ($\text{m/s}$ for velocity and $\text{V/m}$ for field) yields the unit $\text{m}^2\text{V}^{-1}\text{s}^{-1}$.

A wire of resistance $16\,\Omega$ is stretched so that its length becomes twice its original length. Calculate its new resistance. [2 marks]

Answer $64\,\Omega$ 📝

Explanation

When a wire is stretched, its volume ($V = A \times l$) remains constant. If new length $l’ = 2l$, the new area becomes $A’ = A/2$.
Initial resistance $R = \rho \frac{l}{A} = 16\,\Omega$.
New resistance $R’ = \rho \frac{l’}{A’} = \rho \frac{2l}{A/2} = 4 \left(\rho \frac{l}{A}\right)$.
$R’ = 4 \times 16\,\Omega = 64\,\Omega$.
(Shortcut: For stretching to $n$ times the length, $R_{new} = n^2 R_{old}$. Here $n=2$, so $2^2 \times 16 = 64$).

In a potentiometer arrangement, a cell of EMF $1.25\text{ V}$ gives a balance point at $35.0\text{ cm}$ length of the wire. If the cell is replaced by another cell and the balance point shifts to $63.0\text{ cm}$, what is the EMF of the second cell? [3 marks]

Answer $2.25\text{ V}$ 📝

Explanation

The principle of the potentiometer states that the EMF is directly proportional to the balancing length ($\varepsilon \propto l$). Therefore, comparing two cells gives the formula:
$\frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1}{l_2}$
Given: $\varepsilon_1 = 1.25\text{ V}$, $l_1 = 35.0\text{ cm}$, $l_2 = 63.0\text{ cm}$.
$\frac{1.25}{\varepsilon_2} = \frac{35.0}{63.0}$
$\varepsilon_2 = \frac{1.25 \times 63.0}{35.0} = 2.25\text{ V}$.

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