Current Electricity in One Shot for JEE 2026

Mastering Current Electricity in One Shot for JEE 2026 is essential for students, as it explores the flow of electric charge which forms the bedrock of circuit analysis. This topic is fundamental for understanding how electronic components function and interact, providing the necessary knowledge for solving complex problems in competitive exams. Check the comprehensive breakdown of definitions, microscopic models, circuit laws, and AC fundamentals to ensure total exam readiness.

Current Electricity Fundamentals

Understanding the movement of charges is the first step in mastering circuit dynamics. This section breaks down the two primary ways we measure charge flow: as a total macroscopic quantity and as a localized vector quantity.

Electric Current (I)

Electric current is the rate of flow of charge through any conductor's cross-section.

• Instantaneous Current: I = dq / dt

• Total Charge Flow: q = ∫ I dt

• Unit: Ampere (A). Charge is in Coulombs (C).

• Nature: Current is a scalar quantity; its direction does not follow vector addition laws.

Current Density (J)

Current density is the current flowing per unit cross-sectional area.

• For uniform current: J = I / A

• For non-uniform current: I = ∫ J ⋅ dA

• Unit: Ampere per square meter (A/m²).

• Nature: Current density is a vector quantity, aligned with the current flow direction.

Microscopic Model of Current and Ohm's Law

In a current-carrying wire, the electric field (E) and current density (J) align with the conventional current direction, while electrons drift opposite to the electric field.

Key Microscopic Formulas

1. Current and Drift Velocity: I = n A e v_d

• n: Electron density, A: Area, e: Electron charge, v_d: Drift velocity.

2. Current Density and Drift Velocity: J = n e v_d

3. Ohm's Law (Vector Form): E = ρ J or J = σ E

• E: Electric field in conductor (not zero in electrodynamics).

• ρ: Resistivity, σ: Conductivity (σ = 1/ρ).

4. Mobility (μ): μ = v_d / E (Drift velocity per unit electric field).

5. Resistivity and Relaxation Time (τ): ρ = m / (n e² τ)

• m: Electron mass, τ: Relaxation time (avg. time between collisions).

6. Potential Difference and Electric Field: For uniform conductor, V = E L.

7. Resistance (Macroscopic Ohm's Law): V = I R.

• R = ρ L / A. L is length along current, A is area perpendicular to current.

Worked Example 1: Calculating Charge Flow

Problem: Current I(t) = 0.02t + 0.01 A. Find charge from t=1s to t=2s.

Solution: q = ∫[1, 2] (0.02t + 0.01) dt = [0.01t² + 0.01t] |_[1, 2] = 0.04 C.

Worked Example 2: Calculating Mobility

Problem: Conductor L=2m, A=2mm² (2x10⁻⁶ m²), I=1.6A, V=2V. Electron density n=5x10²⁸ m⁻³. Find μ.

Solution: μ = (I * L) / (n * A * e * V) = (1.6 * 2) / (5x10²⁸ * 2x10⁻⁶ * 1.6x10⁻¹⁹ * 2) = 2 * 10⁻⁴ m²/Vs.

Special Case: Resistance of a Hollow Sphere

For a hollow conducting sphere (inner a, outer b) with radial current I:

R = (ρ / 4π) * (1/a - 1/b).

Variation of Resistance with Temperature

Resistance of most conductors increases with temperature.

• Formula: R_T = R₀ (1 + αΔT)

• R_T: Resistance at T, R₀: Resistance at reference T₀.

• α: Temperature coefficient of resistance, ΔT = T - T₀.

• Crucial Pedagogical Instruction: R₀ must be the resistance at 0°C. Always relate back to 0°C.

Worked Example: Finding R₀ and α

Problem: Resistance 440 Ω at 50°C, 480 Ω at 100°C. Find R₀.

Solution:

1. 480 = R₀ (1 + 100α)

2. 440 = R₀ (1 + 50α)
   Dividing (1) by (2) gives α = 1/500 °C⁻¹. Substituting α into (1) yields R₀ = 400 Ω.

