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#LCR

5 public questions tagged with this topic.

Transformer, AC Generator and LC Oscillations

A series LCR circuit has \( R = 70 \, \Omega \), \( X_L = 40 \, \Omega \), \( X_C = 20 \, \Omega \). What is the impedan

**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). Z = √(R² + (X_L - X_C)²) . Z = √(70² + (40 - 20)²) = √(4900 + 400) = √(5300) ≈ 72.8 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 72.8 Ω, consistent with

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

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Resonance in LCR Circuit and Q-Factor

A series LCR circuit with \( R = 50 \, \Omega \) is at resonance with a \( 230 \, \text{V} \) (rms) source. What is the

**Condition for resistive behavior** X_L = X_C, so ωL=1/ωC, ω²=1/LC, this is resonance, impedance purely resistive Z=R, current maximum, power maximum P= V²/R, inductive and capacitive reactances equal magnitude opposite sign cancel, circuit appears resistive. At resonance, Z = R = 50 Ω . RMS current: I = (V/R) = (230/50) = 4.6 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 4.6 A, consistent with phasor analysis and resonance condition X_L = X_C.

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

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AC Fundamentals - RMS, Average and Peak Values

A series LCR circuit with \( R = 40 \, \Omega \), \( X_L = 50 \, \Omega \), \( X_C = 30 \, \Omega \) has a \( 120 \, \te

**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. Z = √(R² + (X_L - X_C)²) = √(40² + (50 - 30)²) = √(1600 + 400) = √(2000) ≈ 44.72 Ω . RMS current: I = (V/Z) = (120/44.72) ≈ 2.68 A . Power: P = I² R = (2.68)² × 40 ≈ 287.3 W . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L -

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

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AC Fundamentals - RMS, Average and Peak Values

A series LCR circuit with \( R = 100 \, \Omega \), \( X_L = 130 \, \Omega \), \( X_C = 70 \, \Omega \) has a \( 300 \, \

**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. Z = √(R² + (X_L - X_C)²) = √(100² + (130 - 70)²) = √(10000 + 3600) = √(13600) ≈ 116.62 Ω . RMS current: I = (V/Z) = (300/116.62) ≈ 2.573 A . Power: P = I² R = (2.573)² × 100 ≈ 661.8 W . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P =

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

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