#energy loss

Transformer, AC Generator and LC Oscillations

What is the primary energy loss mechanism in a real transformer that reduces its efficiency?

**AC generator** converts mechanical to electrical, emf e = N B A ω sin ωt, maximum e₀ = N B A ω, frequency = rotation frequency, transformer cannot work on DC because steady flux no induction, LC oscillations energy swaps between ½ L I² and ½ Q²/C at ω₀=1/√(LC). In a real transformer, energy losses occur due to resistance in the windings ( I² R losses), flux leakage, eddy currents, and hysteresis. Among these, resistance in the windings (copper loss) is a primary mechanism, converting electrical energy into heat. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms =

Ref: NCERT > Physics Book > Alternating Currents > Transformer, AC Generator and LC Oscillations

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Sharing of Charges, Common Potential and Applications

A \( 2 \, \mu\text{F} \) capacitor charged to \( 400 \, \text{V} \) is connected to an uncharged \( 6 \, \mu\text{F} \)

**Common potential** after connection is weighted average of initial potentials by capacitances. Energy loss ΔU = ½ C₁ C₂ (V₁-V₂)²/(C₁+C₂) dissipated as heat and spark, always positive unless V₁=V₂, explaining why energy reduces after sharing. Initial energy: U_i = (1/2) × 2 × 10⁻⁶ × (400)² = 0.16 J . Charge: Q = 2 × 10⁻⁶ × 400 = 8 × 10⁻⁴ C . Total C = 2 + 6 = 8 μF , V = (8 × 10⁻⁴/8 × 10⁻⁶) = 100 V . Final energy: U_f = (1/2) × 8 × 10⁻⁶ × (100)² = 0.04 J . Loss: U_i - U_f =

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Sharing of Charges, Common Potential and Applications

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Sharing of Charges, Common Potential and Applications

A \( 6 \, \mu\text{F} \) capacitor charged to \( 150 \, \text{V} \) is connected to an uncharged \( 9 \, \mu\text{F} \)

**Common potential** after connection is weighted average of initial potentials by capacitances. Energy loss ΔU = ½ C₁ C₂ (V₁-V₂)²/(C₁+C₂) dissipated as heat and spark, always positive unless V₁=V₂, explaining why energy reduces after sharing. Initial energy: U_i = (1/2) × 6 × 10⁻⁶ × (150)² = 0.0675 J . Charge: Q = 6 × 10⁻⁶ × 150 = 9 × 10⁻⁴ C . Total C = 6 + 9 = 15 μF , V = (9 × 10⁻⁴/15 × 10⁻⁶) = 60 V . Final energy: U_f = (1/2) × 15 × 10⁻⁶ × (60)² = 0.027 J . Loss: U_i - U_f =

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Sharing of Charges, Common Potential and Applications

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Sharing of Charges, Common Potential and Applications

A \( 2 \, \mu\text{F} \) capacitor charged to \( 200 \, \text{V} \) loses how much energy when connected to an uncharged

**Applications** include charge sharing for voltage division, Van de Graaff generator accumulates charge on spherical dome to high potential V = k Q/R, up to MV, using belt to transport charge, capacitance of sphere C=4π ε₀ R ≈10 pF for R=0.1 m, so Q= C V ≈10⁻⁸ C for 1000 V. Initial energy: U_i = (1/2) × 2 × 10⁻⁶ × (200)² = 0.04 J . Charge: Q = 2 × 10⁻⁶ × 200 = 4 × 10⁻⁴ C . Total C = 2 + 3 = 5 μF , V = (4 × 10⁻⁴/5 × 10⁻⁶) = 80 V . Final energy: U_f =

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Sharing of Charges, Common Potential and Applications

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Sharing of Charges, Common Potential and Applications

A \( 5 \, \mu\text{F} \) capacitor charged to \( 120 \, \text{V} \) is connected to an uncharged \( 15 \, \mu\text{F} \)

**Common potential** after connection is weighted average of initial potentials by capacitances. Energy loss ΔU = ½ C₁ C₂ (V₁-V₂)²/(C₁+C₂) dissipated as heat and spark, always positive unless V₁=V₂, explaining why energy reduces after sharing. Initial energy: U_i = (1/2) × 5 × 10⁻⁶ × (120)² = 0.036 J . Charge: Q = 5 × 10⁻⁶ × 120 = 6 × 10⁻⁴ C . Total C = 5 + 15 = 20 μF , V = (6 × 10⁻⁴/20 × 10⁻⁶) = 30 V . Final energy: U_f = (1/2) × 20 × 10⁻⁶ × (30)² = 0.009 J . Loss: U_i - U_f =

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Sharing of Charges, Common Potential and Applications

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Sharing of Charges, Common Potential and Applications

A \( 3 \, \mu\text{F} \) capacitor charged to \( 200 \, \text{V} \) is connected to an uncharged \( 9 \, \mu\text{F} \)

**Applications** include charge sharing for voltage division, Van de Graaff generator accumulates charge on spherical dome to high potential V = k Q/R, up to MV, using belt to transport charge, capacitance of sphere C=4π ε₀ R ≈10 pF for R=0.1 m, so Q= C V ≈10⁻⁸ C for 1000 V. Initial energy: U_i = (1/2) × 3 × 10⁻⁶ × (200)² = 0.06 J . Charge: Q = 3 × 10⁻⁶ × 200 = 6 × 10⁻⁴ C . Total C = 3 + 9 = 12 μF , V = (6 × 10⁻⁴/12 × 10⁻⁶) = 50 V . Final energy: U_f =

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Sharing of Charges, Common Potential and Applications