Equivalent Thermal Coefficient (α_eq)

When resistors combine, their equivalent also has a thermal coefficient.

Combination of Resistors and Batteries

• For two resistors in parallel: R_eq = (R₁R₂) / (R₁ + R₂)

• Current Divider Rule: I₁ = I_total * (R₂ / (R₁ + R₂))

Wheatstone Bridge

• Condition for Balance: R₁ / R₄ = R₂ / R₃.

• Result: If balanced, potential at bridge points are equal, and no current flows through the central resistor (it can be removed).

Combination of Batteries

(Memory Tip: The logic for an equivalent property is to "remove the many and replace with one" (दो को हटाओ एक लगाओ).)

Worked Example: Resistance of a Wire Bent into a Shape (Tetrahedron)

Problem: Wire of total resistance R bent into a triangular pyramid (6 identical edges). Find R_eq between adjacent vertices A and B.

Solution: Each edge resistance r = R/6. The circuit forms a balanced Wheatstone bridge. Between A and B, there's one direct r path, and two parallel 2r paths (from series r+r).

R_eq = (r || (2r || 2r)) = (r || r) = r/2.

Substituting r = R/6, R_eq = (R/6)/2 = R/12.

Worked Example: Comparing Series and Parallel Cell Configurations

Problem: Cell 1 (E₁=1V, r₁=2Ω), Cell 2 (E₂=2V, r₂=1Ω), R_ext=6Ω. Compare current in series (I₁) vs. parallel (I₂).

Solution:

1. Series: E_eq = 3V, r_eq = 3Ω. I₁ = E_eq / (R_ext + r_eq) = 3 / (6+3) = 1/3 A.

2. Parallel: 1/r_eq = 1/2 + 1/1 = 3/2 => r_eq = 2/3 Ω. E_eq = (1/2 + 2/1) * (2/3) = 5/3 V. I₂ = E_eq / (R_ext + r_eq) = (5/3) / (6 + 2/3) = (5/3) / (20/3) = 1/4 A.
   Ratio I₁ / I₂ = (1/3) / (1/4) = 4/3.

Worked Example: Equivalent Resistance with an Unknown Resistor

Problem: R, 2R, and X are connected. R_eq between A and B is X. Find X.

Solution: The (2R + X) series combination is parallel with R.

X = [ R * (2R + X) ] / [ R + (2R + X) ]

X(3R + X) = 2R² + RX

X² + 2RX - 2R² = 0

Using quadratic formula, X = R(-1 ± √3). Since resistance is positive, X = R(√3 - 1).

Worked Example: Equivalent Resistance of a Circular Wire with a Chord

Problem: Wire λ (Ω/m) bent into a circle of radius r. Another wire of length 2r connects as a diameter. Find R_eq.

Solution: Three parallel paths: top arc, diameter, bottom arc.

• Path 1 (90° arc): R₁ = λ * (πr/2)

• Path 2 (Straight wire): R₂ = λ * 2r

• Path 3 (270° arc): R₃ = λ * (3πr/2)
  1/R_eq = 1/R₁ + 1/R₂ + 1/R₃ = 2/(λπr) + 1/(2λr) + 2/(3λπr)
  1/R_eq = (12 + 3π + 4) / (6λπr) = (16 + 3π) / (6λπr)
  R_eq = (6λπr) / (16 + 3π).

Worked Example: Identical Cells: Current in Series vs. Parallel

Problem: Two identical cells (E, r) connected to 6Ω. Current is same for series and parallel. Find r.

Solution:

1. Series: I_series = 2E / (6 + 2r)

2. Parallel: I_parallel = E / (6 + r/2)
   Equating I_series = I_parallel: 2 / (6 + 2r) = 1 / (6 + r/2)
   12 + r = 6 + 2r => r = 6 Ω.

Worked Example: Potential Difference in a Modified Wheatstone Bridge

Problem: Wheatstone bridge (R₁, R₂, R₃, R₄ all R). R₃ increases by 10%. Source 40V. Find V_A - V_B.

