The Series LCR Circuit

All Three in Series: The Setup

Connect R, L and C in series to $v = v_m\sin\omega t$ (NCERT Fig. 7.10). Kirchhoff's loop rule gives

$$L\frac{di}{dt} + iR + \frac{q}{C} = v$$

We solve it with phasors. The crucial series-circuit fact: the same current flows through every element at every instant — same amplitude, same phase. Write it as

$$i = i_m\sin(\omega t + \phi)$$

where $\phi$ is the phase of the current relative to the source voltage — the unknown we want, along with $i_m$.

Each element's voltage phasor obeys its own rule from Sections 1-3:

• $\vec{V}_R$ (length $i_mR$): parallel to $\vec{I}$ (in phase),
• $\vec{V}_L$ (length $i_mX_L$): 90 degrees ahead of $\vec{I}$,
• $\vec{V}_C$ (length $i_mX_C$): 90 degrees behind $\vec{I}$.

Since $\vec{V}_L$ and $\vec{V}_C$ lie along the same line in opposite directions, they combine into a single phasor of magnitude $|v_{Cm} - v_{Lm}|$.

Impedance and Phase Angle

The source phasor is the (vector) sum $\vec{V} = \vec{V}_L + \vec{V}_R + \vec{V}_C$. With $\vec{V}_R$ and the combined $(\vec{V}_C + \vec{V}_L)$ as the two perpendicular sides, Pythagoras gives

$$v_m^2 = (i_mR)^2 + (i_mX_C - i_mX_L)^2$$

$$\boxed{i_m = \frac{v_m}{Z}, \qquad Z = \sqrt{R^2 + (X_C - X_L)^2}}$$

Z is the impedance (ohm) — the AC generalisation of resistance. The phase angle comes from the impedance triangle (sides R and $X_C - X_L$, hypotenuse Z):

$$\tan\phi = \frac{X_C - X_L}{R}$$

The three regimes:

Condition	$\phi$	Circuit behaves	Current
$X_C > X_L$	positive	capacitively	leads the voltage
$X_L > X_C$	negative	inductively	lags the voltage
$X_C = X_L$	zero	purely resistively	in phase (resonance!)

The Voltage 'Paradox' and How to Read LCR Problems

A strange feature of series AC circuits: the rms voltages across the elements can algebraically add to MORE than the source voltage. Nothing is wrong — rms voltages across R, L and C are not in phase with one another, so they add as phasors, not as numbers:

$$V = \sqrt{V_R^2 + (V_C - V_L)^2}$$

(NCERT Example 7.6 below makes this concrete: 151 V + 160 V across R and C, from a 220 V source!)

Problem-solving drill (works every time):

1. Compute $X_L = \omega L$ and $X_C = 1/\omega C$ at the source frequency.
2. $Z = \sqrt{R^2 + (X_C - X_L)^2}$, then $I = V/Z$ (rms throughout, or peaks throughout — never mix!).
3. Element voltages: $V_R = IR$, $V_L = IX_L$, $V_C = IX_C$.
4. Phase: $\tan\phi = (X_C - X_L)/R$; sign tells lead/lag.

[JEE Tip] An RL or RC circuit is just LCR with one reactance set to zero — same machinery: $Z_{RL} = \sqrt{R^2 + X_L^2}$, $Z_{RC} = \sqrt{R^2 + X_C^2}$. One formula family, every series problem.

Solved Examples

Example 1: NCERT's RC circuit and the paradox (NCERT Example 7.6)

A 200 $\Omega$ resistor and a 15.0 $\mu$F capacitor are in series across a 220 V, 50 Hz source. (a) Find the current. (b) Find the rms voltages across R and C. Is their algebraic sum more than the source voltage? Resolve the paradox.

Solution:

1. $X_C = 212\ \Omega$ (Section 3), so $Z = \sqrt{200^2 + 212^2} = 291.5\ \Omega$.
2. (a) $I = V/Z = 220/291.5 = 0.755$ A.
3. (b) $V_R = IR = 0.755 \times 200 = 151$ V; $V_C = IX_C = 0.755 \times 212 = 160.3$ V.
4. Algebraic sum $= 311.3$ V $> 220$ V! Resolution: $V_R$ and $V_C$ are 90 degrees out of phase, so they add as phasors: $\sqrt{151^2 + 160.3^2} = 220$ V. Books balanced.

Example 2: A full LCR workout [JEE Numerical]

A series circuit has R = 40 $\Omega$, $X_L = 70\ \Omega$, $X_C = 40\ \Omega$ on a 200 V (rms) source. Find Z, the current, and the phase relationship.

