Physics · Alternating Current · NEET
In a series circuit there is only one path, so the same current I flows through R, L and C at every instant. But each element reacts differently to that current. The resistor voltage (VR) stays in step with the current. The inductor voltage (VL) reaches its peak a quarter cycle BEFORE the current (it leads by 90 degrees). The capacitor voltage (VC) peaks a quarter cycle AFTER the current (it lags by 90 degrees). So the three voltages peak at different times and cannot simply be added like numbers - they must be added as phasors (arrows).
Because VL and VC are exactly opposite in phase (180 degrees apart). VL points up, VC points down on the phasor diagram. When two things point in opposite directions, you subtract to get the net effect. So the net reactance is X = XL - XC. That is why the combined opposition is Z = sqrt(R^2 + (XL - XC)^2), not R + XL + XC. If XL = XC they fully cancel and the circuit behaves like a pure resistor - this special case is called resonance.
Yes. This surprises many students. Because VL and VC point in opposite directions and cancel each other, each one alone can be much bigger than the source voltage V, yet the phasor sum V = sqrt(VR^2 + (VL - VC)^2) still equals the source. For example VL = 100 V and VC = 40 V can appear across a source of only about 80 V. Never assume VR + VL + VC equals the source voltage - it does not.
No. Impedance Z is the TOTAL opposition to AC current, and it includes both resistance R (from the resistor) and reactance X = XL - XC (from the inductor and capacitor). Resistance dissipates energy as heat; reactance only stores and returns energy. Z combines them using Pythagoras: Z = sqrt(R^2 + (XL - XC)^2), with unit ohm. Current amplitude is i0 = V0 / Z.
An inductor 20 mH, a capacitor 50 uF and a resistor 40 ohm are connected in series across a source of emf V = 10 sin 340t. The power loss in the AC circuit is:
To an AC supply of 220 V at 50 Hz, a resistor of 20 ohm, a capacitor of reactance 25 ohm and an inductor of reactance 45 ohm are connected in series. The current and the phase angle are respectively:
An AC voltage V = 220 sin(2 x 10^3 t) V is applied to a series LCR circuit with L = 10 mH, C = 25 uF, R = 100 ohm. The current amplitude in the circuit is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is a circuit where a resistor (R), an inductor (L) and a capacitor (C) are connected one after another in a single loop with an AC source. The same current flows through all three, but each element handles the voltage differently, so their voltages add up as phasors, not as plain numbers.
Z = sqrt(R^2 + (XL - XC)^2), where XL = omega*L is the inductive reactance and XC = 1/(omega*C) is the capacitive reactance. The current amplitude is i0 = V0/Z, and the phase angle satisfies tan(phi) = (XL - XC)/R.
When XL = XC the net reactance is zero, so Z = R (its minimum value) and the current is maximum. The current and voltage are in phase (phi = 0). This condition is called resonance, and it happens at the resonant frequency f0 = 1/(2*pi*sqrt(LC)).
It depends on which reactance is bigger. If XL is greater than XC the circuit is inductive and the current LAGS the voltage. If XC is greater than XL the circuit is capacitive and the current LEADS the voltage. If XL = XC (resonance) the current is exactly in phase with the voltage.
NEET regularly asks direct numericals on impedance, current amplitude, power loss and phase angle in a series LCR circuit (seen in 2016, 2023, 2025 and 2026). One formula, Z = sqrt(R^2 + (XL - XC)^2), plus i0 = V0/Z and cos(phi) = R/Z, solves most of them quickly.