Solution: R₃_new = 1.1R.

• Potential at A: V_A = 20V (since R₁=R₂=R).

• Potential at B: V_B = (V_total * R₄) / (R₄ + R₃_new) (voltage divider assuming B is from 0V to R₄).
  V_B = (40 * R) / (R + 1.1R) = 40R / 2.1R = 40 / 2.1 V.
  V_A - V_B = 20 - (40 / 2.1) = (42 - 40) / 2.1 = 2 / 2.1 V ≈ 0.95 V.

Worked Example: Unbalanced Wheatstone Bridge Analysis

Problem: Unbalanced bridge (3Ω, 6Ω, 3Ω, 3Ω, 6Ω). Find R_eq.

Solution: Use the Point Potential Method (Nodal Analysis).

1. Assign potentials V, 0, x, y to nodes.

2. Apply KCL at nodes x and y to get equations for x and y in terms of V.

• 5x - 2y = V

• -2x + 5y = 2V

3. Solve for x and y. Then calculate total current I = (V-x)/R_ax + (V-y)/R_ay.

4. R_eq = V / I. The result is R_eq = 21/5 Ω.

Resistance Change from Stretching a Wire

When a wire is stretched, its volume (Volume = A * L) remains constant.

Derivation of the Key Relationship

R = ρL/A. Since A = Volume/L, R = ρL / (Volume/L) = (ρ/Volume) * L².

Thus, R ∝ L² (as ρ and Volume are constant).

Application based on Magnitude of Change

1. For Large Changes: If L_new = n * L_old, then R_new = R_old * n².

• Doubling length (n=2) makes resistance four times original (4R_old).

2. For Small Percentage Changes: ΔR/R ≈ 2 * (ΔL/L).

• Percentage change in resistance is twice the percentage change in length.

Worked Example: Stretched Wire Bent into a Circle

Problem: Metal wire R=3 Ω elongated to double length, then bent into a circle. Find R_eq between two points subtending 60°.

Solution:

1. New Resistance: R_new = R_initial * (2)² = 3 Ω * 4 = 12 Ω.

2. Arc Resistances: 60° arc is 1/6th of total.

• Arc 1 (60°): R₁ = (1/6) * 12 Ω = 2 Ω.

• Arc 2 (300°): R₂ = (5/6) * 12 Ω = 10 Ω.

3. Equivalent Resistance: Arcs in parallel.
   R_eq = (R₁ × R₂) / (R₁ + R₂) = (2 × 10) / (2 + 10) = 20 / 12 = 5/3 Ω.

Electric Power and Electric Energy

Electric Power (P)

• Universal Formula: P = V × I (Applies to any device).

• Resistor Specific Formulas: Using V=IR, P = I²R or P = V²/R. (Safer to use I²R for resistors).

Efficiency (η)

• Universal Definition: η = (Power_out / Power_in) × 100.

• Battery Efficiency: η = [R / (R + r)] × 100.

Electric Energy (E)

• Formula: E = Power × time (for constant power), E = ∫P dt (for variable power).

• Units:

1. Joule (J): SI unit.

2. Kilowatt-hour (kWh): Commercial unit ("unit" of electricity). 1 kWh = 3.6 × 10⁶ J.

3. Calorie (cal): 1 cal ≈ 4.2 J.

Bulb-Related Problems

A light bulb is a resistor. Use rated power (P_rated) and voltage (V_rated) to find its constant resistance:

R = V_rated² / P_rated. Then treat it as a simple resistor.

Worked Example 1: Energy in a Mobile Phone Battery

Problem: Battery 4.2V, 5800 mAh. Find stored energy.

Solution: E = V × (I × t) = 4.2 V × (5800 × 10⁻³ A × 3600 s). Calculate in Joules.

Worked Example 2: Efficiency of a Motor

Problem: Motor 100V, draws 1A. Efficiency 91.6%. Find power loss in cal/s.

