Current Electricity: CBSE Class 12 Physics NCERT

The Flow of Charge: Electric Current and Drift Velocity

At the heart of current electricity is the concept of electric current, which is simply the rate of flow of electric charge through a conductor. Conventionally, the direction of electric current is taken as the direction of flow of positive charges. However, in metallic conductors, it's the electrons (negative charges) that actually move. These free electrons do not flow in a straight line; instead, they undergo frequent collisions with the positive ions of the conductor. In the absence of an external electric field, their motion is random, and the net current is zero. When an electric field is applied, these electrons experience a force and start drifting in a direction opposite to the field. This average velocity with which free electrons drift under the influence of an electric field is called drift velocity ($v_d$). Even though $v_d$ is typically very small (of the order of $10^{-4}$ m/s), the electric current is established almost instantaneously throughout the circuit due to the propagation of the electric field at nearly the speed of light. The relationship between current ($I$), drift velocity ($v_d$), charge of an electron ($e$), number density of free electrons ($n$), and cross-sectional area of the conductor ($A$) is given by $I = n e A v_d$. This equation is crucial for understanding how microscopic electron movement translates into macroscopic current.

Key Concepts and Definitions

Electric Current (I)

The rate of flow of electric charge through any cross-section of a conductor. $I = \frac{\Delta Q}{\Delta t}$. Its S.I. unit is Ampere (A).

Drift Velocity ($v_d$)

The average velocity attained by charged particles (like electrons) in a material due to an electric field. It is typically very small, in the order of $10^{-4}$ m/s.

Current Density (J)

The current per unit cross-sectional area of a conductor, $J = \frac{I}{A}$. It's a vector quantity, its direction being the direction of current flow. S.I. unit: A/m$^2$.

Ohm's Law

States that the current flowing through a conductor is directly proportional to the potential difference across its ends, provided physical conditions (like temperature) remain constant. $V = IR$.

Resistance (R)

The opposition offered by a conductor to the flow of electric current. Its S.I. unit is Ohm ($\Omega$). It depends on the material, length, cross-sectional area, and temperature of the conductor.

Resistivity ($\rho$)

An intrinsic property of a material, representing its ability to resist electric current. $R = \rho \frac{L}{A}$. S.I. unit: Ohm-meter ($\Omega$ m).

Conductivity ($\sigma$)

The reciprocal of resistivity, representing a material's ability to conduct electric current. $\sigma = \frac{1}{\rho}$. S.I. unit: Siemens per meter (S/m) or $\Omega^{-1}$m$^{-1}$.

Electromotive Force (EMF, $\mathcal{E}$)

The work done per unit charge by the source (e.g., battery) to move a charge from its negative terminal to its positive terminal inside the source. It is the maximum potential difference a source can provide when no current is drawn from it.

