#circuit analysis

Doping, Charge Carriers and Conductivity

The direction of conventional current in a forward-biased diode is:

**Charge carriers** in n-type majority electrons, minority holes, in p-type majority holes, minority electrons, minority concentration reduced due to recombination n_e n_h = n_i², doping increases majority, conductivity σ = e(n_e μ_e + n_h μ_h), increases with doping. Donor ionization energy small ~0.01 eV, electrons easily promoted to conduction band at room temperature. In forward bias, current flows from p-side (positive) to n-side (negative) conventionally, as indicated by the diode symbol’s arrow, due to the flow of majority carriers. Substituting values gives P-side to n-side, which matc

Ref: NCERT > Physics Book > Electronic Devices > Doping, Charge Carriers and Conductivity

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LCR Series Circuit - Impedance and Phasor Diagram

What is the primary factor determining the resonant frequency in a series LCR circuit?

**LCR series impedance** Z = √(R² + (X_L - X_C)²), R resistance (Ω), X_L=ωL, X_C=1/ωC, phase angle φ = tan⁻¹((X_L-X_C)/R), current I_rms = V_rms/Z, voltage across R in phase with I, across L leads by 90°, across C lags by 90°, phasor diagram vector sum V = √(V_R² + (V_L - V_C)²). The resonant frequency in a series LCR circuit is determined by the inductance ( L ) and capacitance ( C ), given by f₀ = (1/2π √(L C)) . Resistance affects damping but not the resonant frequency itself. Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms

Ref: NCERT > Physics Book > Alternating Currents > LCR Series Circuit - Impedance and Phasor Diagram

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AC Through Inductor - Inductive Reactance

A \( 70 \, \text{mH} \) inductor is connected to a \( 230 \, \text{V} \), \( 60 \, \text{Hz} \) source. What is the peak

**AC through inductor** voltage leads current by 90°, V = L dI/dt, V(t)=V_peak sin(ωt+90°), I(t)=I_peak sin ωt, instantaneous power P= V I =½ V_peak I_peak sin2ωt, average zero over cycle because energy stored in magnetic field ½ L I² returned to source each quarter cycle, no net dissipation. X_L = ω L , ω = 2π × 60 = 376.8 rad/s . L = 70 × 10⁻³ H . X_L = 376.8 × 0.07 = 26.38 Ω . RMS current: I = (V/X_L) = (230/26.38) ≈ 8.72 A . Peak current: i_m = √(2) I = 1.414 × 8.72 ≈ 12.33 A . Applying X_L

Ref: NCERT > Physics Book > Alternating Currents > AC Through Inductor - Inductive Reactance

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Displacement Current and Ampere-Maxwell Law

A capacitor in a circuit has a conduction current of \( 2.5 \, \text{A} \) in the wires. What is the displacement curren

**Displacement current** I_d = ε₀ dΦ_E/dt, Φ_E = ∫ E·dA electric flux (V·m), ε₀=8.85×10⁻¹² F/m, ensures continuity of current in charging capacitor where conduction current stops between plates, I_d equals conduction current in wires, 3 A conduction ⇒ 3 A displacement, maintaining Ampere's law ∮ B·dl = μ₀(I_c+I_d). The document states that in a charging capacitor, the displacement current between the plates equals the conduction current in the wires, so i_d = 2.5 A . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 2.5 A, illustrating EM wave tr

Ref: NCERT > Physics Book > Electromagnetic Waves > Displacement Current and Ampere-Maxwell Law

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Sharing of Charges, Common Potential and Applications

Three capacitors \( 5 \, \text{pF} \), \( 10 \, \text{pF} \), and \( 20 \, \text{pF} \) are in parallel. What is the tot

**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. C = 5 + 10 + 20 = 35 pF . 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, result 35 pF follows, reflecting potential-capacitance relations.

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

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Potentiometer, Conductivity and Special Cases

A Wheatstone bridge has \( R_1 = 18 \, \Omega \), \( R_2 = 36 \, \Omega \), \( R_3 = 12 \, \Omega \). What is \( R_4 \)

**Conductivity** σ = 1/ρ = n e² τ/m (S/m), τ relaxation time, m electron mass. Resistivity deviation at high fields occurs when τ depends on E or n changes due to impact ionization, breaking Ohm's law, seen in varistors, gas discharge. Balance condition: (R₁/R₂) = (R₃/R₄) . Substitute: (18/36) = (12/R₄) . Solve: 0.5 = (12/R₄) ⇒ R₄ = (12/0.5) = 24 Ω . 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 and P = I²R, evaluation yields 24 Ω,