Solution:

1. P_in = V × I = 100W.

2. P_out = η × P_in = 0.916 × 100 = 91.6W.

3. P_loss = P_in - P_out = 8.4W (or 8.4 J/s).

4. P_loss (cal/s) = 8.4 / 4.2 = 2 cal/s.

Worked Example 3: N Identical Bulbs in Series

Problem: N identical bulbs (rated P at V) in series to V. Total power drawn?

Solution:

1. Resistance of one bulb: R = V² / P.

2. Total series resistance: R_eq = N × R.

3. Total power: P_total = V² / R_eq = V² / (N × R) = V² / (N × (V²/P)) = P/N.

• Power per bulb: P_one_bulb = P_total / N = P/N².

Circuit Analysis Problems

Worked Example 1: Balanced Wheatstone Bridge

Problem: 9V battery (1Ω internal r). External circuit: 1Ω, 2Ω, 2Ω, 4Ω in bridge with 1Ω middle resistor. Find heat in external circuit in 1 min.

Solution:

1. Bridge is balanced: 1Ω/2Ω = 2Ω/4Ω = 1/2. Middle 1Ω resistor can be removed.

2. External R_eq: Top branch 1+2=3Ω. Bottom branch 2+4=6Ω. R_ext = (3×6)/(3+6) = 2Ω.

3. Total current: R_total = R_ext + r_internal = 2Ω + 1Ω = 3Ω. I = V / R_total = 9V / 3Ω = 3A.

4. Heat generated: H = I² × R_ext × t = (3A)² × 2Ω × 60s = 1080 J.

Maximum Power Transfer Theorem

• Statement: For a DC source with internal resistance r, maximum power is delivered to external load R when R = r.

• Condition: R = r.

Worked Example: Power Line Efficiency

Problem: Power line (2Ω resistance) delivers 1 kW at 250 V. Find efficiency.

Solution:

1. P_out = 1000W, V_out = 250V.

2. Line current: I = P_out / V_out = 1000 / 250 = 4A.

3. Power loss in line: P_loss = I² × R_line = 4² × 2 = 32W.

4. Input power: P_in = P_out + P_loss = 1000 + 32 = 1032W.

5. Efficiency: η = (P_out / P_in) × 100 = (1000 / 1032) × 100.

Conversion of Galvanometer

Conversion to Ammeter

• Method: Connect a low shunt (S) resistance in parallel with the galvanometer.

• Formula: S = (IG × RG) / (I - IG)

• IG: Full-scale deflection current, RG: Galvanometer resistance.

• I: Desired max current range.

• Ideal Ammeter: Zero resistance.

Conversion to Voltmeter

• Method: Connect a high resistance (R) in series with the galvanometer.

• Formula: V = IG × (RG + R)

• V: Max voltage range.

• Ideal Voltmeter: Infinite resistance.

Instruments in Circuits: A Key Strategy

Replace the instrument with its corresponding resistance.

Worked Examples on Instrument Conversion & Circuits

Example 1: Ammeter Conversion

Problem: Convert galvanometer (RG = 100 Ω, IG = 1 mA) to ammeter range I = 5 mA. Find shunt (S).

Solution: S = (1mA × 100Ω) / (5mA - 1mA) = 100 / 4 = 25 Ω.

Example 2: Ammeter in a Circuit

Problem: 20V source, 150.4Ω resistor in series with parallel (ammeter 240Ω, shunt 10Ω). Find ammeter reading.

Solution:

1. Parallel R_eq (ammeter + shunt): R_p = (240×10)/(240+10) = 9.6Ω.

2. Total resistance: R_total = 150.4 + 9.6 = 160Ω.

3. Total current: I_total = 20V / 160Ω = 1/8 A.

4. Ammeter current (current divider): I_ammeter = I_total × [10 / (240 + 10)] = (1/8) × (10/250) = 1/200 A = 5 mA.

Example 3: Voltmeter in a Circuit

Problem: Two 100Ω resistors in series with 9V. Voltmeter (400Ω) in parallel across one 100Ω. Find voltmeter reading.

