Alternating Current
RMS values, reactance, LCR circuits, resonance, power factor & transformer
AC Quantities and Reactance
Alternating Voltage, RMS Values and AC Through R, L, CTopic 1
An alternating current reverses direction periodically, with voltage and current varying sinusoidally: $V = V_0\sin(\omega t)$ and $I = I_0\sin(\omega t)$. The peak values $V_0$ and $I_0$ describe the maximum, but since the average of a sine over a cycle is zero, we need a more useful measure of the 'effective' value.
That measure is the root-mean-square (rms) value, the equivalent DC value that would dissipate the same heat. For a sinusoid, $V_{rms} = \dfrac{V_0}{\sqrt{2}}$ and $I_{rms} = \dfrac{I_0}{\sqrt{2}}$. The mains supply quoted as $220\ \text{V}$ is the rms value, so its peak is actually about $311\ \text{V}$ — a classic NEET fact. AC meters always read rms values.
How a component responds to AC depends on the component. Across a pure resistor, current and voltage stay in phase, and Ohm's law holds as usual with $R$. Across a pure inductor, the current lags the voltage by $90^{\circ}$, and the opposition is the inductive reactance $X_L = \omega L$, which grows with frequency — an inductor blocks high-frequency AC.
Across a pure capacitor, the current leads the voltage by $90^{\circ}$, and the opposition is the capacitive reactance $X_C = \dfrac{1}{\omega C}$, which falls with frequency — a capacitor blocks DC (infinite reactance at zero frequency) but passes high-frequency AC easily. Remembering these phase relationships and how each reactance varies with frequency is the foundation for the whole chapter.
| Element | Opposition / phase |
|---|---|
| RMS value | $V_{rms} = V_0/\sqrt{2}$ |
| Resistor | $R$; $V$ and $I$ in phase |
| Inductor | $X_L = \omega L$; $I$ lags by $90^{\circ}$ |
| Capacitor | $X_C = 1/\omega C$; $I$ leads by $90^{\circ}$ |
The peak voltage of an AC source is $311\ \text{V}$. Find its rms value.
Show solution
$V_{rms} = \dfrac{V_0}{\sqrt{2}} = \dfrac{311}{1.414} \approx 220\ \text{V}$ — the standard mains value.
Find the reactance of a $0.1\ \text{H}$ inductor at angular frequency $\omega = 100\ \text{rad/s}$.
Show solution
$X_L = \omega L = 100\times0.1 = 10\ \Omega$.
The rms value of a sinusoidal voltage of peak $V_0$ is:
In a pure resistor, the current and voltage are:
Inductive reactance $X_L$ equals:
Capacitive reactance varies with frequency as:
In a pure inductor, the current:
NEET tip: Mains $220\ \text{V}$ is rms; peak $\approx 311\ \text{V}$. Inductor: $X_L = \omega L$ (current lags); capacitor: $X_C = 1/\omega C$ (current leads). $X_L$ rises and $X_C$ falls with frequency.
Series LCR Circuit and ResonanceTopic 2
When a resistor, inductor and capacitor are connected in series across an AC source, the total opposition is the impedance $Z = \sqrt{R^{2} + (X_L - X_C)^{2}}$. Because the inductor and capacitor produce opposite phase shifts, their reactances subtract, and the current is $I_{rms} = \dfrac{V_{rms}}{Z}$. The voltage leads or lags the current by a phase angle $\phi$ given by $\tan\phi = \dfrac{X_L - X_C}{R}$.
The most important behaviour is resonance. As frequency changes, $X_L$ rises while $X_C$ falls, and at one special frequency they become equal, $X_L = X_C$. At this point the reactances cancel, the impedance drops to its minimum value $Z = R$, and the current reaches its maximum — the circuit behaves as if purely resistive.
The resonant frequency follows from $\omega L = 1/\omega C$, giving $\omega_0 = \dfrac{1}{\sqrt{LC}}$ or $f_0 = \dfrac{1}{2\pi\sqrt{LC}}$. This is the basis of tuning a radio: adjusting the capacitor changes $f_0$ so the circuit responds strongly to just one broadcast frequency. The sharpness of the resonance peak is described by the quality factor $Q$ — a high $Q$ gives a narrow, selective response.
Phasor diagrams make these relationships clear: $R$, $X_L$ and $X_C$ are drawn as vectors, with $V_R$ along the current, $V_L$ leading by $90^{\circ}$ and $V_C$ lagging by $90^{\circ}$. At resonance the $V_L$ and $V_C$ phasors are equal and opposite, leaving only $V_R$ — a picture NEET often expects you to interpret rather than calculate from scratch.
| Quantity | Expression |
|---|---|
| Impedance | $Z = \sqrt{R^{2} + (X_L - X_C)^{2}}$ |
| Phase angle | $\tan\phi = \dfrac{X_L - X_C}{R}$ |
| Resonance condition | $X_L = X_C$ |
| Resonant frequency | $\omega_0 = \dfrac{1}{\sqrt{LC}}$ |
A series LCR circuit has $R = 30\ \Omega$, $X_L = 60\ \Omega$ and $X_C = 20\ \Omega$. Find the impedance.
Show solution
$Z = \sqrt{R^{2} + (X_L - X_C)^{2}} = \sqrt{30^{2} + (60 - 20)^{2}} = \sqrt{900 + 1600} = \sqrt{2500} = 50\ \Omega$.
What is the impedance of a series LCR circuit at resonance?
Show solution
At resonance $X_L = X_C$, so $Z = \sqrt{R^{2} + 0} = R$. The impedance is minimum and equals just the resistance, giving maximum current.
