Adding the voltages in a series LCR circuit.

VR, VL and VC phasors of a series LCR circuit added head to tail into the 220 V supply phasor, beside the plain arithmetic sum that overshoots it, with the impedance triangle.

Load R = 200 Ω, 15 μF at 50 Hz and read VR + VC = 311 V but phasor sum 220 V; then set L so the current lags the supply by 45° (XL − XC = R, ±2°)

Adding VR, VL and VC as plain numbers to get the supply voltage.

• In a series LCR circuit the same current i = im sin(ωt + φ) flows through R, L and C at every instant. φ is the phase of the current relative to the source voltage.
• Draw the current phasor first. The resistor's voltage vRm = imR is along it, the inductor's vLm = imXL is π/2 ahead of it, and the capacitor's vCm = imXC is π/2 behind it.
• VL and VC point in opposite directions, so they combine into one phasor of size |vCm − vLm|. The source phasor V is the hypotenuse of a right triangle with sides VR and that difference: vm² = vRm² + (vCm − vLm)².
• So im = vm/√(R² + (XC − XL)²) = vm/Z, where Z = √(R² + (XC − XL)²) is the impedance, in ohm.
• The phase angle is given by tan φ = (XC − XL)/R. R, (XC − XL) and Z form the impedance triangle.
• If XC > XL the circuit is mainly capacitive and the current leads the voltage; if XL > XC it is mainly inductive and the current lags.
• The phasor method gives the steady-state behaviour only. Right after switching on, a transient part is also present; it dies away with time.
• Example 7.6: 200 Ω and 15.0 μF in series on 220 V, 50 Hz. XC = 212.3 Ω, Z = 291.67 Ω, I = 0.755 A, VR = 151 V and VC = 160.3 V.
• Those two add to 311.3 V, more than 220 V. There is no paradox: VR and VC are 90° apart, so they add as √(VR² + VC²) = 220 V.

Teaching this idea? Put the simulation on your own page or LMS with this code.