#current

In an ideal AC circuit with only a capacitor, what is the relationship between the rates of change of voltage and curren

**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 capacitor, I = C (dV/dt) , meaning the current is directly proportional to the rate of change of voltage. Conversely, the rate of change of current relates to the second derivative of voltage, but the primary relationship is that current depends on (dV/dt) . 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

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Power in AC Circuits - Power Factor and Wattless Current

What happens to the current in an AC circuit containing only an inductor when the source frequency approaches zero?

**Wattless current** occurs in pure inductor or capacitor, I_rms non-zero but average power zero because φ=±90°, cos φ=0, energy oscillates between source and field, no dissipation, used in choke coil to limit current without heating, unlike resistor where power dissipated. In a purely inductive circuit, X_L = ω L , and I = (V/X_L) . As frequency ( f ) approaches zero, ω = 2π f also approaches zero, making X_L very small. Thus, the current increases significantly, approaching a maximum limited only by resistance (which is zero in an ideal inductor). Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X

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

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AC Fundamentals - RMS, Average and Peak Values

What happens to the current in a purely inductive AC circuit when the frequency of the source decreases?

**Peak current** I_peak = V_peak/R for resistor, I_rms = V_rms/R, V_peak = √2 V_rms, for 200 V rms, V_peak=282.8 V, I_peak=282.8/80=3.535 A, rms I=200/80=2.5 A, average over complete cycle zero because positive and negative halves cancel. In a purely inductive circuit, X_L = ω L , and current I = (V/X_L) . Decreasing frequency reduces ω , lowering X_L , which increases the current since I is inversely proportional to X_L . 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 increases, consistent with phasor ana

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

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AC Through Resistor - Phasor and Power

In an AC circuit with only a resistor, how does the current behave relative to the applied voltage?

**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 a purely resistive AC circuit, the current and voltage oscillate in phase, meaning they reach their peak, zero, and minimum values simultaneously. This occurs because a resistor does not introduce any phase shift between voltage and current. 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 g

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

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Production of EM Waves and Hertz Experiment

A parallel plate capacitor with plate area \( A = 0.02 \, \text{m}^2 \) and separation \( d = 5 \, \text{mm} \) is conne

**Production of EM waves** requires accelerated charge, oscillating LC circuit produces changing E and B, antenna radiates when charge accelerates, frequency determined by L and C, f=1/(2π√(LC)). Hertz used spark gap with inductor and capacitor, produced ~10⁸ Hz radio waves, detected with loop, confirmed transverse nature, reflection, refraction, polarization, speed c. Displacement current i_d = ε₀ (d Φ_E/dt) . For a capacitor, i_d = i . Given i = 2 A , we have (d Φ_E/dt) = (i/ε₀) = (2/8.85 × 10⁻¹²) ≈ 2.26 × 10¹¹ Vm/s . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ =

Ref: NCERT > Physics Book > Electromagnetic Waves > Production of EM Waves and Hertz Experiment

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Potentiometer, Conductivity and Special Cases

A copper wire carries \( 2 \, \text{A} \) with a drift speed of \( 8 \times 10^{-5} \, \text{m/s} \). If \( n = 8.5 \tim

**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. Drift speed: v_d = (I/n e A) . Rearrange: A = (I/n e v_d) . Substitute: A = (2/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 8 × 10⁻⁵) . Calculate: A = (2/1.088 × 10⁵) ≈ 1.84 × 10⁻⁵ m² . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT],