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#electrical theory

3 public questions tagged with this topic.

Power in AC Circuits - Power Factor and Wattless Current

In an AC circuit with a series LCR combination, why does the power dissipated depend only on the resistive component?

**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. Power dissipation in an AC circuit ( P = I² R cos Φ ) occurs only through resistance, as inductors and capacitors store and release energy without converting it to heat. The reactive components affect the current and phase, but only R dissipates power. 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 Capacitor - Capacitive Reactance

In an AC circuit with only a capacitor, why does the current lead the voltage?

**Capacitor average power** zero over complete cycle because P=V I =½ V_peak I_peak sin2ωt average zero, energy stored in field, not dissipated, unlike resistor. For 15 μF, 60 Hz, X_C=176.8 Ω, V_rms=110 V, I_rms=0.622 A, illustrating lower C higher X_C. In a capacitor, current flows to charge or discharge the plates before the voltage across it changes. This causes the current to reach its maximum 90° ahead of the voltage, as the voltage builds up after the current starts flowing. 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 Capacitor - Capacitive Reactance

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Kirchhoff's Laws and Combination of Resistors

In a circuit with multiple loops, what does Kirchhoff’s junction rule imply about the currents at a node?

**Total voltage drop** across series equals source voltage because loop rule Σ V = ε, with internal resistance r included V = ε - I r. For 4 Ω,8 Ω,16 Ω parallel, 1/R_p =1/4+1/8+1/16=7/16, R_p=16/7≈2.29 Ω, then total resistance with internal 3 Ω is 5.29 Ω, current I=18/5.29≈3.4 A. Kirchhoff’s junction rule states that the sum of currents entering a node equals the sum leaving it, ensuring charge conservation, as charge cannot accumulate at a point in a steady-state circuit. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r

Ref: NCERT > Physics Book > Current Electricity > Kirchhoff's Laws and Combination of Resistors

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