Applying Ohm's Law and Kirchhoff's Rules for Circuit Analysis

1. Ohm's Law: The Basic Relationship — Ohm's Law, $V = IR$, is the cornerstone of circuit analysis. It states that the potential difference ($V$) across a resistor is directly proportional to the current ($I$) flowing through it, where $R$ is the constant of proportionality, known as resistance. This law applies to ohmic materials. Remember, this law is not universal; it fails for non-ohmic devices like semiconductors. For series combinations of resistors, the equivalent resistance is $R_{eq} = R_1 + R_2 + ...$, and the current is the same through each resistor. For parallel combinations, the reciprocal of equivalent resistance is the sum of reciprocals: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + ...$, and the potential difference across each resistor is the same.
2. Kirchhoff's First Rule (Current Rule or Junction Rule) — This rule is based on the conservation of charge. It states that the algebraic sum of currents entering any junction (or node) in an electrical circuit is equal to the algebraic sum of currents leaving that junction. Mathematically, $\Sigma I_{in} = \Sigma I_{out}$. This means charge doesn't accumulate at a junction; whatever current flows in, must flow out.
3. Kirchhoff's Second Rule (Voltage Rule or Loop Rule) — Based on the conservation of energy, this rule states that the algebraic sum of changes in potential around any closed loop in an electrical circuit must be zero. Mathematically, $\Sigma \Delta V = 0$. When traversing a loop: if you move from negative to positive terminal of a source, EMF is taken as positive; if from positive to negative, it's negative. For a resistor, if you traverse in the direction of current, the potential drop ($IR$) is negative; if opposite to current, it's positive. Careful application of sign conventions is crucial.
4. Applying Kirchhoff's Rules in Steps — 1. Assign currents to each branch, making sure to apply KCL at junctions (e.g., if $I_1$ and $I_2$ enter, $I_1+I_2$ leaves). 2. Choose closed loops and a direction of traversal for each loop (clockwise or counter-clockwise). 3. Apply KVL to each chosen loop, carefully noting potential changes across resistors ($\pm IR$) and EMF sources ($\pm \mathcal{E}$). 4. Solve the resulting system of linear equations to find the unknown currents or voltages. For $N$ unknowns, you'll need $N$ independent equations.

Solved Problems for Better Understanding

• Example 1: Resistors in Combination Question: Three resistors, $R_1 = 10 \Omega$, $R_2 = 20 \Omega$, and $R_3 = 30 \Omega$, are connected first in series and then in parallel to a 12V battery. Calculate the equivalent resistance and the total current drawn from the battery in both cases. Solution: Case 1: Series Connection Step 1: Calculate equivalent resistance for series combination. $R_{series} = R_1 + R_2 + R_3 = 10 \Omega + 20 \Omega + 30 \Omega = 60 \Omega$. Step 2: Calculate total current using Ohm's Law. $I_{series} = \frac{V}{R_{series}} = \frac{12 \text{ V}}{60 \Omega} = 0.2 \text{ A}$. Case 2: Parallel Connection Step 1: Calculate equivalent resistance for parallel combination. $\frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{10} + \frac{1}{20} + \frac{1}{30}$ $\frac{1}{R_{parallel}} = \frac{6 + 3 + 2}{60} = \frac{11}{60} \implies R_{parallel} = \frac{60}{11} \approx 5.45 \Omega$. Step 2: Calculate total current using Ohm's Law. $I_{parallel} = \frac{V}{R_{parallel}} = \frac{12 \text{ V}}{(60/11) \Omega} = \frac{12 \times 11}{60} = \frac{11}{5} = 2.2 \text{ A}$. Final Answer: For series connection: Equivalent resistance = $60 \Omega$, Total current = $0.2 \text{ A}$. For parallel connection: Equivalent resistance = $5.45 \Omega$, Total current = $2.2 \text{ A}$.
• Example 2: Applying Kirchhoff's Laws Question: In the circuit below, use Kirchhoff's laws to find the current flowing through each resistor ($R_1$, $R_2$). Assume $E_1 = 10V$, $E_2 = 4V$, $R_1 = 2\Omega$, $R_2 = 1\Omega$, $R_3 = 3\Omega$. (Circuit Description: A left loop with $E_1$, $R_1$, $R_3$. A right loop with $E_2$, $R_2$, $R_3$. $R_3$ is common to both loops. Assume current $I_1$ flows from $E_1$ through $R_1$, $I_2$ from $E_2$ through $R_2$, and $I_3$ flows through $R_3$ from top to bottom (junction between $R_1,R_2$ to junction below $R_3$).) Solution: Step 1: Assign currents and apply Kirchhoff's Junction Rule (KCL). Let $I_1$ be the current from $E_1$ through $R_1$. Let $I_2$ be the current from $E_2$ through $R_2$. Let $I_3$ be the current through $R_3$. At the junction above $R_3$: $I_1 + I_2 = I_3$ (Equation 1) Step 2: Apply Kirchhoff's Loop Rule (KVL) to the left loop (containing $E_1, R_1, R_3$). Traverse clockwise. Starting from the negative terminal of $E_1$: $+E_1 - I_1 R_1 - I_3 R_3 = 0$ $10 - I_1(2) - I_3(3) = 0 \implies 2I_1 + 3I_3 = 10$ (Equation 2) Step 3: Apply KVL to the right loop (containing $E_2, R_2, R_3$). Traverse clockwise. Starting from the negative terminal of $E_2$: $+E_2 - I_2 R_2 - I_3 R_3 = 0$ $4 - I_2(1) - I_3(3) = 0 \implies I_2 + 3I_3 = 4$ (Equation 3) Step 4: Solve the system of equations. Substitute $I_3 = I_1 + I_2$ from (1) into (2): $2I_1 + 3(I_1 + I_2) = 10 \implies 5I_1 + 3I_2 = 10$ (Equation 4) Substitute $I_3 = I_1 + I_2$ from (1) into (3): $I_2 + 3(I_1 + I_2) = 4 \implies 3I_1 + 4I_2 = 4$ (Equation 5) Multiply Equation (4) by 4 and Equation (5) by 3: $(5I_1 + 3I_2 = 10) \times 4 \implies 20I_1 + 12I_2 = 40$ $(3I_1 + 4I_2 = 4) \times 3 \implies 9I_1 + 12I_2 = 12$ Subtract the second modified equation from the first: $(20I_1 + 12I_2) - (9I_1 + 12I_2) = 40 - 12$ $11I_1 = 28 \implies I_1 = \frac{28}{11} \text{ A} \approx 2.55 \text{ A}$ Substitute $I_1$ back into Equation (5): $3(\frac{28}{11}) + 4I_2 = 4 \implies \frac{84}{11} + 4I_2 = 4$ $4I_2 = 4 - \frac{84}{11} = \frac{44 - 84}{11} = -\frac{40}{11}$ $I_2 = -\frac{10}{11} \text{ A} \approx -0.91 \text{ A}$ The negative sign for $I_2$ indicates that the assumed direction for $I_2$ was opposite to its actual direction. Now, find $I_3$: $I_3 = I_1 + I_2 = \frac{28}{11} + (-\frac{10}{11}) = \frac{18}{11} \text{ A} \approx 1.64 \text{ A}$. Final Answer: Current through $R_1$ ($I_1$) = $\frac{28}{11} \text{ A}$. Current through $R_2$ ($I_2$) = $-\frac{10}{11} \text{ A}$ (meaning $10/11$ A in the opposite direction of what was assumed). Current through $R_3$ ($I_3$) = $\frac{18}{11} \text{ A}$.

