A series LCR circuit with \( R = 100 \, \Omega \), \( L = 1 \, \text{H} \), \( C = 1 \, \mu\text{F} \) is at resonance.
**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 Z = R . Given: R = 100 Ω . Impedance Z = 100 Ω . 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 100 Ω, consistent with phasor analysis and resonance condition X_L = X_C.
Ref: NCERT > Physics Book > Alternating Currents > Resonance in LCR Circuit and Q-Factor