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Sharing of Charges, Common Potential and Applications

A \( 6 \, \mu\text{F} \) capacitor charged to \( 100 \, \text{V} \) is connected to an uncharged \( 18 \, \mu\text{F} \)

**Common potential** after connection is weighted average of initial potentials by capacitances. Energy loss ΔU = ½ C₁ C₂ (V₁-V₂)²/(C₁+C₂) dissipated as heat and spark, always positive unless V₁=V₂, explaining why energy reduces after sharing. Initial energy: U_i = (1/2) × 6 × 10⁻⁶ × (100)² = 0.03 J . Charge: Q = 6 × 10⁻⁶ × 100 = 6 × 10⁻⁴ C . Total C = 6 + 18 = 24 μF , V = (6 × 10⁻⁴/24 × 10⁻⁶) = 25 V . Final energy: U_f = (1/2) × 24 × 10⁻⁶ × (25)² = 0.0075 J . Loss: U_i - U_f =

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Sharing of Charges, Common Potential and Applications

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Energy Stored in Capacitor and Energy Density

An isolated charged capacitor is connected across a resistor momentarily. Why does the energy stored in the capacitor de

**Work done charging capacitor** is integral ∫ V dQ = ∫ Q/C dQ = Q²/2C, stored as electrostatic energy. When capacitor discharges, energy released as heat or work, explaining spark when shorted, energy proportional to V². When a charged capacitor is connected across a resistor, it discharges through the resistor, forming an RC circuit. The charge Q on the capacitor decreases exponentially ( Q(t) = Q₀ e⁻t/RC ). After a long time ( t to ∞ ), Q to 0 , so the energy U = (Q²/2C) to 0 . The energy is dissipated as heat in the resistor during discharge, leaving no energy stored

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Energy Stored in Capacitor and Energy Density

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Sharing of Charges, Common Potential and Applications

A \( 8 \, \mu\text{F} \) capacitor charged to \( 80 \, \text{V} \) is connected to an uncharged \( 24 \, \mu\text{F} \)

**Sharing of charges** when charged capacitor C₁ at V₁ connected to uncharged C₂, total charge Q = C₁ V₁ conserved, common potential V_common = Q/(C₁+C₂) = C₁ V₁/(C₁+C₂), final charges Q₁' = C₁ V_common, Q₂' = C₂ V_common. For 4 μF at 100 V (Q=4×10⁻⁴ C) connected to 4 μF uncharged, V_common=4×10⁻⁴/8×10⁻⁶=50 V. Initial energy: U_i = (1/2) × 8 × 10⁻⁶ × (80)² = 0.0256 J . Charge: Q = 8 × 10⁻⁶ × 80 = 6.4 × 10⁻⁴ C . Total C = 8 + 24 = 32 μF , V = (6.4 × 10⁻⁴/32 × 10⁻⁶) = 20 V . Final energy:

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Sharing of Charges, Common Potential and Applications

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Sharing of Charges, Common Potential and Applications

A \( 4 \, \mu\text{F} \) capacitor charged to \( 150 \, \text{V} \) is connected to an uncharged \( 12 \, \mu\text{F} \)

**Common potential** after connection is weighted average of initial potentials by capacitances. Energy loss ΔU = ½ C₁ C₂ (V₁-V₂)²/(C₁+C₂) dissipated as heat and spark, always positive unless V₁=V₂, explaining why energy reduces after sharing. Initial energy: U_i = (1/2) × 4 × 10⁻⁶ × (150)² = 0.045 J . Charge: Q = 4 × 10⁻⁶ × 150 = 6 × 10⁻⁴ C . Total C = 4 + 12 = 16 μF , V = (6 × 10⁻⁴/16 × 10⁻⁶) = 37.5 V . Final energy: U_f = (1/2) × 16 × 10⁻⁶ × (37.5)² = 0.01125 J . Loss: U_i - U_f =

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Sharing of Charges, Common Potential and Applications

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Sharing of Charges, Common Potential and Applications

A \( 4 \, \mu\text{F} \) capacitor charged to \( 300 \, \text{V} \) is connected to an uncharged \( 8 \, \mu\text{F} \)

**Sharing of charges** when charged capacitor C₁ at V₁ connected to uncharged C₂, total charge Q = C₁ V₁ conserved, common potential V_common = Q/(C₁+C₂) = C₁ V₁/(C₁+C₂), final charges Q₁' = C₁ V_common, Q₂' = C₂ V_common. For 4 μF at 100 V (Q=4×10⁻⁴ C) connected to 4 μF uncharged, V_common=4×10⁻⁴/8×10⁻⁶=50 V. Initial energy: U_i = (1/2) × 4 × 10⁻⁶ × (300)² = 0.18 J . Charge: Q = 4 × 10⁻⁶ × 300 = 1.2 × 10⁻³ C . Total C = 4 + 8 = 12 μF , V = (1.2 × 10⁻³/12 × 10⁻⁶) = 100 V . Final energy:

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Sharing of Charges, Common Potential and Applications

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Electrical Power, Energy and Heating Effect

Why is high voltage used in power transmission lines to reduce energy loss?

**Power dissipation** in resistor converts electrical energy to heat, P = V²/R inversely proportional to R for fixed V, directly proportional for fixed I. For battery with internal r, power wasted internally = I² r, useful power = I² R, efficiency η = R/(R+r). Power loss in lines is Plₒₛₛ = I² R . For a given power ( P = V I ), increasing V reduces I ( I = P / V ), significantly lowering I² R losses. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r

Ref: NCERT > Physics Book > Current Electricity > Electrical Power, Energy and Heating Effect

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