#RL circuit

AC Generator, Back EMF and Time Duration of EMF

A coil with a high self-inductance is connected to a battery. Why does the current take time to reach its maximum value?

**Circular loop deformed into straight wire** in field B=0.12 T radius 16 cm area πr²=0.0804 m² flux 0.00965 Wb drops to zero in 0.6 s e=0.0161 V, illustrating flux change due to area change induces emf, even without B change, area deformation changes Φ = B A cosθ. High self-inductance induces a back emf that opposes the increase in current, slowing the rate at which the current builds up to its steady-state value. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀ N²A/l, M = e/(dI/dt) and

Ref: NCERT > Physics Book > Electromagnetic Induction > AC Generator, Back EMF and Time Duration of EMF

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AC Through Resistor - Phasor and Power

In an AC circuit with a resistor and inductor in series, what determines the magnitude of the phase difference between v

**Resistor in AC** behaves as DC, no reactance, impedance Z=R, current follows voltage exactly, average power over cycle V_rms I_rms, for 200 V rms, 80 Ω, P=500 W? Actually 200²/80=500 W, peak current √2×2.5=3.535 A, average power ½ V_peak I_peak. In an RL series circuit, the phase angle Φ = tan⁻¹ ( (X_L/R) ) . The magnitude of this angle depends on the ratio of inductive reactance ( X_L = ω L ) to resistance ( R ), as it reflects the relative contributions of inductance and resistance. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z,

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 inductor, what happens to the impedance if the frequency decr

**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 RL series circuit, impedance Z = √(R² + X_L²) , where X_L = ω L . Decreasing frequency reduces ω , lowering X_L , which decreases the impedance since R remains constant. 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 decreases, consistent with phasor analysis and resonance cond

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

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Power in AC Circuits - Power Factor and Wattless Current

In an AC circuit with a resistor and inductor in series, what happens to the power factor if the inductance is increased

**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. In an RL series circuit, power factor cos Φ = (R/Z) , where Z = √(R² + X_L²) and X_L = ω L . Increasing inductance increases X_L , which increases Z , reducing cos Φ (closer to 0), as the circuit becomes more inductive. 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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