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#capacitance increase

2 public questions tagged with this topic.

AC Through Resistor - Phasor and Power

In an AC circuit with a capacitor and resistor in series, how does the phase difference between voltage and current chan

**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 RC series circuit, the phase angle Φ = tan⁻¹ ( (X_C/R) ) , where X_C = (1/ω C) . Increasing capacitance decreases X_C , reducing the phase angle, meaning the current leads the voltage by a smaller angle. 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 Resistor - Phasor and Power

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Capacitor with Dielectric and Effect of Inserting Slab

Why does the capacitance of a parallel plate capacitor increase when a dielectric slab is inserted between the plates wh

**Effect of dielectric** is to reduce effective field due to polarization, bound surface charges opposite to free charges, net field E = E₀ - E_p = E₀/K. Capacitance C = Q/V = K ε₀ A/d for full filling, partial filling C = ε₀ A/(d - t + t/K) for slab thickness t. When a dielectric slab (with K > 1 ) is inserted while the capacitor remains connected to a battery (constant voltage V ), the electric field decreases due to polarization ( E = E₀/K ), and the capacitance increases ( C' = K C ). This happens because the dielectric reduces the field,

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Capacitor with Dielectric and Effect of Inserting Slab

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