For a series RLC circuit driven by voltage v = v m sinωt, the current is given by i = i m sin(ωt + φ) and the power fact
Power factor cosφ=R/Z, φ phase between V and I, P=VI cosφ.
Ref: NCERT Physics Class 12 Chapter 4 Moving Charges and Magnetism
#phase angle
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A series LCR circuit has \( R = 15 \, \Omega \), \( X_L = 30 \, \Omega \), \( X_C = 45 \, \Omega \). 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) = (45 - 30/15) = 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
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
In an AC circuit with a resistor and inductor in series, what determines the magnitude of the phase difference between v
**Resistor in AC** behaves as DC, no reactance, impedance Z=R, current follows voltage exactly, average power over cycle V_rms I_rms, for 200 V rms, 80 Ω, P=500 W? Actually 200²/80=500 W, peak current √2×2.5=3.535 A, average power ½ V_peak I_peak. In an RL series circuit, the phase angle Φ = tan⁻¹ ( (X_L/R) ) . The magnitude of this angle depends on the ratio of inductive reactance ( X_L = ω L ) to resistance ( R ), as it reflects the relative contributions of inductance and resistance. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z,
Ref: NCERT > Physics Book > Alternating Currents > AC Through Resistor - Phasor and Power
In an AC circuit with a capacitor and resistor in series, how does the phase difference between voltage and current chan
**Resistor in AC** behaves as DC, no reactance, impedance Z=R, current follows voltage exactly, average power over cycle V_rms I_rms, for 200 V rms, 80 Ω, P=500 W? Actually 200²/80=500 W, peak current √2×2.5=3.535 A, average power ½ V_peak I_peak. In an RC series circuit, the phase angle Φ = tan⁻¹ ( (X_C/R) ) , where X_C = (1/ω C) . Increasing capacitance decreases X_C , reducing the phase angle, meaning the current leads the voltage by a smaller angle. 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,
Ref: NCERT > Physics Book > Alternating Currents > AC Through Resistor - Phasor and Power
A series LCR circuit has \( R = 20 \, \Omega \), \( X_L = 40 \, \Omega \), \( X_C = 60 \, \Omega \). What is the phase a
**Impedance behavior at high frequencies** X_L=ωL dominates ∝ f, X_C=1/ωC →0, so Z≈√(R²+X_L²)≈X_L large, current small, circuit inductive, φ→90°, at low frequencies X_C large, Z≈X_C, capacitive, φ→-90°, at intermediate resonance Z minimal =R. tan Φ = (X_C - X_L/R) = (60 - 40/20) = 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
A series LCR circuit has \( R = 15 \, \Omega \), \( X_L = 25 \, \Omega \), \( X_C = 40 \, \Omega \). What is the phase a
**LCR series impedance** Z = √(R² + (X_L - X_C)²), R resistance (Ω), X_L=ωL, X_C=1/ωC, phase angle φ = tan⁻¹((X_L-X_C)/R), current I_rms = V_rms/Z, voltage across R in phase with I, across L leads by 90°, across C lags by 90°, phasor diagram vector sum V = √(V_R² + (V_L - V_C)²). tan Φ = (X_C - X_L/R) = (40 - 25/15) = 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
Ref: NCERT > Physics Book > Alternating Currents > LCR Series Circuit - Impedance and Phasor Diagram
In a purely inductive AC circuit, the current lags the voltage by what phase angle?
**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. For a pure inductor, i = i_m sin (ω t - (π/2)) , while v = v_m sin ω t . Phase difference: Φ = -(π/2) , meaning current lags voltage by (π/2) or 90° . 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 =
Ref: NCERT > Physics Book > Alternating Currents > AC Fundamentals - RMS, Average and Peak Values
A series LCR circuit has \( R = 35 \, \Omega \), \( X_L = 55 \, \Omega \), \( X_C = 75 \, \Omega \). 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) = (75 - 55/35) = (20/35) ≈ 0.571 . Φ = tan⁻¹(0.571) ≈ 29.74° . 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 29.74°, consistent with phasor analysis and resonance condition X_L = X_C.
Ref: NCERT > Physics Book > Alternating Currents > LCR Series Circuit - Impedance and Phasor Diagram
A series LCR circuit has \( R = 30 \, \Omega \), \( X_L = 40 \, \Omega \), \( X_C = 60 \, \Omega \). What is the phase a
**LCR series impedance** Z = √(R² + (X_L - X_C)²), R resistance (Ω), X_L=ωL, X_C=1/ωC, phase angle φ = tan⁻¹((X_L-X_C)/R), current I_rms = V_rms/Z, voltage across R in phase with I, across L leads by 90°, across C lags by 90°, phasor diagram vector sum V = √(V_R² + (V_L - V_C)²). tan Φ = (X_C - X_L/R) = (60 - 40/30) = (20/30) = 0.667 . Φ = tan⁻¹(0.667) ≈ 33.7° . 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 =
Ref: NCERT > Physics Book > Alternating Currents > LCR Series Circuit - Impedance and Phasor Diagram
In an AC circuit with a resistor and inductor in series, what happens to the power factor if the inductance is increased
**Average power in LCR** P_avg = V_rms I_rms cos φ, cos φ = R/Z power factor, φ phase between V and I, tan φ = (X_L - X_C)/R. For R=80 Ω, X_L=100 Ω, X_C=40 Ω, X_L-X_C=60 Ω, Z=√(80²+60²)=100 Ω, cos φ=0.8, V_rms=240 V, I_rms=2.4 A, P=240×2.4×0.8=460.8 W, only R dissipates. In an RL series circuit, power factor cos Φ = (R/Z) , where Z = √(R² + X_L²) and X_L = ω L . Increasing inductance increases X_L , which increases Z , reducing cos Φ (closer to 0), as the circuit becomes more inductive. Applying X_L = ωL, X_C = 1/ωC, Z = √(R²
Ref: NCERT > Physics Book > Alternating Currents > Power in AC Circuits - Power Factor and Wattless Current
What is the effect on the power factor of an LCR series circuit when the frequency is decreased below resonance?
**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. Below resonance, X_C > X_L , making the circuit capacitive. The phase angle Φ = tan⁻¹ ( (X_C - X_L/R) ) is positive, and the power factor cos Φ decreases (less than 1) as the circuit becomes less resistive. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L -
Ref: NCERT > Physics Book > Alternating Currents > Power in AC Circuits - Power Factor and Wattless Current
