A brief note on Series LCR Circuits and Resonance . Read it before a test. Try each question at the end before you open its answer.
The series circuit One current: the same current flows in the resistor, the coil and the capacitor. Phases: V R V_R V R is in step with the current. V L V_L V L is 90 ∘ 90^\circ 9 0 ∘ ahead. V C V_C V C is 90 ∘ 90^\circ 9 0 ∘ behind. Impedance: Z = R 2 + ( X L − X C ) 2 Z = \sqrt{R^2 + (X_L - X_C)^2} Z = R 2 + ( X L − X C ) 2 , in Ω \Omega Ω . Then V = I Z V = IZ V = I Z for peak or for rms values. Phase angle: tan ϕ = X L − X C R \tan\phi = \dfrac{X_L - X_C}{R} tan ϕ = R X L − X C . Positive: inductive, the current lags. Negative: capacitive, the current leads. Reactances: X L = ω L X_L = \omega L X L = ω L and X C = 1 ω C X_C = \dfrac{1}{\omega C} X C = ω C 1 , with ω = 2 π f \omega = 2\pi f ω = 2 π f . Resonance Condition: X L = X C X_L = X_C X L = X C , so ω 0 = 1 L C \omega_0 = \dfrac{1}{\sqrt{LC}} ω 0 = L C 1 and f 0 = 1 2 π L C f_0 = \dfrac{1}{2\pi\sqrt{LC}} f 0 = 2 π L C 1 . At resonance: Z = R Z = R Z = R , I m a x = V R I_{max} = \dfrac{V}{R} I ma x = R V and ϕ = 0 \phi = 0 ϕ = 0 . Quality factor: Q = ω 0 L R = 1 R L C Q = \dfrac{\omega_0L}{R} = \dfrac{1}{R}\sqrt{\dfrac{L}{C}} Q = R ω 0 L = R 1 C L . At resonance V L = V C = Q V V_L = V_C = QV V L = V C = Q V . Bandwidth: Δ ω = R L \Delta\omega = \dfrac{R}{L} Δ ω = L R and Δ f = f 0 Q \Delta f = \dfrac{f_0}{Q} Δ f = Q f 0 . A small R R R gives a narrow peak. Power Average power: P = V r m s I r m s cos ϕ = I r m s 2 R P = V_{rms}I_{rms}\cos\phi = I_{rms}^2R P = V r m s I r m s cos ϕ = I r m s 2 R , in W. Power factor: cos ϕ = R Z \cos\phi = \dfrac{R}{Z} cos ϕ = Z R . It is 1 1 1 at resonance, and 0 0 0 for a pure coil or capacitor. Apparent power: V r m s I r m s V_{rms}I_{rms} V r m s I r m s , in VA. A low power factor needs a larger current for the same power, and heats the wires more. A worked example A series circuit has R = 36 Ω R = 36\ \Omega R = 36 Ω , X L = 105 Ω X_L = 105\ \Omega X L = 105 Ω and X C = 28 Ω X_C = 28\ \Omega X C = 28 Ω . It is joined to a source of 255 255 255 V (rms). Find the current, the power factor and the average power.
Impedance. Z = ( 36 Ω ) 2 + ( 77 Ω ) 2 = 85 Ω Z = \sqrt{(36\ \Omega)^2 + (77\ \Omega)^2} = 85\ \Omega Z = ( 36 Ω ) 2 + ( 77 Ω ) 2 = 85 Ω . Current. I r m s = 255 V 85 Ω = 3.0 A I_{rms} = \dfrac{255\ \text{V}}{85\ \Omega} = 3.0\ \text{A} I r m s = 85 Ω 255 V = 3.0 A . Power factor. cos ϕ = 36 Ω 85 Ω = 0.42 \cos\phi = \dfrac{36\ \Omega}{85\ \Omega} = 0.42 cos ϕ = 85 Ω 36 Ω = 0.42 . Power. P = I 2 R = ( 3.0 A ) 2 × 36 Ω = 324 P = I^2R = (3.0\ \text{A})^2 \times 36\ \Omega = 324 P = I 2 R = ( 3.0 A ) 2 × 36 Ω = 324 W. Practise with new numbers
Check yourself A series circuit has R = 9 Ω R = 9\ \Omega R = 9 Ω and a net reactance of 12 Ω 12\ \Omega 12 Ω . What is the impedance, and what is the power factor? Stuck? Show a hint Show the answerA coil of 2.0 2.0 2.0 mH and a capacitor of 0.50 μ F 0.50\ \mu\text{F} 0.50 μ F are in series. At what angular frequency is the circuit in resonance? Stuck? Show a hint Show the answerDoes a pure capacitor use power from an AC source? Give the reason. Stuck? Show a hint Show the answer