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Alternating Current: LCR, Resonance & Transformers

Apply concepts from Alternating Current: LCR, Resonance & Transformers to problem-solving. Focus on numerical practice, shortcuts, and real-world applications.

3.7%50 minPhase 2 · APPLICATIONMCQ + Numerical

Concept Core

AC Voltage and Current

An alternating voltage is described by V(t) = V0V_{0} sin(ω\omegat), where V0V_{0} is the peak (amplitude) voltage and ω\omega = 2π\pif is the angular frequency. The corresponding current depends on the circuit element. Peak value V0V_{0} is the maximum instantaneous voltage.
The root-mean-square (RMS) value VrmsV_{rms} = V02\frac{V_{0}}{\sqrt{2}} is the equivalent DC voltage that dissipates the same power in a resistor.
Similarly, IrmsI_{rms} = I02\frac{I_{0}}{\sqrt{2}}.
For household AC (India): VrmsV_{rms} = 220 V, so V0V_{0} = 220
2\sqrt{2} = 311 V, f = 50 Hz, ω\omega = 100*π\pi rad/s.

Phase Relations and Phasors

In a pure resistor, voltage and current are in phase. In a pure inductor, voltage leads current by π2\frac{\pi}{2} (remember: ELI — voltage E leads current I in inductor L). In a pure capacitor, current leads voltage by π2\frac{\pi}{2} (remember: ICE — current I leads voltage E in capacitor C). Phasor diagrams represent sinusoidal quantities as rotating vectors, making phase relationships visual. The projection of the phasor on the vertical axis gives the instantaneous value.

Reactance and Impedance

Inductive reactance: XLX_{L} = ω\omegaL = 2π\pifL (increases with frequency — inductors oppose rapid changes).
Capacitive reactance: XCX_{C} = 1/(ω\omegaC) = 1/(2π\pifC) (decreases with frequency — capacitors pass high frequencies). Both have units of ohms. Impedance Z is the total opposition to AC flow. For a series LCR circuit:

Z = R2+(XLXC\sqrt{R^{2} + ({X}_{L} - {X}_{C}}2)

The phase angle between voltage and current: tan(ϕ\phi) = (XLX_{L} - XCX_{C})/R. If XLX_{L} > XCX_{C}, the circuit is inductive (voltage leads). If XCX_{C} > XLX_{L}, the circuit is capacitive (current leads).

Series LCR Circuit

Series LCR Circuit Diagram:

Series LCR circuit with AC source

In a series LCR circuit driven by V = V0V_{0} sin(ω\omegat), the current I = V0V_{0}/Z * sin(ω\omegat - ϕ\phi).
The voltage across each element: VRV_{R} = IR (in phase with I), VLV_{L} = I
XLX_{L} (leads I by π2\frac{\pi}{2}), VCV_{C} = I
XCX_{C} (lags I by π2\frac{\pi}{2}).
The phasor sum: V = VR2+(VLVC\sqrt{{V}_{R}^2 + ({V}_{L} - {V}_{C}}2). Note that VLV_{L} and VCV_{C} can individually exceed the source voltage V (voltage magnification).

Phasor Diagram for Series LCR Circuit (XLX_{L} > XCX_{C}):

Phasor diagram showing $V_{R}$, $V_{L}$, $V_{C}$ and resultant V with phase angle $\phi$

Resonance

At resonance, XLX_{L} = XCX_{C}, so ω0\omega_{0}*L = 1/(ω0\omega_{0}*C), giving the resonant frequency:

ω0\omega_{0} = 1/LC\sqrt{LC}, or f0f_{0} = 1/(2*π\pi*LC\sqrt{LC})

Resonance Curve — Impedance vs Frequency:

Resonance curve showing current vs frequency for different R values in a series LCR circuit

At resonance: Z = R (minimum impedance), current is maximum (I0I_{0} = V0V_{0}/R), phase angle ϕ\phi = 0 (voltage and current in phase), and the circuit behaves as a pure resistor.
The quality factor Q = ω0\omega_{0}L/R = 1/(ω0\omega_{0}CR) = (1/R)L/C\sqrt{L/C} measures the sharpness of resonance.
Higher Q means a narrower resonance peak and greater voltage magnification: VLV_{L} = VCV_{C} = Q*V at resonance.
Bandwidth = ω0\omega_{0}/Q = R/L.

Power in AC Circuits

Instantaneous power p(t) = VI oscillates at 2ω\omega.
Average power: P = VrmsV_{rms} * IrmsI_{rms} * cos(ϕ\phi), where cos(ϕ\phi) is the power factor.
For pure R: cos(ϕ\phi) = 1, P = Vrms2V_{rms}^{2}/R.
For pure L or C: cos(ϕ\phi) = 0, P = 0 (wattless current). The current component Icos(ϕ\phi) is the active (power) component; Isin(ϕ\phi) is the reactive (wattless) component.
Apparent power = VrmsV_{rms} * IrmsI_{rms} (in VA).
Real power = apparent power x power factor.
Wattless current = IrmsI_{rms} * sin(ϕ\phi) dissipates no power but increases the current drawn from the source.

Transformers

Transformer — Primary and Secondary Coils:

Transformer with primary and secondary coils on an iron core

A transformer transfers AC energy between circuits through mutual induction.
For an ideal transformer: VsVp\frac{V_{s}}{V_{p}} = NsNp\frac{N_{s}}{N_{p}} = k (turns ratio), and IsIp\frac{I_{s}}{I_{p}} = NpNs\frac{N_{p}}{N_{s}} = 1/k (power conservation: VpV_{p}IpI_{p} = VsV_{s}IsI_{s}). Step-up: NsN_{s} > NpN_{p} (increases voltage, decreases current). Step-down: NsN_{s} < NpN_{p} (decreases voltage, increases current). Real transformer losses: copper losses (I2I^{2}R in windings), iron/core losses (eddy currents and hysteresis in core), flux leakage.
Efficiency = PoutputPinput\frac{P_{output}}{P_{input}} x 100%. Laminated soft iron cores minimize eddy current losses. Transformers only work with AC — DC produces no changing flux.

The key problem-solving concept is impedance analysis using phasors: draw the phasor diagram, identify phase relationships, and compute Z and ϕ\phi to find current and power.


Key Testable Concept

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