Solution:

1. Parallel R_eq (voltmeter + 100Ω): R_p = (400×100)/(400+100) = 80Ω.

2. Total resistance: R_total = 100Ω + 80Ω = 180Ω.

3. Total current: I = 9V / 180Ω = 1/20 A.

4. Voltmeter reading: V_reading = I × R_p = (1/20 A) × 80Ω = 4 V.

Meter Bridge Experiment

Operates on the principle of a balanced Wheatstone bridge.

Core Principle: At Balance Point

For resistances R and S in gaps, balance at length x: R * (100 - x) = S * x.

Worked Example 1 (JEE Main)

Problem: Meter bridge R₁, R₂ balanced at 40 cm. Add 16Ω parallel to R₂, null point shifts to 50 cm. Find R₁, R₂.

Solution:

1. Initial: R₁ * 60 = R₂ * 40 => 3R₁ = 2R₂ (Eq 1).

2. Modified: R₂' = (16R₂) / (16+R₂). Null at 50 cm. R₁ * 50 = R₂' * 50 => R₁ = R₂' (Eq 2).

3. Substitute (1) into (2): 2R₂/3 = (16R₂)/(16+R₂) => 2(16+R₂) = 3*16 => 32 + 2R₂ = 48 => R₂ = 8 Ω.

4. From (1): 3R₁ = 2*8 => R₁ = 16/3 Ω.

Potentiometer (Precautionary Overview for a Deleted Topic)

Instructor's Note: Though officially deleted from JEE syllabus, questions have appeared. Study these formulas as a precaution.

Experiment 1: Comparison of Two EMFs

E₁ / E₂ = L₁ / L₂ (L₁, L₂ are balance lengths for E₁, E₂).

Experiment 2: Internal Resistance of a Cell

r = [(L₁ - L₂) / L₂] * R (L₁: balance length without shunt R; L₂: with shunt R).

Parallel Plate Capacitor

Consists of two plates (area A, separation d).

• Capacitance (C): C = ε₀A / d

• Charge (Q): Q = CV

• Electric Field (E): E = σ / ε₀ = Q / (Aε₀) = V / d

• Potential Difference (V): V = Ed

• Energy Stored (U): U = ½CV² = Q² / (2C) = ½QV

• Energy Density (u): u = ½ε₀E²

• Force (F) between plates: F = Q² / (2Aε₀)

Capacitor with Battery

When connected to a battery V, it charges instantaneously to Q = CV with V across it.

Series and Parallel Combination of Capacitors

Formulas are the opposite of resistors.

• In Series: Charge (Q) is the same.

• In Parallel: Voltage (V) is the same.

• Charge Division in Parallel: Q₁ = [C₁ / (C₁ + C₂)] * Q_total. (Memory Tip: C ज्यादा तो Q ज्यादा).

• Wheatstone Bridge for Capacitors: If C₁ * C₃ = C₂ * C₄, the bridge is balanced and central capacitor can be removed.

Capacitors with Dielectrics

Inserting a dielectric (K) affects properties. Q_induced on dielectric surfaces.

• Capacitance: C_dielectric = K * C_air = A * K * ε₀ / d. (Always increases C). For metal, K = ∞.

• Electric Field: E = Q / (A * K * ε₀).

• Induced Charge: Q_induced = Q * (1 - 1/K). (Frequently tested).

• Induced Electric Field: E_induced = (Q / Aε₀) * (1 - 1/K).

Combinations of Dielectrics

(Memory Tip: Series is like two roads end-to-end; Parallel is like two separate parallel roads.)

Special Case: Inserting a Metal Slab (Series)

Thickness t metal slab in d gap: C_eq = A * ε₀ / (d - t). (K of metal is ∞).

Advanced Dielectric Configurations

For complex setups where one slab partially covers others, virtually split the top slab to identify series/parallel combinations of paths.

Sharing of Charge Between Capacitors

When two charged capacitors (C₁V₁, C₂V₂) are connected, charge flows until a common potential is reached.

Comparative Structure: Connecting Charged Capacitors

(Memory Tip: "Like" terminals: less conflict, V₁-V₂. "Unlike" terminals: more conflict, V₁+V₂.)

