Class 12 Physics · NCERT chapter 7 · 8 topics · free to read
AC Voltage & Current
Alternating current reverses direction periodically, described by sinusoidal functions. RMS values let us compare AC with DC on an equal energy footing — your 220 V mains is actually the RMS value, not the peak.
V_ rms = V_0 2
Sinusoidal AC
AC voltage and current are described by sinusoidal functions of time with amplitude, angular frequency, and phase.
An alternating voltage is produced when a coil rotates in a uniform magnetic field. The instantaneous voltage and current are sinusoidal functions of time.
v = V_0 ( t) and i = I_0 ( t + )
Here V_0 is the peak (maximum) voltage, I_0 is the peak current, = 2 f is the angular frequency, and is the phase difference between voltage and current.
V_0 = peak voltage (amplitude) — the maximum value the voltage reaches
= 2 f = 2 /T where f = frequency (Hz) and T = time period
In India: f = 50 Hz, so = 100 rad/s and T = 0.02 s
In USA: f = 60 Hz — this is why Indian appliances sometimes buzz on American outlets
One complete cycle: voltage goes from 0 → V_0 → 0 → -V_0 → 0
RMS Values
Root Mean Square (RMS) value is the steady DC value that would produce the same heating effect as the given AC. This is what voltmeters and ammeters actually read.
The RMS value of AC is defined as the square root of the mean of the square of the instantaneous values over one complete cycle.
V_ rms = V_0 2 0.707 V_0
I_ rms = I_0 2 0.707 I_0
When we say 'household supply is 220 V', that IS the RMS value — the peak is V_0 = 220 2 311 V
AC ammeter and voltmeter always read RMS values, not peak values
RMS value = that DC value which produces the same heating in the same resistance in the same time
For sinusoidal AC only: V_ rms = V_0/ 2 . For other waveforms, this formula changes
Quick recall
The peak voltage of household AC supply in India is approximately?
V_0 = V_ rms 2 = 220 1.414 311 V. The voltage actually swings between +311 V and -311 V, but we quote the RMS value of 220 V.
Mean & Average Values
The average value of AC over a full cycle is zero. The useful average is taken over a half cycle, and the ratio of RMS to average is the form factor.
Over one full cycle, the average value of ( t) is zero — the positive and negative halves exactly cancel. So we define the average over a half cycle to get a meaningful number.
V_ avg = 2V_0 0.637 V_0 (half-cycle average)
I_ avg = 2I_0 0.637 I_0
The form factor is the ratio of RMS to average value. For a sinusoidal wave, it is a fixed constant.
Full-cycle average of sinusoidal AC = 0 (positive and negative cancel)
Half-cycle average = 2V_0/ 0.637 V_0
RMS value ( 0.707 V_0 ) > Average value ( 0.637 V_0 ) > Full-cycle average (0)
Form factor for sine wave = /(2 2 ) 1.11 — always greater than 1
AC Through Pure Resistor
In a pure resistor, voltage and current are perfectly in phase — they rise and fall together. The resistor is the only component that actually dissipates power as heat.
P = V_ rms I_ rms = I_ rms ^2 R
V and I in Phase
When AC passes through a pure resistor, the current follows the voltage exactly — no lead, no lag. The phase difference is zero.
Apply v = V_0 ( t) across a pure resistor R . By Ohm's law, the instantaneous current is simply i = v/R .
v = V_0 ( t) i = V_0 R ( t) = I_0 ( t)
Both v and i have the same phase t — they reach their peaks, zeros, and troughs at exactly the same instants. The phase difference = 0 .
Phase difference between V and I: = 0 (in phase)
Peak current: I_0 = V_0/R
Ohm's law works at every instant: v(t) = i(t) R
Phasor diagram: both V and I phasors are along the same direction
Power in Pure R
A resistor converts electrical energy to heat. The average power equals the product of RMS voltage and RMS current, with power factor = 1.
Instantaneous power is p = vi = V_0 I_0 ^2( t) . Averaging over one cycle using ^2( t) = 1/2 gives the average power.
Power is always positive — energy is always absorbed (converted to heat)
Power factor = 0° = 1 — maximum power transfer
All three DC power formulas ( P = VI , P = I^2R , P = V^2/R ) work with RMS values
A 100 W bulb on 220 V AC draws I_ rms = P/V = 100/220 0.45 A
AC Through Pure Inductor
An inductor resists changes in current — it forces the current to lag behind the voltage by exactly π/2. It stores energy in its magnetic field but returns it all back, dissipating zero average power.
