#current

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 analysis

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 gives It is in phase with

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 = c/f,

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],