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#electrical circuits

18 public questions tagged with this topic.

AC Generator, Back EMF and Time Duration of EMF

A coil with high self-inductance is suddenly disconnected from a battery. The resulting high voltage spike is due to wha

**Circular loop deformed into straight wire** in field B=0.12 T radius 16 cm area πr²=0.0804 m² flux 0.00965 Wb drops to zero in 0.6 s e=0.0161 V, illustrating flux change due to area change induces emf, even without B change, area deformation changes Φ = B A cosθ. The rapid current drop induces a large back emf ( ε = -L (dI/dt) ) due to self-inductance, causing a voltage spike. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀ N²A/l, M = e/(dI/dt) and U = ½

Ref: NCERT > Physics Book > Electromagnetic Induction > AC Generator, Back EMF and Time Duration of EMF

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Transformer, AC Generator and LC Oscillations

What is the effect on the impedance of a series LCR circuit when the frequency is increased beyond the resonant frequenc

**Transformer principle** alternating current in primary creates changing flux linking secondary, emf induced e = -N dΦ/dt, flux same through both, so V∝N. For N_p=550, N_s=1100, V_p=110 V rms, V_s= V_p×N_s/N_p=220 V, step-up factor 2, efficiency

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

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Resonance in LCR Circuit and Q-Factor

A \( 60 \, \text{mH} \) inductor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) source. What is the peak

**At resonance** V_L = I X_L = I X_C = V_C, may be larger than source voltage Q times, Q-factor = ω₀ L/R =1/(ω₀ C R)= V_L/V = V_C/V, measures sharpness, higher Q sharper resonance, bandwidth Δω = R/L = ω₀/Q, resonant frequency independent of R. X_L = ω L , ω = 2π × 60 = 376.8 rad/s . L = 60 × 10⁻³ H . X_L = 376.8 × 0.06 = 22.61 Ω . RMS current: I = (V/X_L) = (110/22.61) ≈ 4.87 A . Peak current: i_m = √(2) I = 1.414 × 4.87 ≈ 6.88 A . Applying X_L = ωL,

Ref: NCERT > Physics Book > Alternating Currents > Resonance in LCR Circuit and Q-Factor

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LCR Series Circuit - Impedance and Phasor Diagram

A series LCR circuit has \( R = 10 \, \Omega \), \( X_L = 20 \, \Omega \), \( X_C = 30 \, \Omega \). What is the phase a

**LCR example** R=100 Ω, X_L=130 Ω, X_C=70 Ω, X_L-X_C=60 Ω, Z=√(100²+60²)=116.6 Ω, V_rms=300 V, I_rms=2.573 A, power P= I_rms² R =662 W? Actually P= V_rms I_rms cos φ, cos φ=R/Z=0.857, P=300×2.573×0.857=661.6 W, illustrating power factor. tan Φ = (X_C - X_L/R) = (30 - 20/10) = 1 . Φ = tan⁻¹(1) = 45° . Since X_C > X_L , current leads voltage. 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 45°, consistent with phasor analysis and resonance condition X_L = X_C.

Ref: NCERT > Physics Book > Alternating Currents > LCR Series Circuit - Impedance and Phasor Diagram

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Resonance in LCR Circuit and Q-Factor

In a series LCR circuit at resonance, the power dissipated is \( 4800 \, \text{W} \) with \( R = 3 \, \Omega \). What is

**At resonance** V_L = I X_L = I X_C = V_C, may be larger than source voltage Q times, Q-factor = ω₀ L/R =1/(ω₀ C R)= V_L/V = V_C/V, measures sharpness, higher Q sharper resonance, bandwidth Δω = R/L = ω₀/Q, resonant frequency independent of R. At resonance, Z = R , and power P = I² R . 4800 = I² × 3 . I² = (4800/3) = 1600 . I = √(1600) = 40 A . 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

Ref: NCERT > Physics Book > Alternating Currents > Resonance in LCR Circuit and Q-Factor

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AC Through Inductor - Inductive Reactance

A \( 21 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) AC source. What is th

**Inductive reactance calculation** for 60 mH, 50 Hz, X_L=2π×50×0.06=18.85 Ω, V_rms=220 V, I_rms=220/18.85=11.67 A, for 70 mH, 50 Hz, X_L=21.99 Ω, I=110/21.99=5 A, showing X_L ∝ f L, higher f or L increases opposition. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 21 × 10⁻⁶ F . X_C = (1/314 × 21 × 10⁻⁶) ≈ 151.6 Ω . RMS current: I = (V/X_C) = (230/151.6) ≈ 1.517 A . 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 1.517

Ref: NCERT > Physics Book > Alternating Currents > AC Through Inductor - Inductive Reactance

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AC Through Inductor - Inductive Reactance

A \( 150 \, \Omega \) resistor is connected to a \( 300 \, \text{V} \) (peak) AC supply. What is the rms current in the

**Inductive reactance calculation** for 60 mH, 50 Hz, X_L=2π×50×0.06=18.85 Ω, V_rms=220 V, I_rms=220/18.85=11.67 A, for 70 mH, 50 Hz, X_L=21.99 Ω, I=110/21.99=5 A, showing X_L ∝ f L, higher f or L increases opposition. Peak voltage: v_m = 300 V . RMS voltage: V = (v_m/√(2)) = (300/1.414) ≈ 212.1 V . RMS current: I = (V/R) = (212.1/150) ≈ 1.414 A . 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 1.414 A, consistent with phasor analysis and resonance condition X_L = X_C.

Ref: NCERT > Physics Book > Alternating Currents > AC Through Inductor - Inductive Reactance

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AC Fundamentals - RMS, Average and Peak Values

A \( 226.3 \, \text{V} \) (peak) AC source is connected to a \( 80 \, \Omega \) resistor. What is the average power cons

**Average power** in resistor P_avg = V_rms I_rms = V_rms²/R = I_rms² R, for 254.6 V peak, V_rms =180 V, R=90 Ω, P=180²/90=360 W, for 226.3 V peak, V_rms=160 V, R=80 Ω, P=320 W, illustrating rms use for power. RMS voltage: V = (v_m/√(2)) = (226.3/1.414) ≈ 160 V . RMS current: I = (V/R) = (160/80) = 2 A . Average power: P = I² R = 2² × 80 = 320 W . 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 320 W, consistent

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

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Combination of Capacitors - Series and Parallel

Five capacitors of \( 30 \, \mu\text{F} \) 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) = (1/30) + (1/30) + (1/30) + (1/30) + (1/30) = (5/30) . C = (30/5) = 6 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 6 µF follows, reflecting potential-capacitance relations.

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

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Capacitance of Parallel Plate and Spherical Capacitor

Two capacitors \( 10 \, \mu\text{F} \) and \( 20 \, \mu\text{F} \) are in parallel. What is the total capacitance?

**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. C = C₁ + C₂ = 10 + 20 = 30 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 30 µF follows, reflecting potential-capacitance relations.

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

Five capacitors of \( 10 \, \mu\text{F} \) each are in series. What is the equivalent capacitance?

**Series combination** of capacitors has same charge Q on each because connected end-to-end, single path for charge flow, induced charges equal, total voltage V = Σ V_i = Q Σ 1/C_i, so 1/C_eq = Σ 1/C_i. Different potential differences V_i = Q/C_i inversely proportional to C_i, smaller C gets larger V. (1/C) = (1/10) + (1/10) + (1/10) + (1/10) + (1/10) = (5/10) . C = (10/5) = 2 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 2 µF follows, reflecting potential-capacitance relations.

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

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Potentiometer, Conductivity and Special Cases

A Wheatstone bridge has \( R_1 = 25 \, \Omega \), \( R_2 = 50 \, \Omega \), \( R_3 = 20 \, \Omega \). What is \( R_4 \)

**Potentiometer** measures potential difference without drawing current, using null deflection, principle V ∝ l, l balance length, accurate because no I r drop. Potential drop across resistor V = I R arises because electric field does work on charges, energy converted to heat, maintaining E = -dV/dx along wire. Balance condition: (R₁/R₂) = (R₃/R₄) . Substitute: (25/50) = (20/R₄) . Solve: 0.5 = (20/R₄) ⇒ R₄ = (20/0.5) = 40 Ω . 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 and P = I²R, evaluation yields 40 Ω,

Ref: NCERT > Physics Book > Current Electricity > Potentiometer, Conductivity and Special Cases

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