RC Circuits

Contain a resistor (R) and capacitor (C).

RC Charging Circuit

(Uncharged capacitor, R, DC source V)

• Charge q(t): q(t) = CV * (1 - e^(-t/RC)) (Memorize).

• CV: Steady-state charge. RC: Time constant τ.

• Current i(t): i(t) = (V/R) * e^(-t/RC).

RC Discharging Circuit

(Charged capacitor q_not connected across R)

• Charge q(t): q(t) = q_not * e^(-t/RC).

(Memory Tip: 1 - e^(-t/τ) for charging/increasing quantities; e^(-t/τ) for discharging/decreasing quantities.)

Circuit Analysis Shortcuts for Capacitors (Jugaad)

Worked Examples: RC Circuits

Problem 1 (Steady State Ammeter Reading): In a circuit, find ammeter reading in steady state.

Solution: Capacitor acts as broken wire. Redraw circuit removing capacitor branch. Solve simplified circuit. If 10V source, 1Ω resistor, two 8Ω parallel resistors, then ammeter is 1A.

Problem 2 (Time-dependent Voltage): 1µF capacitor, 100V battery, 100Ω resistor. Time for capacitor voltage to reach 50V?

Solution: Vc(t) = V_battery * (1 - e^(-t/RC)).

50 = 100 * (1 - e^(-t/RC)) => 0.5 = 1 - e^(-t/RC) => e^(-t/RC) = 0.5.

-t/RC = ln(0.5) = -ln(2) => t = RC * ln(2) = (100 * 10⁻⁶) * ln(2) s.

Problem 3 (Steady State Charge): Find steady-state charge on 50µF capacitor in a circuit.

Solution: At steady state, capacitor is broken wire. Find voltage across capacitor terminals (V_c). Then Q = C * V_c. If 6V battery, three 2kΩ resistors, then V_c = 2V, Q = 50µF * 2V = 100µC.

LR Circuits and Inductors

Self-Inductance of a Solenoid

L = μ₀ n² * Volume (where n = turns per unit length).

If material with μ_r inserted, L' = μ_r * L.

Inductor Behavior in a Circuit

• V = L (di/dt) (Voltage across inductor).

• Inductor opposes any change in current. (Memory Tip: Inductor resists change – "neither let the current live nor let it die").

• If di/dt = 0 (constant current), V = 0.

• Energy Stored: U = ½ LI².

Combination of Inductors

Combine like resistors:

• Series: L_eq = L₁ + L₂ + ...

• Parallel: 1/L_eq = 1/L₁ + 1/L₂ + ...

LR Charging and Discharging Circuits

LR Charging Circuit

(Inductor, R, DC source V)

• Current i(t): i(t) = (V/R) * (1 - e^(-Rt/L)) (Memorize).

• V/R: Steady-state current. L/R: Time constant τ.

LR Discharging Circuit

(Inductor i_not connected across R)

• Current i(t): i(t) = i_not * e^(-Rt/L).

Circuit Analysis Shortcuts for Inductors (Jugaad)

Inductor behavior is the opposite of a capacitor.

Worked Examples: LR Circuits

Problem 1 (Initial and Final State): Find ammeter reading at t=0 and t=∞.

Solution:

• At t=0: Inductor acts as broken wire. Solve circuit with open inductor branches.

• At t=∞: Inductor acts as conducting wire. Solve circuit with shorted inductor branches.

Problem 2 (Energy Density): Inductor (L=10mH) with 10V DC, R=10Ω. Energy density when current reaches 1/e of max value.

Solution:

1. Max current i_not = V/R = 10V/10Ω = 1A. Current i = i_not/e = 1/e A.

2. Energy stored U = ½ L i² = ½ (10 * 10⁻³ H) * (1/e)² = 5 * 10⁻³ / e² J.

3. Inductance L = μ₀ n² * Volume. So, L / Volume = μ₀ n².

4. Energy Density = U / Volume = (½ L i²) / Volume = ½ (L/Volume) i² = ½ (μ₀ n²) (1/e)² = (μ₀ n²) / (2e²).

Alternating Current

1. AC Source: The AC Generator

Coil (N turns, area A) rotates with ω in magnetic field B.