Exam Corner: Common Pitfalls & Tips

1. Sign Conventions in Kirchhoff's Laws: This is the most frequent source of error. For KVL, remember: potential increase when going from negative to positive terminal of a battery (take EMF as positive), and decrease (negative $IR$) when going across a resistor in the direction of current. Vice versa for opposite directions.
2. Distinguishing EMF and Terminal Voltage: EMF ($\mathcal{E}$) is the maximum potential difference a source can provide. Terminal voltage ($V$) is the actual potential difference across the terminals when current is drawn ($V = \mathcal{E} - Ir$, where $r$ is internal resistance). Remember, $V < \mathcal{E}$ when current is supplied, and $V > \mathcal{E}$ when current is drawn in by an external source.
3. Limitations of Ohm's Law: Do not apply Ohm's Law universally. It's valid only for ohmic conductors under constant physical conditions (especially temperature). Materials like semiconductors (diodes, transistors) are non-ohmic.
4. Wheatstone Bridge: Learn the balanced condition for a Wheatstone bridge ($P/Q = R/S$) and how to derive it using KVL or by showing zero current in the galvanometer arm. Practice applying it to simplify complex circuits.
5. Potentiometer vs. Voltmeter: Understand their distinct principles and applications. A potentiometer measures EMF accurately because it draws no current from the source at null point, unlike a voltmeter which always draws a small current.

Practice Questions with Solutions

• Q: A wire of resistance $4 \Omega$ is stretched to twice its original length. Assuming that its volume remains unchanged, what will be its new resistance? A: Step 1: Understand the relationship between resistance, length, and area. Resistance $R = \rho \frac{L}{A}$, where $\rho$ is resistivity, $L$ is length, and $A$ is cross-sectional area. Step 2: Relate initial and final states using volume conservation. Original length $L_0$, original area $A_0$. Original volume $V_0 = A_0 L_0$. New length $L_1 = 2L_0$. New area $A_1$. New volume $V_1 = A_1 L_1$. Since volume remains unchanged, $A_0 L_0 = A_1 L_1 = A_1 (2L_0) \implies A_1 = \frac{A_0}{2}$. Step 3: Calculate the new resistance. Initial resistance $R_0 = \rho \frac{L_0}{A_0} = 4 \Omega$. New resistance $R_1 = \rho \frac{L_1}{A_1} = \rho \frac{2L_0}{A_0/2} = \rho \frac{4L_0}{A_0} = 4 \left( \rho \frac{L_0}{A_0} \right) = 4 R_0$. $R_1 = 4 \times 4 \Omega = 16 \Omega$. Final answer: The new resistance will be $16 \Omega$.
• Q: Calculate the current flowing through the $6 \Omega$ resistor in the given circuit if the internal resistance of the battery is negligible. (Circuit Description: A 12V battery is connected to a parallel combination of a $4 \Omega$ resistor and a series combination of $6 \Omega$ and $3 \Omega$ resistors. The parallel combination is connected across the battery.) A: Step 1: Identify the combinations of resistors. The $6 \Omega$ and $3 \Omega$ resistors are in series. Their equivalent resistance is $R_{s} = 6 \Omega + 3 \Omega = 9 \Omega$. Step 2: Calculate the equivalent resistance of the parallel combination. This $R_s$ is in parallel with the $4 \Omega$ resistor. So, $R_{eq} = \left( \frac{1}{4} + \frac{1}{9} \right)^{-1} = \left( \frac{9+4}{36} \right)^{-1} = \frac{36}{13} \Omega$. Step 3: Calculate the total current drawn from the battery. Total current $I_{total} = \frac{V}{R_{eq}} = \frac{12 \text{ V}}{(36/13) \Omega} = \frac{12 \times 13}{36} = \frac{13}{3} \text{ A}$. Step 4: Calculate the current through the $6 \Omega$ resistor. The voltage across the parallel combination is 12V. So, the voltage across the series combination ($6 \Omega$ and $3 \Omega$) is also 12V. The current through this series branch ($I_{branch}$) is $I_{branch} = \frac{V}{R_s} = \frac{12 \text{ V}}{9 \Omega} = \frac{4}{3} \text{ A}$. Since the $6 \Omega$ resistor is in series with the $3 \Omega$ resistor within this branch, the current through the $6 \Omega$ resistor is the same as $I_{branch}$. Final answer: The current flowing through the $6 \Omega$ resistor is $\frac{4}{3} \text{ A}$ (approximately $1.33 \text{ A}$).
• Q: A cell of EMF $2 \text{ V}$ and internal resistance $0.1 \Omega$ is connected to a $3.9 \Omega$ external resistance. What is the terminal voltage across the external resistor? A: Step 1: Calculate the total resistance of the circuit. Total resistance $R_{total} = R_{external} + R_{internal} = 3.9 \Omega + 0.1 \Omega = 4.0 \Omega$. Step 2: Calculate the total current flowing in the circuit. Current $I = \frac{\text{EMF}}{R_{total}} = \frac{2 \text{ V}}{4.0 \Omega} = 0.5 \text{ A}$. Step 3: Calculate the terminal voltage across the external resistor. The terminal voltage across the external resistor is $V_{terminal} = I \times R_{external}$. $V_{terminal} = 0.5 \text{ A} \times 3.9 \Omega = 1.95 \text{ V}$. Final answer: The terminal voltage across the external resistor is $1.95 \text{ V}$.