X_L = L
Voltage Leads Current by π/2
In a pure inductor, the voltage across it leads the current through it by 90°. The opposition to AC is called inductive reactance X_L.
Apply v = V_0 ( t) across a pure inductor L . The back-EMF of the inductor equals L di/dt , which must equal the applied voltage.
v = L di dt i = V_0 L ( t - 2 )
The current lags behind the voltage by /2 (90°). When voltage is at its peak, current is passing through zero — and vice versa.
Phase difference: = + /2 (voltage leads current by 90°)
Peak current: I_0 = V_0/( L) = V_0/X_L
Inductive reactance: X_L = L = 2 f L (unit: ohm)
X_L increases with frequency — inductor blocks high-frequency AC more
At DC ( f = 0 ): X_L = 0 — inductor is a short circuit for DC
Zero Average Power
A pure inductor absorbs energy from the source during one quarter cycle (storing it in the magnetic field) and returns it during the next quarter cycle. Net energy consumed = zero.
Instantaneous power: p = vi = V_0 I_0 ( t) ( t - /2) . Using the identity, this simplifies to p = -(V_0 I_0/2) (2 t) .
P_ avg = V_ rms I_ rms = V_ rms I_ rms 90° = 0
Power oscillates at frequency 2 — positive in one quarter, negative in the next
Positive power = energy stored in magnetic field ( U = 1 2 LI^2 )
Negative power = energy returned to the source
Net power over one full cycle = 0 — the inductor is a 'power borrower', not a consumer
Power factor: = 90° = 0
AC Through Pure Capacitor
A capacitor in AC lets current flow by charging and discharging its plates. Current leads voltage by π/2 — the exact opposite of an inductor. Like the inductor, it stores energy (in the electric field) and returns it, consuming zero net power.
X_C = 1 C
Current Leads Voltage by π/2
In a pure capacitor, current leads voltage by 90°. The opposition to AC is called capacitive reactance X_C, which decreases with frequency.
Apply v = V_0 ( t) across a pure capacitor C . The charge on the capacitor is q = Cv , and the current is i = dq/dt .
i = C dv dt = C V_0 ( t) = I_0 ( t + 2 )
The current leads the voltage by /2 (90°). When voltage is zero (capacitor uncharged), current is at its maximum (charging fastest). When voltage is at peak (fully charged), current is zero.
Phase difference: = - /2 (current leads voltage by 90°)
Peak current: I_0 = V_0 C = V_0/X_C
Capacitive reactance: X_C = 1/( C) = 1/(2 f C) (unit: ohm)
X_C decreases with frequency — capacitor passes high-frequency AC easily
At DC ( f = 0 ): X_C = — capacitor blocks DC completely
X_L = L — increases with f
X_C = 1/ C — decreases with f
V leads I by 90°
I leads V by 90°
Short circuit at DC ( X_L = 0 )
Open circuit at DC ( X_C = )
Blocks high frequency
Passes high frequency
Stores energy in B-field
Stores energy in E-field
Zero average power
Zero average power
Zero Average Power
Like the inductor, a pure capacitor is lossless. It stores energy in the electric field between its plates and returns it to the source every quarter cycle.
Instantaneous power: p = vi = (V_0 I_0 / 2) (2 t) . This oscillates symmetrically about zero — equal positive and negative halves.
P_ avg = V_ rms I_ rms 90° = 0
Energy stored in capacitor: U = 1 2 CV^2 — stored in the electric field
Energy is alternately stored and released every quarter cycle
Power factor: = (-90°) = 0
The current flowing through the capacitor (called 'wattless current') does no useful work
Quick recall
A 10 μF capacitor is connected to 220 V, 50 Hz AC. Find the capacitive reactance and RMS current.
X_C = 1/(2 50 10 10^ -6 ) = 1/( 10^ -3 ) 318 . Current I_ rms = V_ rms /X_C = 220/318 0.69 A. Average power = 0 (pure capacitor).
LCR Series Circuit
When L, C, and R are connected in series with AC, the voltages across them add as phasors (not simply). The total opposition is called impedance Z, and the phase angle φ determines whether the circuit is inductive or capacitive.
