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#resonance

47 public questions tagged with this topic.

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

A \( 200 \, \text{V} \) (rms) AC source is connected to a series LCR circuit with \( R = 20 \, \Omega \) at resonance. W

**Resonance in LCR** occurs when X_L = X_C, ω₀ =1/√(LC), f₀=1/(2π√(LC)), impedance Z=R minimum, current maximum I₀=V/R, circuit behaves as if only resistance because inductive and capacitive voltages equal opposite cancel, net reactance zero, phase φ=0°, power factor 1, current in phase with voltage. At resonance, Z = R = 20 Ω . RMS current: I = (V/R) = (200/20) = 10 A . Power: P = I² R = 10² × 20 = 2000 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 2000

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

A series LCR circuit with \( R = 100 \, \Omega \), \( L = 1 \, \text{H} \), \( C = 1 \, \mu\text{F} \) is at resonance.

**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, X_L = X_C , so Z = R . Given: R = 100 Ω . Impedance Z = 100 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 100 Ω, consistent with phasor analysis and resonance condition X_L = X_C.

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

A series LCR circuit with \( R = 50 \, \Omega \) is at resonance with a \( 250 \, \text{V} \) (rms) source. What is the

**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 = 50 Ω . RMS current: I = (V/R) = (250/50) = 5 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 5 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

A series LCR circuit with \( R = 40 \, \Omega \) is at resonance with a \( 200 \, \text{V} \) (rms) source. What is the

**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 = 40 Ω . RMS current: I = (V/R) = (200/40) = 5 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 5 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

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

What is the role of resonance in the operation of a radio receiver?

**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. In a radio receiver, resonance in an LCR circuit is used to tune to a specific frequency. By adjusting the capacitance or inductance, the circuit’s resonant frequency matches the desired station’s signal, amplifying it while filtering out others. 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

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

Why does the voltage across an inductor in a series LCR circuit potentially exceed the source voltage at resonance?

**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, current is maximized ( I = (V/R) ), and the voltage across the inductor ( V_L = I X_L ) can exceed the source voltage if X_L > R . This is possible because V_L and V_C are out of phase with the source voltage, and their vector sum with V_R equals the source voltage. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L -

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

Why does the current amplitude in a series LCR circuit peak at the resonant frequency?

**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, X_L = X_C , so the impedance Z = √(R² + (X_L - X_C)²) reduces to R , the minimum possible value. Since I = (V/Z) , the current amplitude peaks when Z is minimized. 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 Because impedance is minimum

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

What is the condition for maximum power dissipation in a series LCR circuit?

**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. Maximum power dissipation in a series LCR circuit occurs at resonance, where X_L = X_C , making the impedance equal to the resistance ( Z = R ). At this point, the power factor ( cos Φ ) is 1, and power ( P = I² R ) is maximized. Applying X_L

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

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

A series LCR circuit with \( R = 60 \, \Omega \) is at resonance with a \( 240 \, \text{V} \) (rms) source. What is the

**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 = 60 Ω . RMS current: I = (V/R) = (240/60) = 4 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 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