• Q: A potentiometer wire of length 1 m has a resistance of $10 \Omega$. It is connected in series with a resistance of $R$ and a $2 \text{ V}$ battery. If a primary cell of EMF $1.0 \text{ V}$ is balanced across $40 \text{ cm}$ of the potentiometer wire, calculate the value of $R$. A: Step 1: Calculate the potential gradient of the potentiometer wire. Let $\phi$ be the potential gradient (potential drop per unit length). At balance, the EMF of the cell equals the potential drop across the balanced length: $EMF_{cell} = \phi \times L_{balance}$. $1.0 \text{ V} = \phi \times 0.40 \text{ m} \implies \phi = \frac{1.0}{0.40} = 2.5 \text{ V/m}$. Step 2: Relate potential gradient to the main circuit. The potential gradient is also given by $\phi = \frac{I \times R_{potentiometer}}{L_{potentiometer}}$, where $I$ is the current from the main battery. So, $2.5 \text{ V/m} = \frac{I \times 10 \Omega}{1 \text{ m}} \implies I = \frac{2.5}{10} = 0.25 \text{ A}$. Step 3: Apply Ohm's Law to the main circuit. The total resistance in the main circuit is $R_{total} = R_{potentiometer} + R = 10 \Omega + R$. The current in the main circuit is $I = \frac{EMF_{battery}}{R_{total}}$. $0.25 \text{ A} = \frac{2 \text{ V}}{10 \Omega + R}$. Step 4: Solve for $R$. $0.25 (10 + R) = 2 \implies 2.5 + 0.25R = 2 \implies 0.25R = 2 - 2.5 = -0.5$. This result indicates an issue with the problem statement or my interpretation. Let's recheck logic. If $EMF_{cell} = 1.0V$ is balanced by $40cm$, the potential drop across $40cm$ of the wire is $1.0V$. The total potential drop across the 1m wire is $\phi \times 1m = 2.5 \text{ V/m} \times 1 \text{ m} = 2.5 \text{ V}$. This cannot be, as the driving battery is only 2V. The potential drop across the potentiometer wire must be less than or equal to the battery voltage. A potential gradient of 2.5 V/m means total drop across 1m is 2.5V, which exceeds the source EMF of 2V. This is an invalid setup. Let's assume the question meant EMF of primary cell is $0.4 V$ instead of $1.0 V$. Recalculate with $EMF_{cell} = 0.4 V$ and balance length $40cm$. $EMF_{cell} = \phi \times L_{balance} \implies 0.4 \text{ V} = \phi \times 0.40 \text{ m} \implies \phi = 1.0 \text{ V/m}$. Now, $I = \frac{\phi \times L_{potentiometer}}{R_{potentiometer}} = \frac{1.0 \text{ V/m} \times 1 \text{ m}}{10 \Omega} = 0.1 \text{ A}$. $I = \frac{EMF_{battery}}{R_{total}} = \frac{2 \text{ V}}{10 \Omega + R}$. $0.1 = \frac{2}{10 + R} \implies 0.1(10 + R) = 2 \implies 1 + 0.1R = 2 \implies 0.1R = 1 \implies R = 10 \Omega$. Assuming a typo in the question for the primary cell EMF, for it to be solvable with the given battery voltage, $R = 10 \Omega$. Final answer: Assuming the primary cell EMF was $0.4 \text{ V}$ (consistent with a $2 \text{ V}$ main battery and $40 \text{ cm}$ balance length), the value of $R$ is $10 \Omega$. (If primary cell EMF is indeed $1.0V$, the circuit parameters are incorrectly stated as potential drop across potentiometer cannot exceed supply EMF).

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