
Alternating Current, RLC Circuits & Transformers
Detailed theory, RMS values, impedance, phase, resonance, AC power, transmission and transformer principles.
Alternating voltage and current: the underlying idea
An alternating current periodically changes magnitude and direction, unlike steady direct current flowing in one direction. A sinusoidal source is commonly described by v(t)=V₀ sin(ωt+φ), where V₀ is peak voltage, ω angular frequency and φ initial phase. One complete cycle takes T=2π/ω, and the ordinary frequency f=1/T is measured in hertz. The Indian household AC system is nominally 50 Hz, meaning a complete voltage cycle repeats fifty times each second; the voltage changes sign twice in each complete cycle.
Alternating voltage and current are not necessarily in phase in circuits containing capacitors and inductors. A source voltage may be sinusoidal while the current is delayed, advanced or distorted by nonlinear loads. For introductory examination problems, assume an ideal sinusoidal source and linear circuit elements unless specified otherwise. Alternating current can transport electrical energy despite having zero arithmetic average current over a full ideal cycle because instantaneous power may have a nonzero positive average. Confusing the average current with RMS current produces wrong heating calculations.
Peak, average and RMS values
The root-mean-square (RMS) value of a periodic quantity is the square root of the mean of its square over a full cycle. For a pure sinusoid, V_rms=V₀/√2 and I_rms=I₀/√2. RMS current is useful because it produces the same average Joule heating in a fixed resistor as a direct current of equal magnitude. The average of a pure sine current over a complete cycle is zero, while its RMS value is positive.
The average of the magnitude |sin| over a complete cycle is 2/π times the peak, a different concept from RMS and from signed cycle average. The nominal 230 V AC supply value, when used in standard simplified examples, refers to RMS voltage rather than peak; its corresponding ideal sinusoidal peak would be about 325 V. Real supply values vary and tolerances matter in engineering. Be sure to read whether a problem states peak, maximum, effective or RMS value before computing energy or power.
Pure resistor in an AC circuit
For an ideal resistor R connected to v(t)=V₀ sinωt, Ohm’s law applies instantaneously: i(t)=v(t)/R=(V₀/R)sinωt. Voltage and current rise and fall together, so their phase difference is zero. The peak current is I₀=V₀/R and RMS current is I_rms=V_rms/R. The average power dissipated is P=I_rms²R=V_rms²/R.
A resistor converts electrical energy into thermal energy regardless of which direction current flows. Its instantaneous power p=i²R is nonnegative, so positive heating is compatible with a zero signed average AC current. If a heating appliance is treated as a constant resistance, use its RMS voltage when calculating normal electrical power, not the peak voltage, because squaring the peak overestimates average heat by a factor of two. Real incandescent filaments may not be perfect fixed resistors when temperature changes.
Pure inductor and inductive reactance
An ideal inductor of inductance L develops an induced emf related to changing current by the self-induction law. In passive circuit notation, applied voltage across the inductor is v=L di/dt. For a sinusoidal driving voltage, the steady-state current lags voltage by a quarter cycle, or 90°, and the inductive reactance is X_L=ωL=2πfL. Reactance is measured in ohms, although an ideal inductor dissipates no average real power.
A pure inductor exchanges energy with the source by building and collapsing its magnetic field: U_L=½LI². The energy flow reverses during each cycle, giving zero average power in the lossless idealization. Increasing frequency increases inductive reactance, making the same RMS voltage drive less current. In practical inductors winding resistance, core loss and stray capacitance alter this simplified behaviour. Do not say the inductor ‘blocks all AC’ or that it behaves as an ordinary resistor with resistance ωL; it presents frequency-dependent opposition with phase shift.
Pure capacitor and capacitive reactance
A capacitor carries charge according to q=Cv and draws current i=C dv/dt in the ideal model. For sinusoidal applied voltage, current leads the voltage by a quarter cycle (90°). The capacitive reactance magnitude is X_C=1/(ωC)=1/(2πfC), expressed in ohms. Its opposition decreases as frequency increases, the opposite trend from an ideal inductor.
The capacitor temporarily stores energy U_C=½CV² in its electric field and returns that energy to the circuit later in the cycle. An ideal capacitor has zero average real power dissipation under a pure sinusoidal steady state, although instantaneous voltage and current are nonzero. At zero frequency under steady DC conditions, an ideal capacitor ultimately acts as an open circuit once fully charged. Real capacitors have leakage and equivalent series resistance, so their behaviour departs from the ideal especially at high frequencies.