The impedance of a series LCR circuit is:
At resonance, $X_L$ and $X_C$ are:
The resonant frequency of an LCR circuit is:
At resonance, the impedance of a series LCR circuit is:
Tuning a radio works by adjusting the circuit's:
NEET tip: Reactances subtract in $Z = \sqrt{R^2 + (X_L - X_C)^2}$. At resonance $X_L = X_C$, so $Z = R$ (minimum) and current is maximum; $\omega_0 = 1/\sqrt{LC}$.
Power and Transformers
Power in AC Circuits and Power FactorTopic 3
In an AC circuit, instantaneous power varies through the cycle, so we use the average power. It is $P = V_{rms}I_{rms}\cos\phi$, where $\phi$ is the phase angle between voltage and current. The factor $\cos\phi$ is called the power factor, and it determines how much of the apparent power actually does useful work.
The two extremes are revealing. In a purely resistive circuit $\phi = 0$, so $\cos\phi = 1$ and all the power is dissipated. In a purely inductive or capacitive circuit $\phi = 90^{\circ}$, so $\cos\phi = 0$ and the average power is zero — energy is stored and returned each cycle without net dissipation. The current in such a case is called a wattless current, a favourite NEET term.
This is why ideal inductors and capacitors consume no power on average, while only resistance dissipates energy. The phase angle, and hence the power factor, comes directly from the LCR analysis: $\cos\phi = \dfrac{R}{Z}$. At resonance $Z = R$, so the power factor is $1$ and the power delivered is maximum.
A low power factor is undesirable in power transmission because it means large currents flow for little useful power, increasing $I^{2}R$ losses in the lines. Industries therefore use capacitors to improve a lagging (inductive) power factor towards unity — a real-world application NEET sometimes references, tying the abstract phase angle to practical efficiency.
| Quantity | Expression |
|---|---|
| Average power | $P = V_{rms}I_{rms}\cos\phi$ |
| Power factor | $\cos\phi = R/Z$ |
| Pure L or C | $\cos\phi = 0$ (wattless) |
An AC circuit has $V_{rms} = 200\ \text{V}$, $I_{rms} = 5\ \text{A}$ and power factor $0.8$. Find the average power.
Show solution
$P = V_{rms}I_{rms}\cos\phi = 200\times5\times0.8 = 800\ \text{W}$.
Why is the average power zero in a pure inductor?
Show solution
The current lags the voltage by $90^{\circ}$, so $\cos\phi = \cos90^{\circ} = 0$. Energy is stored in the magnetic field during one quarter-cycle and returned in the next, with no net dissipation — a wattless current.
The average power in an AC circuit is:
The power factor is:
A wattless current occurs in a:
At resonance, the power factor of an LCR circuit is:
The power factor for a purely resistive circuit is:
NEET tip: Only resistance dissipates power: $P = V_{rms}I_{rms}\cos\phi$, with $\cos\phi = R/Z$. Pure L or C is wattless ($\cos\phi = 0$); at resonance the power factor is 1.
The TransformerTopic 4
The transformer is the device that makes large-scale AC distribution possible, changing an alternating voltage from one level to another using mutual induction. It has two coils — a primary and a secondary — wound on a common soft-iron core. An alternating current in the primary produces a changing flux that the core guides through the secondary, inducing an EMF there.
For an ideal transformer the voltage ratio equals the turns ratio: $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$. A step-up transformer has more turns on the secondary ($N_s > N_p$) and raises the voltage; a step-down transformer has fewer and lowers it. Crucially, a transformer works only on AC — it needs a changing flux and does nothing with steady DC.
Energy is conserved, so in an ideal transformer the power in equals the power out: $V_p I_p = V_s I_s$. This means current and voltage scale oppositely — stepping the voltage up steps the current down, and vice versa. So a step-up transformer that doubles the voltage halves the current, a relationship NEET tests with simple ratios.
This is exactly why power is transmitted at very high voltage. For a given power, a higher voltage means a smaller current, which sharply reduces the $I^{2}R$ heating losses in the transmission lines. Real transformers are highly efficient but not perfect, with small losses from winding resistance (copper loss), eddy currents and hysteresis in the core — losses minimised by laminated cores and good core materials.
| Quantity | Relation (ideal) |
|---|---|
| Voltage ratio | $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$ |
| Power conservation | $V_p I_p = V_s I_s$ |
| Step-up | $N_s > N_p$: $V$ up, $I$ down |
A transformer has $100$ primary turns and $500$ secondary turns. If the primary voltage is $220\ \text{V}$, find the secondary voltage.
Show solution
$V_s = V_p\dfrac{N_s}{N_p} = 220\times\dfrac{500}{100} = 1100\ \text{V}$ — a step-up transformer.
Why is electrical power transmitted at high voltage?
Show solution
For a fixed power, raising the voltage lowers the current ($P = VI$). Lower current means much smaller $I^{2}R$ heating losses in the long transmission lines, making distribution far more efficient.
A transformer works on the principle of:
A transformer can operate with:
In a step-up transformer:
For an ideal transformer, if voltage is stepped up, the current is:
Power is transmitted at high voltage mainly to reduce:
NEET tip: $V_s/V_p = N_s/N_p$ and $V_p I_p = V_s I_s$ — voltage up means current down. Transformers work on AC only; high-voltage transmission cuts $I^{2}R$ losses.
Ready to test yourself?
Attempt the full timed mock tests — Main & Advanced level.
Start Mock Test 1 →