• Induced EMF: V = V₀ sin(ωt).

• Maximum EMF: V₀ = NBAω (Frequently tested).

• Phase Relationship: Flux and EMF are π/2 (90°) out of phase. When flux is max, EMF is zero, and vice-versa.

• ω must be in radians per second.

2. RMS Value and Superposition

The Root Mean Square (RMS) value is the effective value for time-varying current/voltage.

• Sinusoidal Wave (I = I₀ sin(ωt)): I_rms = I₀ / √2.

• Full-Wave Rectified: I_rms = I₀ / √2.

• Half-Wave Rectified: I_rms = I₀ / 2.

• Square Wave: I_rms = I₀.

• Triangular / Sawtooth Wave: I_rms = I₀ / √3.

Superposition of Currents

1. Two AC sources (I₁ = A sin(ωt), I₂ = B sin(ωt + φ)):

• I_net_max = √[ A² + B² + 2AB cos(φ) ].

• I_net_rms = I_net_max / √2.

2. AC and DC (I(t) = A + B sin(ωt)):

• I_rms = √[ A² + (B² / 2) ]. (Frequently tested).

3. RLC Series Circuit

(R, L, C in series to V = V₀ sin(ωt))

• Inductive Reactance: XL = ωL.

• Capacitive Reactance: XC = 1 / (ωC).

• Impedance (Z): Total opposition. Z = √[ R² + (XL - XC)² ].

• Current: I_rms = V_rms / Z. (Voltage given without qualification is RMS).

• Phase Difference (φ): Angle between V and I. tan(φ) = (XL - XC) / R.

• Power Factor (cos φ): cos(φ) = R / Z.

• Average Power: P_avg = I_rms² * R.

4. Resonance in RLC Circuits

Special condition where XL = XC.

• Conditions: XL = XC, Z is minimum (Z = R), I is maximum (I_max = V/R), Power is maximum, cos(φ) = 1 (V and I in phase).

• Resonance Frequency: ω_r = 1 / √[LC].

• Quality Factor (Q): Q = ω_r / Δω = XL / R = XC / R.

Solved Problem: LCR Circuit Power Factor

Problem: LCR circuit with V=220V, f=50Hz, R=60Ω, XL=150Ω, XC=70Ω. Power factor = α/10. Find α.

Solution:

1. Z = √(R² + (XL - XC)²) = √(60² + (150-70)²) = √(60² + 80²) = √10000 = 100 Ω.

2. cos φ = R / Z = 60 / 100 = 0.6.

3. α / 10 = 0.6 => α = 6.

Solved Problem: Inductor with Internal Resistance

Problem: Inductor stores 16J, dissipates 32W (thermal) with 2A (RMS) AC. Find XL / R ratio.

Solution:

1. Energy stored: E = ½ L I² => 16 = ½ L (2)² => L = 8 H.

2. Power dissipated: P = I² R => 32 = (2)² R => R = 8 Ω.

3. Ratio XL / R = (ωL) / R = (2πfL) / R. (Assuming typical f=100Hz for these problems).
   Ratio = (2 * π * 100 * 8) / 8 = 100π.

Identifying Resonance Conditions in LCR Circuits

A circuit is at resonance if any of these are mentioned:

• Maximum current

• Maximum power

• Minimum impedance (Z_min)

• Power factor (cos φ) = 1
  At resonance, Z = R and ω = 1 / √(LC).

Orally Solved Problems (Resonance)

Problem 1 (JEE 2025): LCR circuit. Find ω for maximum current.

Solution: Maximum current means resonance. ω = 1 / √(LC).

Problem 2 (JEE 2025): LCR circuit. Current I₀ at resonance. If R is doubled, new current at resonance?

Solution: At resonance, I = V/R. If R doubles, new current I' = V/(2R) = I₀ / 2.