#capacitor discharge

Rectifiers, Filters and Applications

In a rectifier with a capacitor filter, the capacitor discharges through:

**Rectifier applications** diode must have reverse breakdown voltage higher than peak inverse voltage, centre-tap transformer provides two opposite phase voltages for full-wave, capacitor discharges through load R_L when diode off, drift current in junction is minority carrier motion due to field, diffusion due to gradient, diode conducts when forward biased anode positive. The capacitor charges to the peak voltage and discharges through the load resistor ( R_L ) during the non-conducting half-cycle, smoothing the output voltage. Substituting values gives Load resistor, which matches expected

Ref: NCERT > Physics Book > Electronic Devices > Rectifiers, Filters 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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