Inductor versus capacitor phase: the memory rule
In an ideal inductor current lags the voltage; in an ideal capacitor current leads the voltage. One memory aid is ‘ELI the ICE man’: in an inductor (L) voltage E precedes current I, while in a capacitor C current I precedes voltage E. However, understanding the derivatives is more robust than relying on a mnemonic: voltage across an inductor is proportional to how rapidly current changes, whereas current through a capacitor is proportional to how rapidly its voltage changes.
Phasor diagrams represent sinusoidal quantities as rotating vectors or stationary complex amplitudes with relative phase angles. Choosing current as the reference axis, voltage across a resistor is in phase, voltage across an inductor leads by 90°, and voltage across a capacitor lags by 90°. For questions that add alternating voltages, simply adding peak magnitudes ignores these phase differences and generally gives a wrong result. Vector or phasor addition is the essential mathematical method.
Series RLC circuit and impedance
In a series circuit containing resistor R, ideal inductor L and ideal capacitor C, the same sinusoidal current passes through all three. The resistor voltage is in phase with current, the inductor voltage leads by 90°, and the capacitor voltage lags by 90°. Their reactive contributions oppose each other, so impedance magnitude is Z=√[R²+(X_L−X_C)²]. The RMS current is I_rms=V_rms/Z, and the phase angle of source voltage relative to current satisfies tanφ=(X_L−X_C)/R.
If X_L>X_C the net circuit is inductive and current lags source voltage. If X_C>X_L it is capacitive and current leads. At X_L=X_C the reactive parts cancel in the series voltage sum, but their individual voltages may still be large. Impedance has units of ohms but, unlike simple resistance, includes frequency-dependent phase behaviour. For a combination question, compute ω first, then X_L and X_C, and finally Z; adding R, X_L and X_C arithmetically is incorrect.
Resonance in a series LCR circuit
Series resonance occurs when inductive reactance equals capacitive reactance: ω₀L=1/(ω₀C). Therefore ω₀=1/√(LC), f₀=1/(2π√LC). At this frequency, the ideal series circuit’s impedance is at its minimum value R and current is maximum for fixed source RMS voltage. The source voltage and current are in phase, and power factor is unity.
Resonance does not mean the inductor and capacitor each have zero voltage or zero stored energy. Their reactive voltage phasors can be equal and opposite and may each be large. The sharpness of resonance depends on circuit resistance; in the simple series model quality factor Q=ω₀L/R. Lower resistance produces a sharper response and potentially higher internal circulating energies. Real circuits involve component losses and frequency-dependent parameters, so infinitely sharp resonance is an idealized limit. Be alert to the difference between the source voltage and the voltage across individual reactive components.
Average AC power, phase and power factor
For sinusoidal voltage and current at the same frequency in a linear steady-state circuit, average real power is P=V_rms I_rms cosφ. The factor cosφ is called power factor, the cosine of the phase difference between voltage and current. For pure resistance φ=0 and power factor is 1. In a pure lossless inductive or capacitive branch φ=±90°, so average real power is zero even though current flows and energy is exchanged with the source.
The real power in a series RLC circuit is also I_rms²R, because ideal reactances store and return energy without net loss. Low power factor means that a larger RMS current may be required for a given real power at a given supply voltage, raising conductor losses and equipment ratings. Actual electrical systems contain nonlinear loads and harmonics, making true power-factor analysis more complex than the simple displacement cosine. For UPSC numerical practice with sinusoidal elements, the standard cosφ model is appropriate if the assumptions are explicitly satisfied.
Power transmission and why AC is transformed
A power system delivering fixed real power P through a line at higher voltage typically uses lower current, which reduces resistive heating losses proportional to I²R. This is a central motivation for high-voltage electricity transmission. Alternating voltage is readily stepped up or down with suitable transformers based on electromagnetic induction, making different voltage levels practical for generation, transmission, distribution and utilization.
If current is reduced by a factor of ten through the same resistance, I²R heating falls to one-hundredth. Real systems are three-phase, with impedance, reactive power, insulation requirements, capacitance and grid control, so voltage selection is an engineering optimisation rather than simply making voltage infinite. Modern high-voltage DC transmission also exists and can be advantageous for particular routes; the explanation of transformer usefulness should not be distorted into a claim that only AC can ever transmit electrical energy efficiently.