Solution:

1. $Z = \sqrt{40^2 + (40-70)^2} = \sqrt{1600+900} = 50\ \Omega$.
2. $I = 200/50 = 4$ A.
3. $\tan\phi = (X_C - X_L)/R = -30/40 = -0.75$, so $\phi \approx -37^{\circ}$: $X_L$ dominates — the current lags the voltage by about 37 degrees.

Example 3: Element voltages and the phasor check [JEE Numerical]

A series LCR circuit carries I = 2 A with R = 30 $\Omega$, $X_L = 40\ \Omega$, $X_C = 80\ \Omega$. Find $V_R$, $V_L$, $V_C$ and the source voltage.

Solution:

1. $V_R = 60$ V, $V_L = 80$ V, $V_C = 160$ V.
2. Source: $V = \sqrt{V_R^2 + (V_C - V_L)^2} = \sqrt{60^2 + 80^2} = 100$ V.
3. Note: $V_C = 160$ V alone exceeds the 100 V source — routine in AC, impossible in DC. (Current leads here, since $X_C > X_L$.)

Example 4: Impedance triangle in reverse [JEE Numerical]

A series circuit has Z = 50 $\Omega$ and R = 30 $\Omega$. Find the net reactance and the power factor $\cos\phi$.

Solution:

1. $|X_C - X_L| = \sqrt{Z^2 - R^2} = \sqrt{2500-900} = 40\ \Omega$.
2. $\cos\phi = R/Z = 30/50 = 0.6$.
3. The impedance triangle works both ways — given any two sides, the third and the angle follow.

Example 5: An RL circuit by the same machinery [NEET Numerical]

A coil of R = 30 $\Omega$ and $X_L = 40\ \Omega$ is connected to a 220 V AC source. Find Z and the current.

Solution:

1. RL is LCR with $X_C = 0$: $Z = \sqrt{30^2 + 40^2} = 50\ \Omega$.
2. $I = 220/50 = 4.4$ A, lagging the voltage ($\tan\phi = -X_L/R = -4/3$).
3. The 3-4-5 triangle — examiners' favourite numbers for a reason.

Example 6: Finding the reactances at 50 Hz [NEET Numerical]

In a series LCR circuit, L = 0.2 H, C = 40 $\mu$F, R = 50 $\Omega$, on 50 Hz mains. Compute $X_L$, $X_C$ and Z.

Solution:

1. $X_L = 2\pi \times 50 \times 0.2 = 62.8\ \Omega$.
2. $X_C = \frac{1}{2\pi \times 50 \times 40 \times 10^{-6}} = 79.6\ \Omega$.
3. $Z = \sqrt{50^2 + (79.6-62.8)^2} = \sqrt{2500 + 282} \approx 52.7\ \Omega$ — slightly capacitive (current leads).

Example 7: Why the same current everywhere?

Justify the starting assumption of the phasor solution: that all three series elements carry the same current at every instant.

Solution:

1. In a series loop there is only one path; charge cannot pile up at the junctions between ideal elements.
2. Conservation of charge then forces the instantaneous current to be identical through R, L and C.
3. That common $\vec{I}$ becomes the reference phasor against which each voltage phasor is drawn — the whole construction rests on this.

Example 8: Why $V_L$ and $V_C$ fight

Why do the inductor and capacitor voltage phasors point in opposite directions?

Solution:

1. Relative to the common current, $V_L$ runs 90 degrees ahead and $V_C$ runs 90 degrees behind.
2. Ahead-by-90 and behind-by-90 differ by 180 degrees — they lie along one line, opposed.
3. Hence they partially cancel into $|V_C - V_L|$, and at resonance cancel exactly — the seed of Section 5.

Example 9: Lead or lag at a glance

Without computing anything, state whether the current leads or lags in: (a) $X_C = 100\ \Omega, X_L = 30\ \Omega$; (b) $X_L = 90\ \Omega, X_C = 90\ \Omega$; (c) $X_L = 200\ \Omega, X_C = 50\ \Omega$.

Solution:

1. (a) $X_C > X_L$: net capacitive — current leads.
2. (b) Equal reactances: $\phi = 0$ — in phase (the circuit is at resonance).
3. (c) $X_L > X_C$: net inductive — current lags. The sign of $(X_C - X_L)$ decides; R only sets how much.

Example 10: The DC test on an LCR circuit

The AC source of a series LCR circuit is replaced by a DC battery (long after transients die). What is the steady current?

Solution:

1. For DC, $\omega = 0$: $X_L = 0$ but $X_C = 1/\omega C \to \infty$.
2. The fully charged capacitor blocks the loop completely.
3. Steady current = zero. Any series circuit containing a capacitor passes no steady DC — whatever R and L are.