Z = R^2 + (X_L - X_C)^2
Impedance Z
Impedance combines resistance and net reactance using the Pythagorean theorem (because R is in phase with I, while net reactance is 90° out of phase).
In a series LCR circuit, the same current i = I_0 ( t) flows through all three components. But the voltages across R, L, and C have different phases relative to this current.
V_R = IR is in phase with I
V_L = IX_L leads I by 90°
V_C = IX_C lags I by 90°
V_L and V_C are exactly opposite ( 180° apart) — they partially cancel
Since V_L and V_C are anti-parallel, the net reactive voltage is |V_L - V_C| . This is perpendicular to V_R . The total voltage (source voltage) is the vector sum.
V = V_R^2 + (V_L - V_C)^2 = I R^2 + (X_L - X_C)^2
Z = V I = R^2 + (X_L - X_C)^2 (unit: ohm)
Phase Angle φ
The phase angle tells you whether the circuit behaves more like an inductor or a capacitor. It depends on which reactance is larger.
= X_L - X_C R = L - 1/( C) R
If X_L > X_C ( > 0 ): circuit is inductive — V leads I
If X_C > X_L ( < 0 ): circuit is capacitive — I leads V
If X_L = X_C ( = 0 ): circuit is at resonance — V and I are in phase
ranges from -90° to +90° depending on component values
Quick recall
A series LCR circuit has R = 40 Ω, X_L = 60 Ω, X_C = 30 Ω. Find Z and φ.
Z = 40^2 + (60-30)^2 = 1600 + 900 = 2500 = 50 . = (60-30)/40 = 3/4 , so = ^ -1 (0.75) 36.87° . Circuit is inductive (V leads I).
Resonance in AC Circuits
At resonance, the inductive and capacitive reactances exactly cancel, impedance drops to its minimum (just R), and current reaches its maximum. This is how your radio picks one station out of thousands — by tuning the circuit to resonate at that station's frequency.
_0 = 1 LC
Resonance Condition
Resonance occurs at the frequency where inductive reactance equals capacitive reactance. At this frequency, the circuit is purely resistive.
At resonance, X_L = X_C , which means L = 1/( C) . Solving for the resonant angular frequency:
_0 = 1 LC or f_0 = 1 2 LC
At resonance: Z = R (minimum impedance), so I = V/R (maximum current)
Phase angle = 0 — voltage and current are in phase (purely resistive behavior)
Resonant frequency depends ONLY on L and C — NOT on R
V_L = V_C = IX_L at resonance — these can be individually much larger than the source voltage!
Quality Factor & Sharpness
The quality factor Q measures how sharp the resonance peak is. High Q means the circuit responds strongly at resonance but weakly at other frequencies — perfect for selecting radio stations.
Q = _0 L R = 1 _0 CR = 1 R L C
The bandwidth (or 2 ) is the range of frequencies over which the current remains at or above I_ max / 2 (half-power points).
Q = _0 2 = _0 Bandwidth
High Q: sharp, narrow resonance peak — highly selective (good for radio tuning)
Low Q: broad, flat peak — responds to a wide range of frequencies
Q increases when R decreases — less resistance means sharper resonance
Q also equals V_L/V_ source at resonance — the voltage magnification factor
Bandwidth = R/L — depends on R and L but not C
Quick recall
An LCR circuit has L = 2 H, C = 32 μF, R = 10 Ω. Find resonant frequency and Q factor.
_0 = 1/ 2 32 10^ -6 = 1/ 64 10^ -6 = 1/(8 10^ -3 ) = 125 rad/s. Q = _0 L/R = 125 2/10 = 25 . This is a sharp resonance — voltage across L or C is 25 times the source voltage!
Power in AC Circuits
Only the resistive component of a circuit consumes real power. The power factor cos φ tells you what fraction of the apparent power (V × I) actually does useful work. Industries pay penalties for low power factor because it wastes grid capacity.
P = V_ rms I_ rms
Average Power
The average power in an AC circuit depends on the cosine of the phase angle between voltage and current. This cos φ factor is called the power factor.
For a general AC circuit where voltage leads current by angle , the instantaneous power p = vi oscillates. The time-averaged power over one cycle is:
P = V_ rms I_ rms
Here is the power factor. The product V_ rms I_ rms is called apparent power (unit: VA). The actual power consumed is always less than or equal to apparent power.
