Physics · Alternating Current · NEET
In a SERIES circuit the same current flows through R, L and C at every instant. Only one quantity is common to all three components, and that is the current. So we draw the current phasor I first, along the x-axis, and measure every voltage phase relative to it. The three voltage phasors then sit at fixed angles to I: VR along I, VL 90 degrees ahead, VC 90 degrees behind. This is why NCERT starts the LCR solution by placing I on the reference axis.
Across an inductor the voltage LEADS the current by 90 degrees, so VL is drawn upward (ahead of I). Across a capacitor the voltage LAGS the current by 90 degrees, so VC is drawn downward (behind I). Up and down are exactly 180 degrees apart, so VL and VC are anti-parallel. That is why we combine them by SUBTRACTING their sizes: the net vertical phasor has magnitude |VL - VC|.
You cannot add them like ordinary numbers because they peak at different times (different phases). VR peaks with the current, VL peaks a quarter-cycle earlier, VC a quarter-cycle later. Adding phasors means adding arrows head-to-tail. VR (horizontal) and (VL - VC) (vertical) are at right angles, so their resultant is found by Pythagoras: V = sqrt(VR^2 + (VL - VC)^2). This is smaller than the plain sum VR + VL + VC, which is a very common NEET trap.
The phase angle phi is the angle between the source-voltage phasor V (the hypotenuse) and the current phasor I (the x-axis). From the right triangle, tan(phi) = (VL - VC) / VR. If VL > VC the voltage leads the current (circuit acts inductive, phi positive); if VC > VL the current leads (capacitive, phi negative). At VL = VC the vertical side vanishes and phi = 0 (resonance).
It is the net reactive voltage. It is the part of the source voltage that does NOT sit in phase with the current, so it carries no average power. Only VR (the horizontal side) is in phase with I and does real work. That is why the power factor is cos(phi) = VR / V = R / Z, the ratio of the base to the hypotenuse of the same triangle.
An inductor L, a capacitor C and a resistor R are connected in series to an ac source. The potential differences across L, C and R are 40 V, 10 V and 40 V respectively. The current amplitude is 10*sqrt(2) A. The impedance of the circuit is:
The potential differences across the resistance, capacitance and inductance are 80 V, 40 V and 100 V respectively in a series LCR circuit. The power factor of this circuit is:
To an ac supply of 220 V, a resistor of 20 ohm, a capacitor of reactance 25 ohm and an inductor of reactance 45 ohm are connected in series. The current in the circuit and the phase angle between current and voltage are respectively:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
For a SERIES LCR circuit, always draw the current I as the reference (along the x-axis), because current is common to R, L and C. For a PARALLEL circuit you would take the voltage as reference instead, because voltage is common there.
Two results at once. First, the source voltage magnitude V = sqrt(VR^2 + (VL - VC)^2), which leads to impedance Z = sqrt(R^2 + (XL - XC)^2). Second, the phase angle tan(phi) = (VL - VC)/VR = (XL - XC)/R between the source voltage and the current.
Because they are out of phase and partly cancel. VL and VC point opposite ways, so their scalar readings do not add directly. The source voltage is the phasor (vector-like) sum, which is always the hypotenuse and is smaller than the plain arithmetic sum.
At resonance, when XL = XC (or VL = VC). The vertical side becomes zero, phi = 0, the phasor V lies along I, and the circuit behaves like a pure resistor with Z = R (minimum impedance, maximum current).
NCERT gives two methods: the phasor-diagram method (geometric, faster for NEET) and the analytical method (solving the differential equation). Both give the same V = sqrt(VR^2 + (VL - VC)^2) and Z = sqrt(R^2 + (XL - XC)^2). For NEET the phasor triangle is faster and less error-prone.
