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CalibratePhysics

Chapter 13

AC Circuits

Maharashtra HSC · 10 subtopics · textbook pages 288–305

Formula SheetEvery formula this chapter can be examined on, in one table.

AC Generator — Alternating emf and Current

FormulaSymbols, units and conditions
e=e0sin⁡ωte = e_0\sin\omega t
  • Instantaneous alternating emf. e0e_0 = peak (maximum) value of emf in volt, ω\omega = angular frequency of rotation of the coil in rad/s, tt = time in second.
i=i0sin⁡(ωt+α)i = i_0\sin(\omega t + \alpha)
  • Instantaneous alternating current. i0i_0 = peak current in ampere, α\alpha = phase difference between the current and the emf (α=0\alpha = 0 for a resistor, −π/2-\pi/2 for a pure inductor, +π/2+\pi/2 for an ideal capacitor).
ω=2πf=2πT,f=ω2π,T=1f\omega = 2\pi f = \dfrac{2\pi}{T}, \qquad f = \dfrac{\omega}{2\pi}, \qquad T = \dfrac{1}{f}
  • ff = frequency of the AC supply in hertz (Hz), TT = periodic time in second. Indian mains: f=50f = 50 Hz, so T=0.02T = 0.02 s and ω=100π≈314.2\omega = 100\pi \approx 314.2 rad/s.

Average and RMS Values

FormulaSymbols, units and conditions
iav=2i0π=0.637 i0,eav=0.637 e0i_{av} = \dfrac{2i_0}{\pi} = 0.637\,i_0, \qquad e_{av} = 0.637\,e_0
  • Average (mean) value over one half cycle. Over a full cycle the average of a sinusoidal AC is zero, so the mean is always defined over half a cycle.
irms=i02=0.707 i0,erms=e02=0.707 e0i_{rms} = \dfrac{i_0}{\sqrt2} = 0.707\,i_0, \qquad e_{rms} = \dfrac{e_0}{\sqrt2} = 0.707\,e_0
  • Root-mean-square (effective, virtual) value — the steady DC that would produce the same heat in the same resistance in the same time. This is what an AC ammeter or voltmeter reads.
i0=2 irms=1.414 irms,e0=2 ermsi_0 = \sqrt2\, i_{rms} = 1.414\, i_{rms}, \qquad e_0 = \sqrt2\, e_{rms}
  • Mains supply of 220 V (rms) has e0=1.414×220≈311e_0 = 1.414\times220 \approx 311 V.
H=irms2 R tH = i_{rms}^2\,R\,t
  • Heat produced in a resistance RR (ohm) in time tt (second) by a sinusoidal AC; HH in joule.

AC Voltage Applied to a Resistor

FormulaSymbols, units and conditions
e=e0sin⁡ωt,i=e0Rsin⁡ωt=i0sin⁡ωt,i0=e0Re = e_0\sin\omega t, \qquad i = \dfrac{e_0}{R}\sin\omega t = i_0\sin\omega t, \qquad i_0 = \dfrac{e_0}{R}
  • Purely resistive circuit. Current and emf are in phase (phase difference zero). RR behaves identically for DC and AC.
irms=ermsRi_{rms} = \dfrac{e_{rms}}{R}
  • Form used in every resistive numerical.

AC Voltage Applied to an Inductor

FormulaSymbols, units and conditions
XL=ωL=2πfLX_L = \omega L = 2\pi f L
  • Inductive reactance in ohm (Ω\Omega). LL = self inductance in henry (H). XL∝fX_L \propto f and XL∝LX_L \propto L. For DC, f=0f = 0, so XL=0X_L = 0.
i=i0sin⁡ ⁣(ωt−π2),i0=e0ωL=e0XLi = i_0\sin\!\left(\omega t - \dfrac{\pi}{2}\right), \qquad i_0 = \dfrac{e_0}{\omega L} = \dfrac{e_0}{X_L}
  • Pure inductor. Current lags the emf by π/2\pi/2 rad (90°); equivalently the voltage across LL leads the current by 90°.

AC Voltage Applied to a Capacitor

FormulaSymbols, units and conditions
XC=1ωC=12πfCX_C = \dfrac{1}{\omega C} = \dfrac{1}{2\pi f C}
  • Capacitive reactance in ohm (Ω\Omega). CC = capacitance in farad (F). XC∝1/fX_C \propto 1/f and XC∝1/CX_C \propto 1/C. For DC, f=0f = 0, so XC=∞X_C = \infty.
i=i0sin⁡ ⁣(ωt+π2),i0=e01/ωC=ωC e0=e0XCi = i_0\sin\!\left(\omega t + \dfrac{\pi}{2}\right), \qquad i_0 = \dfrac{e_0}{1/\omega C} = \omega C\,e_0 = \dfrac{e_0}{X_C}
  • Ideal capacitor. Current leads the emf by π/2\pi/2 rad (90°).