Pure R: = 0 , = 1 — maximum power consumed ( P = V_ rms I_ rms )
Pure L: = 90° , = 0 — zero power consumed
Pure C: = -90° , = 0 — zero power consumed
LCR at resonance: = 0 , = 1 — maximum power (same as pure R)
Power factor ranges from 0 to 1 — closer to 1 is better
= R Z = R R^2 + (X_L - X_C)^2
Wattless Current
The component of current that is 90° out of phase with voltage does zero work. It flows through the circuit, heats nothing, but occupies transmission capacity — costing utilities money.
The total current can be resolved into two components relative to the voltage: one in phase ( I ) and one perpendicular ( I ).
I_ wattful = I_ rms (does work)
I_ wattless = I_ rms (does NO work)
Wattless current flows through L and C — it shuttles energy back and forth without consuming it
In pure L or pure C: ALL the current is wattless ( = 90° , so I_ wattful = 0 )
In pure R: ALL current is wattful ( = 0° , so I_ wattless = 0 )
Industries add capacitors to their circuits to reduce and improve power factor — called power factor correction
Transformers
Transformers step voltage up or down using mutual induction between two coils on a shared iron core. They are the reason we can transmit power over hundreds of kilometers without losing most of it to heating — the single most important device in the power grid.
V_s V_p = N_s N_p = I_p I_s
Working Principle
A transformer works on Faraday's law: changing current in the primary coil creates a changing magnetic flux that induces an EMF in the secondary coil. The voltage ratio equals the turns ratio.
Two coils (primary with N_p turns, secondary with N_s turns) are wound on the same laminated soft iron core. AC in the primary creates a changing magnetic flux in the core, which links both coils.
V_s V_p = N_s N_p = k (turns ratio)
For an ideal transformer (no energy losses), input power = output power: V_p I_p = V_s I_s .
V_s V_p = N_s N_p = I_p I_s
Step-up ( N_s > N_p , k > 1 ): voltage increases, current decreases
Step-down ( N_s < N_p , k < 1 ): voltage decreases, current increases
Transformers work ONLY with AC — DC produces constant flux, so no induced EMF in secondary
The iron core channels the magnetic flux so nearly all of it links both coils
Power is conserved: what you gain in voltage, you lose in current (and vice versa)
Energy Losses
Real transformers lose energy through copper losses, eddy currents, hysteresis, and flux leakage. Each loss has a specific engineering solution.
No real transformer is 100% efficient. Energy losses reduce the output power below the input power. Efficiency is = P_ out /P_ in 100 % , typically 90–99% for power transformers.
Copper loss ( I^2R heating in coils) — minimized using thick copper wires with low resistance
Eddy current loss (currents induced in the iron core) — minimized by using laminated core (thin insulated sheets)
Hysteresis loss (energy lost in magnetizing and demagnetizing core each cycle) — minimized using soft iron core (low hysteresis)
Flux leakage (not all flux links both coils) — minimized by winding coils tightly over each other on a closed core
Long-Distance Power Transmission
Transmitting power at high voltage (and low current) dramatically reduces I²R losses in transmission lines. Transformers make this possible by stepping voltage up at the power station and down at your home.
Power transmitted: P = VI . For the same power P , if you increase voltage by factor n , current decreases by factor n . Power loss in transmission wires: P_ loss = I^2 R .
P_ loss = I^2 R = P^2 R V^2
Doubling the transmission voltage reduces the power loss by a factor of FOUR. This is why power stations step up to 11 kV → 132 kV → 400 kV for long-distance lines.
Power loss 1/V^2 — transmitting at 10× voltage reduces loss by 100×
High voltage is dangerous — that's why step-down transformers bring it to 220 V before entering your home
Power loss depends on line resistance R — shorter, thicker wires also help
Quick recall
A power station generates 2 MW at 10 kV. If transmitted at 10 kV, line loss is 20%. What is the loss if stepped up to 100 kV?
Voltage increased by factor 10 → loss decreases by factor 10^2 = 100 . New loss = 20 %/100 = 0.2 % . Stepping up from 10 kV to 100 kV reduces the transmission loss from 20% to just 0.2% — transformers save 99% of the wasted energy!