Ideal transformer, turns ratio and current ratio
An ideal transformer has primary and secondary windings coupled by alternating magnetic flux in a core. With negligible losses and perfect coupling, V_s/V_p=N_s/N_p, where N denotes winding turns. The ideal input and output powers match, V_pI_p=V_sI_s, so I_s/I_p=N_p/N_s. A step-up transformer increases voltage and reduces current; a step-down transformer reduces voltage and increases current.
A transformer works on changing magnetic flux. Applying a constant DC voltage to the primary of a conventional power transformer does not generate a continuing secondary emf after switching transients, and may cause excessive current and heating if the primary winding resistance is low. Practical transformers incur copper, hysteresis, eddy-current and stray losses. Their efficiency is less than 100%, and winding current depends on loading as well as turns ratio. The question ‘does a transformer change frequency?’ has the answer no in the ordinary ideal transformer model.
LC oscillations and exchange of electrical energy
An ideal isolated LC circuit can oscillate because a charged capacitor transfers energy into an inductor’s magnetic field and back again. With no resistance or losses, the natural angular frequency is ω₀=1/√(LC), the same relation encountered in RLC resonance. Initially the capacitor may carry maximum charge and the current be zero; later the capacitor discharges and the inductor current reaches maximum while capacitor energy becomes minimum.
The total electrical energy remains constant in the ideal model, alternating between U_C=q²/(2C) and U_L=½LI². Real circuits contain resistance and other losses, causing the oscillation amplitude to decay unless energy is supplied. This is analogous to a damped mechanical oscillator with mass and spring stiffness. Students should distinguish free oscillation of an LC circuit from driven resonance in a series RLC circuit connected to an external AC source.
AC generators, diodes, rectification and practical waveforms
An alternating-current generator uses rotating coils or magnetic fields to create a changing magnetic flux linkage. Its induced emf is approximately sinusoidal under ideal uniform rotation and field assumptions. Frequency depends on rotational speed and pole configuration. Rectifier circuits use diodes to allow current mainly in one direction, turning an alternating input into pulsating unidirectional output; smoothing capacitors can reduce output ripple.
A half-wave rectifier uses only one half of an applied sinusoidal cycle in the elementary ideal model, whereas a full-wave bridge arrangement uses both half-cycles, reversing their polarity at the load. Rectified voltage is not automatically constant DC without filtering and regulation. Modern inverters perform the opposite functional conversion, producing controlled AC from DC power electronically. Objective questions may ask which component changes direction, frequency or waveform; these are distinct functions and should not be conflated.
How to solve AC circuit questions under exam conditions
Before calculating, write the given frequency f and convert to ω=2πf. Identify whether voltage and current values are peaks or RMS. For a pure resistor use R; for a pure coil compute X_L=ωL; for a capacitor compute X_C=1/(ωC). In a series RLC circuit calculate Z by the square-root formula, then find I_rms=V_rms/Z and real power P=I_rms²R.
Check limiting cases to catch errors: as f→0, an ideal capacitor behaves as an open circuit in steady sinusoidal response, while an ideal inductor’s reactance tends toward zero. As f becomes large the inductive reactance rises and capacitive reactance falls. At resonance X_L=X_C and Z=R. For transformers, use turn ratios together with ideal power conservation, not the same voltage/current ratio for both. Always interpret the phase sign and avoid adding reactive voltage magnitudes arithmetically.
Essential formulas and valid assumptions
All sinusoidal RMS and power-factor expressions assume the appropriate steady-state linear AC circuit model. Transformer ratios quoted below assume negligible losses and ideal coupling.
v=V₀ sinωtT=1/fVrms=V₀/√2Irms=I₀/√2XL=2πfLXC=1/(2πfC)Z=√[R²+(XL−XC)²]tanφ=(XL−XC)/RP=Vrms Irms cosφf₀=1/(2π√LC)Vs/Vp=Ns/NpVpIp=VsIsWorked numericals and practical comparisons
A 50 Hz sinusoid has T=1/50=0.02 s.
For a peak 325 V sinusoid, Vrms=325/√2≈230 V.
A 100 Ω resistor connected to 200 V RMS uses P=200²/100=400 W.
L=0.1 H at 50 Hz gives XL=2π×50×0.1≈31.4 Ω.
