#voltage division

AC Through Resistor - Phasor and Power

A \( 50 \, \Omega \) resistor and \( 20 \, \mu\text{F} \) capacitor are in series with a \( 110 \, \text{V} \), \( 50 \,

**AC through resistor** voltage and current in phase, φ=0°, I = V/R instantaneously, I(t)=I_peak sin ωt, V(t)=V_peak sin ωt, phasor diagram V and I same direction, power instantaneous P = V I = V_peak I_peak sin² ωt, average P_avg = V_rms I_rms = V_rms²/R, always positive, energy dissipated as heat. X_C = (1/ω C) = (1/314 × 20 × 10⁻⁶) ≈ 159.2 Ω . Z = √(R² + X_C²) = √(50² + 159.2²) ≈ 166.6 Ω . RMS current: I = (V/Z) = (110/166.6) ≈ 0.66 A . Voltage across resistor: V_R = I R = 0.66 × 50 ≈ 33 V . Applying X_L

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

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Sharing of Charges, Common Potential and Applications

A \( 8 \, \mu\text{F} \) capacitor charged to \( 50 \, \text{V} \) is connected to an uncharged \( 8 \, \mu\text{F} \) c

**Common potential** after connection is weighted average of initial potentials by capacitances. Energy loss ΔU = ½ C₁ C₂ (V₁-V₂)²/(C₁+C₂) dissipated as heat and spark, always positive unless V₁=V₂, explaining why energy reduces after sharing. Initial charge: Q = 8 × 10⁻⁶ × 50 = 4 × 10⁻⁴ C . Total capacitance: 8 + 8 = 16 μF . Final voltage: V = (Q/C) = (4 × 10⁻⁴/16 × 10⁻⁶) = 25 V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V² and common potential V = Q_total/C_total,

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Sharing of Charges, Common Potential and Applications

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Combination of Capacitors - Series and Parallel

In a series combination of capacitors, why does each capacitor have the same charge but different potential differences?

**Series combination** of capacitors has same charge Q on each because connected end-to-end, single path for charge flow, induced charges equal, total voltage V = Σ V_i = Q Σ 1/C_i, so 1/C_eq = Σ 1/C_i. Different potential differences V_i = Q/C_i inversely proportional to C_i, smaller C gets larger V. In a series combination, capacitors are connected end-to-end, forming a single path for charge flow. When a voltage is applied, the same charge Q accumulates on each capacitor (as charge conservation ensures the same Q passes through each during charging). However, the potential difference across

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Combination of Capacitors - Series and Parallel

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Resistance, Resistivity and Ohm's Law

In a circuit with two resistors in series, why does the voltage divide proportionally to their resistances?

**Resistivity** depends on material and temperature, not geometry. For metallic conductor, ρ ≈10⁻⁸ Ω·m for copper. Given ρ=4×10⁻⁸ Ω·m, l=2 m, A=π r², R calculation uses R=ρ l/A. Stretching wire conserves volume V = l A = l' A', so A' = A l/l', new R' = ρ l'²/V. Voltage drop across each resistor is V = I R . With the same current in series, V ∝ R , so the total voltage splits in proportion to the resistances (e.g., V₁ / V₂ = R₁ / R₂ ). Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0,

Ref: NCERT > Physics Book > Current Electricity > Resistance, Resistivity and Ohm's Law

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