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#capacitors

39 public questions tagged with this topic.

AC Fundamentals - RMS, Average and Peak Values

A \( 50 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is th

**RMS value** I_rms = I_peak/√2, V_rms = V_peak/√2 for sinusoidal AC, significance rms gives equivalent DC value producing same heating power P = I_rms² R, average power over cycle, instruments measure rms, average over full cycle zero, half-cycle average 2 I_peak/π, peak = √2 rms. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 50 × 10⁻⁶ F . X_C = (1/314 × 50 × 10⁻⁶) ≈ 63.7 Ω . RMS current: I = (V/X_C) = (220/63.7) ≈ 3.45 A . Peak current: i_m = √(2) I = 1.414 × 3.45 ≈ 4.88 A . Applying X_L =

Ref: NCERT > Physics Book > Alternating Currents > AC Fundamentals - RMS, Average and Peak Values

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AC Through Capacitor - Capacitive Reactance

A \( 40 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the c

**Capacitor average power** zero over complete cycle because P=V I =½ V_peak I_peak sin2ωt average zero, energy stored in field, not dissipated, unlike resistor. For 15 μF, 60 Hz, X_C=176.8 Ω, V_rms=110 V, I_rms=0.622 A, illustrating lower C higher X_C. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 40 × 10⁻⁶ F . X_C = (1/314 × 40 × 10⁻⁶) ≈ 79.6 Ω . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms I_rms cosφ, V_s/V_p = N_s/N_p, calculation gives 79.6 Ω, consistent with phasor analysis

Ref: NCERT > Physics Book > Alternating Currents > AC Through Capacitor - Capacitive Reactance

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Power in AC Circuits - Power Factor and Wattless Current

What is the role of a capacitor in improving the power factor of an AC circuit with an inductive load?

**Power factor** cos φ = R/Z, 0≤cos φ≤1, at resonance cos φ=1 maximum power, pure L or C cos φ=0 zero average power, current wattless because energy stored returned. Instantaneous power when current maximum in pure L: I=I_peak, V=0 because V leads 90°, so P= V I =0 at that instant, average zero. An inductive load causes the current to lag the voltage, reducing the power factor. Adding a capacitor in parallel introduces a leading current that counteracts the lagging current, bringing the net phase difference closer to zero, thus improving the power factor. Applying X_L = ωL, X_C = 1/ωC, Z = √(R

Ref: NCERT > Physics Book > Alternating Currents > Power in AC Circuits - Power Factor and Wattless Current

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AC Through Capacitor - Capacitive Reactance

A \( 15 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the c

**Capacitor average power** zero over complete cycle because P=V I =½ V_peak I_peak sin2ωt average zero, energy stored in field, not dissipated, unlike resistor. For 15 μF, 60 Hz, X_C=176.8 Ω, V_rms=110 V, I_rms=0.622 A, illustrating lower C higher X_C. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 15 × 10⁻⁶ F . X_C = (1/314 × 15 × 10⁻⁶) ≈ 212 Ω . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms I_rms cosφ, V_s/V_p = N_s/N_p, calculation gives 212 Ω, consistent with phasor analysis

Ref: NCERT > Physics Book > Alternating Currents > AC Through Capacitor - Capacitive Reactance

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Combination of Capacitors - Series and Parallel

Two capacitors of \( 15 \, \text{pF} \) each are connected in series. What is the equivalent capacitance?

**Parallel combination** has same voltage V across each, charges Q_i = C_i V, total Q = Σ Q_i = V Σ C_i, so C_eq = Σ C_i, sum of capacitances. For five 10 μF in series, 1/C=5/10=0.5, C_eq=2 μF, much smaller than individual, while parallel would be 50 μF. (1/C) = (1/15) + (1/15) = (2/15) . C = (15/2) = 7.5 pF . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V² and common potential V = Q_total/C_total, result 7.5 pF follows, reflecting potential-capacitance relations.

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Combination of Capacitors - Series and Parallel

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Sharing of Charges, Common Potential and Applications

When two identical capacitors, one charged and one uncharged, are connected in parallel, why does the total energy decre

**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. When a charged capacitor ( C , charge Q , voltage V ) is connected in parallel with an uncharged capacitor ( C ), the total capacitance becomes 2C , and the charge redistributes to a final voltage V' = Q/(2C) = V/2 . Initial energy is U_i = (Q²/2C) , while final energy

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

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Combination of Capacitors - Series and Parallel

Four capacitors of \( 5 \, \mu\text{F} \) each are in series. What is the equivalent capacitance?

