#phase difference

Power in AC Circuits - Power Factor and Wattless Current

What is the significance of the phase difference being 90° in a purely reactive AC 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. A 90° phase difference (in purely inductive or capacitive circuits) means the power factor ( cos 90° = 0 ) is zero, indicating no average power is dissipated. Energy oscillates between the source and the reactive element without being converted to heat. 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

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AC Through Resistor - Phasor and Power

In an AC circuit with a resistor and capacitor in series, why can the sum of individual voltages exceed the source volta

**Current relative to voltage** in resistor in phase, φ=0°, power factor cos φ=1, maximum power, unlike inductor/capacitor where average power zero due to 90° phase shift, explaining why resistor heats while pure L/C does not. In an RC series circuit, the voltages across the resistor and capacitor are 90° out of phase. The algebraic sum of their magnitudes can exceed the source voltage, but the source voltage is the vector sum ( V = √(V_R² + V_C²) ), accounting for the phase difference. 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,

Ref: NCERT > Physics Book > Alternating Currents > AC Through Resistor - Phasor and Power

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AC Through Resistor - Phasor and Power

In an AC circuit with a series combination of resistor and capacitor, what happens to the phase difference if the freque

**Current relative to voltage** in resistor in phase, φ=0°, power factor cos φ=1, maximum power, unlike inductor/capacitor where average power zero due to 90° phase shift, explaining why resistor heats while pure L/C does not. In an RC series circuit, Φ = tan⁻¹ ( (X_C/R) ) , where X_C = (1/ω C) . As frequency ( ω ) approaches infinity, X_C approaches zero, making Φ approach 0°, so the circuit becomes nearly resistive. 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 It approaches 0°, consistent with

Ref: NCERT > Physics Book > Alternating Currents > AC Through Resistor - Phasor and Power

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Damped Oscillations

Which feature of SHM explains why two particles with identical amplitude and frequency may not reach their extreme posit

**Effect of damping** is gradual amplitude reduction while period remains nearly constant for light damping. Mechanical energy decreases as work done against damping force, E(t) = ½ k A(t)² decaying exponentially, and motion ceases without external energy input, distinguishing from ideal undamped SHM. Different phase constants ( Φ ) shift the oscillation cycles, causing particles to reach extremes at different times despite equal amplitude and frequency. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result Difference in phase constants follows, reflecting SHM

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Displacement, Velocity and Acceleration in SHM

In SHM, why does the particle’s velocity lead its displacement by \( \pi/2 \) radians?

**Sinusoidal description** links amplitude A and angular frequency ω to instantaneous values. Given x(t) = A cos(ωt), max acceleration ω²A quantifies force requirement F_max = m ω²A, and velocity at arbitrary x is v = ±ω√(A² - x²) from energy conservation. Displacement ( x = A cos (ω t + Φ) ) and velocity ( v = -ω A sin (ω t + Φ) ) differ by π/2 radians because the cosine and sine functions are shifted by this phase, reflecting their derivative relationship. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² =

Ref: NCERT > Physics Book > Oscillations > Displacement, Velocity and Acceleration in SHM

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Superposition and Interference of Waves

Two waves \( y_1 = 4 \sin (6x - 18t) \) and \( y_2 = 4 \sin (6x - 18t + \frac{\pi}{3}) \) interfere. What is the amplitu

**Wave addition** governed by phase difference determines resultant intensity ∝ A². Phase arises from path difference Δ = (2π/λ)·Δx, and resultant formula captures interference condition quantitatively for NCERT problems. Amplitude: A = 2a cos (Φ/2) , a = 4 m , Φ = (π/3) . A = 2 × 4 cos (π/6) = 8 × (√(3)/2) ≈ 6.93 m . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 6.9 m, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Superposition and Interference of Waves

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Superposition and Interference of Waves

Two waves \( y_1 = 4 \sin (5x - 15t) \) and \( y_2 = 4 \sin (5x - 15t + \pi) \) interfere. What is the amplitude of the

**Superposition principle** states resultant displacement equals algebraic sum of individual waves, y = y₁ + y₂. For coherent waves with phase difference φ, resultant amplitude A = √(a₁² + a₂² + 2a₁a₂ cosφ), equal amplitudes give A = 2a cos(φ/2), constructive when φ = 2nπ, destructive when φ = (2n+1)π. Amplitude: A = 2a cos (Φ/2) , a = 4 m , Φ = π . A = 2 × 4 cos (π/2) = 8 × 0 = 0 m (destructive interference). Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 0 m, illustrating frequency-length-speed interdependence and quantization by boundaries

Ref: NCERT > Physics Book > Waves > Superposition and Interference of Waves

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Wave Equation and Displacement Relation

What is the phase difference between two particles separated by half a wavelength in a progressive wave?

**Displacement relation** encodes λ = 2π/k and f = ω/2π. Comparing given equation y = a sin(kx - ωt) with standard form yields k and ω, hence λ = 2π/k and v = ω/k, essential for identifying propagation characteristics and phase. For a progressive wave y = a sin(kx - ω t) , the phase is kx - ω t . If two particles are separated by λ/2 , then Δ x = λ/2 , and k = (2π/λ) , so Δ Φ = k Δ x = (2π/λ) × (λ/2) = π radians. Using v = fλ and standing-wave condition fₙ = n v/(2L)

Ref: NCERT > Physics Book > Waves > Wave Equation and Displacement Relation

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Superposition and Interference of Waves

What happens to the amplitude of two identical waves undergoing destructive interference when their phase difference is

**Superposition principle** states resultant displacement equals algebraic sum of individual waves, y = y₁ + y₂. For coherent waves with phase difference φ, resultant amplitude A = √(a₁² + a₂² + 2a₁a₂ cosφ), equal amplitudes give A = 2a cos(φ/2), constructive when φ = 2nπ, destructive when φ = (2n+1)π. For two identical waves with phase difference Φ = π , the resultant amplitude is A = 2a cos(Φ/2) = 2a cos(π/2) = 0 , leading to complete cancellation. Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields Becomes zero, illustrating frequency-length-sp

Ref: NCERT > Physics Book > Waves > Superposition and Interference of Waves

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PHYSICS

What is the phase difference corresponding to a path difference of 5lambda/4 in a double-slit experiment?

Given: What is the phase difference corresponding to a path difference of 5lambda/4 in a double-slit experiment? Formula: Phase difference phi = 2π/lambda Δ. Substitution & Calculation: For Δ = 5lambda/4, phi = 2π/lambda · 5lambda/4 = 5π/2 . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27), Chapter: Wave Optics (Latest NCERT 2026-27), Topic: Interference, phase difference φ = (2π/λ)Δ, path difference 5λ/4 and double-slit experi

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