Lesson 6 of 10 · 6 min
Series LCR circuit and impedance
NCERT §7.6, §7.6.1
Riya puts the 200 Ω resistor and 15.0 μF capacitor in series on the 220 V mains. Her meter reads 151 V across the resistor and 160.3 V across the capacitor. That adds to 311.3 V from a 220 V supply. Is the meter broken?
The lesson in notes
In short
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.