**Parallel combination** has same voltage V across each, charges Q_i = C_i V, total Q = Σ Q_i = V Σ C_i, so C_eq = Σ C_i, sum of capacitances. For five 10 μF in series, 1/C=5/10=0.5, C_eq=2 μF, much smaller than individual, while parallel would be 50 μF. (1/C) = (1/5) + (1/5) + (1/5) + (1/5) = (4/5) . C = (5/4) = 1.25 μF . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V² and common potential V = Q_total/C_total, result 1.25 µF follows, reflecting potential-capacitance relations.

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Combination of Capacitors - Series and Parallel

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Sharing of Charges, Common Potential and Applications

A \( 5 \, \mu\text{F} \) capacitor charged to \( 200 \, \text{V} \) is connected to an uncharged \( 5 \, \mu\text{F} \)

**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 charge: Q = 5 × 10⁻⁶ × 200 = 10⁻³ C . Total capacitance: 5 + 5 = 10 μF . Final voltage: V = (Q/C) = (10⁻³/10 × 10⁻⁶) = 100 V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC

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

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Capacitance of Parallel Plate and Spherical Capacitor

A parallel plate capacitor with capacitance \( 80 \, \text{pF} \) has a dielectric (\( K = 5 \), thickness \( d/5 \)) in

**Spherical conductor capacitance** C = 4π ε₀ R, R radius (m), potential V = Q/C = Q/(4π ε₀ R)=k Q/R. For R=2 cm=0.02 m, Q=2×10⁻⁸ C, V=9×10⁹×2×10⁻⁸/0.02=9000 V, showing high voltage for small sphere with modest charge. Potential difference: V = E₀ ( (4d/5) ) + (E₀/K) ( (d/5) ) = E₀ d ( (4/5) + (1/5 × 5) ) . V = E₀ d ( (4/5) + (1/25) ) = E₀ d × (21/25) . C = (Q/V) = (Q/(21/25) V₀) = (25/21) × 80 ≈ 95.24 pF . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Capacitance of Parallel Plate and Spherical Capacitor

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Combination of Capacitors - Series and Parallel

Four capacitors of \( 15 \, \mu\text{F} \) each are in series. What is the equivalent capacitance?

**Charge conservation in series** explains same Q: when battery charges first plate, it induces -Q on second plate of same capacitor, which comes from next capacitor's plate, etc., so all have same magnitude Q, potential divides as per 1/C. (1/C) = (1/15) + (1/15) + (1/15) + (1/15) = (4/15) . C = (15/4) = 3.75 μF . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V² and common potential V = Q_total/C_total, result 3.75 µF follows, reflecting potential-capacitance relations.

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Combination of Capacitors - Series and Parallel

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Combination of Capacitors - Series and Parallel

Why does the total energy stored in a system of capacitors connected in series decrease if one of the capacitors is remo

**Charge conservation in series** explains same Q: when battery charges first plate, it induces -Q on second plate of same capacitor, which comes from next capacitor's plate, etc., so all have same magnitude Q, potential divides as per 1/C. In series, the equivalent capacitance Cₑq decreases when a capacitor is removed ( (1/Cₑq) = sum (1/C_i) ). With a battery maintaining constant voltage V , the energy stored is U = (1/2) Cₑq V² . A smaller Cₑq reduces U , as energy is directly proportional to capacitance under constant V . Additionally, removing a capacitor redistributes charges, potentially

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Combination of Capacitors - Series and Parallel

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Combination of Capacitors - Series and Parallel

In a series combination of capacitors with different dielectric materials between their plates, why do capacitors with h

**Parallel combination** has same voltage V across each, charges Q_i = C_i V, total Q = Σ Q_i = V Σ C_i, so C_eq = Σ C_i, sum of capacitances. For five 10 μF in series, 1/C=5/10=0.5, C_eq=2 μF, much smaller than individual, while parallel would be 50 μF. In series, the charge Q on each capacitor is the same. Capacitance is C = (K ε₀ A/d) , where K is the dielectric constant. A higher K increases C . Since V = (Q/C) , a larger C (due to higher K ) results in a smaller V . Thus, capacitors with higher dielectric constants

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Combination of Capacitors - Series and Parallel

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