C=100 μF at 50 Hz gives XC=1/(2π×50×100×10⁻⁶)≈31.8 Ω.
R=6 Ω, XL=8 Ω and XC=0: Z=√(36+64)=10 Ω.
In prior RL case, cosφ=R/Z=6/10=0.6.
L=0.1 H, C=100 μF: f0=1/(2π√(0.1×0.0001))≈50.3 Hz.
Np=100, Ns=500 with Vp=24 V: ideal Vs=24×500/100=120 V.
If current reduces from 20 A to 5 A at unchanged line resistance, I²R loss becomes (5/20)²=1/16 the original.
UPSC previous-paper concept analyses
The following paraphrase indexed NDA General Ability Test Physics topics. They do not reproduce official wording or answer choices, and are not represented as confirmed EPFO/APFC previous-year questions. Confirm the original year and paper through UPSC’s archive.
Step-up transformers and grid distribution
Indexed UPSC NDA General Ability questions discuss why step-up transformers are used during electricity transmission. Raising voltage allows reduced current for a given delivered power and therefore smaller I²R line loss.
Official UPSC papersIdentifying alternating-current device principles
Past UPSC General Science questions test the distinction between generators and motors: electromagnetic induction produces generator emf, whereas a current-carrying conductor in a magnetic field experiences motor force.
Official UPSC papersElectric power expression in a resistive circuit
NDA questions ask which expressions can represent electrical power. For an Ohmic resistor P=VI=I²R=V²/R; for general AC circuits real power additionally depends on phase and power factor.
Official UPSC papers20 original MCQs, answers and explanations
- 50 s
- 0.02 s
- 2 s
- 100 s
Reveal answer with explanation
Answer B. T=1/50=0.02 s.
- V0
- V0/2
- V0/√2
- √2V0
Reveal answer with explanation
Answer C. The mean of sin² over a cycle is 1/2.
- zero
- I0
- I0/√2
- infinite
Reveal answer with explanation
Answer A. Positive and negative half cycles cancel.
- 90° out
- 180° out
- in phase
- random
Reveal answer with explanation
Answer C. Instantaneous i=v/R.
- Vrms Irms
- 2Vrms Irms always
- zero
- Irms/R
Reveal answer with explanation
Answer A. Power factor is unity for ideal resistance.
- 1/(ωL)
- ωL
- L/ω
- ω/L
Reveal answer with explanation
Answer B. XL=ωL.
- ωC
- C/ω
- 1/(ωC)
- ω/C
Reveal answer with explanation
Answer C. XC=1/(ωC).
- increase
- decrease
- zero always
- independent
Reveal answer with explanation
Answer A. XL∝f.
- increase
- decrease
- unaffected
- negative
Reveal answer with explanation
Answer B. XC∝1/f.
- leads voltage 90°
- lags voltage 90°
- in phase
- always zero
Reveal answer with explanation
Answer B. Voltage leads the current for pure L.
- leads voltage 90°
- lags voltage 90°
- always zero
- in phase
Reveal answer with explanation
Answer A. Current leads for pure C.
- R+XL+XC
- √[R²+(XL−XC)²]
- R²+XL²
- XL/XC
Reveal answer with explanation
Answer B. Reactance voltages are 180° apart as phasors.
- XL=XC
- R=XL+XC
- f=0
- L=0
Reveal answer with explanation
Answer A. Reactive contributions cancel.
- 2π√LC
- 1/(2π√LC)
- 2πLC
- √(L/C)
Reveal answer with explanation
Answer B. ω0=1/√LC.
- zero always
- R
- XL+XC
- infinite always
Reveal answer with explanation
Answer B. Net reactance is zero.
- Vrms Irms cosφ
- Vrms Irms sinφ
- V0I0 always
- zero always
Reveal answer with explanation
Answer A. Power factor is cosφ.
- increases proportionally
- decreases according to inverse turns ratio
- unchanged always
- zero
Reveal answer with explanation
Answer B. Ideal power equality makes current ratio inverse voltage ratio.
- yes
- no
- only if voltage high
- always at 50 Hz
Reveal answer with explanation
Answer B. Secondary emf requires changing coupled flux.
- half
- quarter
- double
- unchanged
Reveal answer with explanation
Answer B. Power loss is proportional to current squared.
- ½CV²
- CV
- C/V
- V²/C²
Reveal answer with explanation
Answer A. Electrical field energy is ½CV².
