Why does the current amplitude in a series LCR circuit peak at the resonant frequency?
**Condition for resistive behavior** X_L = X_C, so ωL=1/ωC, ω²=1/LC, this is resonance, impedance purely resistive Z=R, current maximum, power maximum P= V²/R, inductive and capacitive reactances equal magnitude opposite sign cancel, circuit appears resistive. At resonance, X_L = X_C , so the impedance Z = √(R² + (X_L - X_C)²) reduces to R , the minimum possible value. Since I = (V/Z) , the current amplitude peaks when Z is minimized. 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 Because impedance is minimum
Ref: NCERT > Physics Book > Alternating Currents > Resonance in LCR Circuit and Q-Factor