Series LCR Circuit

FormulaSymbols, units and conditions
eR=iR,eL=iXL,eC=iXCe_R = iR, \qquad e_L = iX_L, \qquad e_C = iX_C
  • Voltages across the three elements; the same current ii flows through all three.
e=eR2+(eL−eC)2e = \sqrt{e_R^2 + (e_L - e_C)^2}
  • Resultant applied emf from the phasor diagram — never the plain sum eR+eL+eCe_R+e_L+e_C.
Z=R2+(XL−XC)2=R2+(ωL−1ωC)2Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + \left(\omega L - \dfrac{1}{\omega C}\right)^2}
  • Impedance in ohm (Ω\Omega): the total effective opposition of RR, LL and CC in series. Defined as the ratio of rms voltage to rms current.
e=iZ,irms=ermsZ,i0=e0Ze = iZ, \qquad i_{rms} = \dfrac{e_{rms}}{Z}, \qquad i_0 = \dfrac{e_0}{Z}—
tan⁡ϕ=XL−XCR,ϕ=tan⁡−1 ⁣(XL−XCR)\tan\phi = \dfrac{X_L - X_C}{R}, \qquad \phi = \tan^{-1}\!\left(\dfrac{X_L - X_C}{R}\right)
  • ϕ\phi = phase angle by which the emf leads the current. Positive when XL>XCX_L > X_C (inductance dominated), negative when XL<XCX_L < X_C (capacitance dominated), zero when XL=XCX_L = X_C.
Admittance=1Z\text{Admittance} = \dfrac{1}{Z}
  • SI unit ohm−1^{-1} or siemens (S).
Δt=ϕω\Delta t = \dfrac{\phi}{\omega}
  • Time lag between maximum voltage and maximum current, with ϕ\phi in radian.

Power in an AC Circuit

FormulaSymbols, units and conditions
P=ei(instantaneous power)P = ei \quad \text{(instantaneous power)}—
Pav=erms irms=e0i02=irms2R=erms2RP_{av} = e_{rms}\,i_{rms} = \dfrac{e_0 i_0}{2} = i_{rms}^2 R = \dfrac{e_{rms}^2}{R}
  • Purely resistive circuit. Also called the apparent power in this case. The last form gives the resistance of a lamp or heater from its wattage rating and supply voltage.
Pav=0(pure inductor, ideal capacitor)P_{av} = 0 \quad \text{(pure inductor, ideal capacitor)}
  • Because ϕ=90∘\phi = 90^\circ and cos⁡90∘=0\cos 90^\circ = 0. The current is then called idle or wattless current.
Pav=erms irmscos⁡ϕ=e0i02cos⁡ϕP_{av} = e_{rms}\,i_{rms}\cos\phi = \dfrac{e_0 i_0}{2}\cos\phi
  • Series LCR circuit — the true power. Depends on the rms values and on the phase difference.
cos⁡ϕ=Paverms irms=true powerapparent power=RZ=RR2+(XL−XC)2\cos\phi = \dfrac{P_{av}}{e_{rms}\,i_{rms}} = \dfrac{\text{true power}}{\text{apparent power}} = \dfrac{R}{Z} = \dfrac{R}{\sqrt{R^2+(X_L-X_C)^2}}
  • Power factor: resistance divided by impedance. cos⁡ϕ=1\cos\phi = 1 for a purely resistive circuit, cos⁡ϕ=0\cos\phi = 0 for a pure inductor or ideal capacitor.
∫0Tsin⁡2ωt dt=T2,∫0Tsin⁡ωt cos⁡ωt dt=0,∫0Tsin⁡2ωt dt=0\int_0^T \sin^2\omega t\,dt = \dfrac{T}{2}, \qquad \int_0^T \sin\omega t\,\cos\omega t\,dt = 0, \qquad \int_0^T \sin 2\omega t\,dt = 0
  • The three standard integrals every power derivation in this chapter uses.

LC Oscillations

FormulaSymbols, units and conditions
UE=12q2C,UB=12LI2U_E = \dfrac{1}{2}\dfrac{q^2}{C}, \qquad U_B = \dfrac{1}{2}LI^2
  • Energy stored in the electric field of the capacitor and in the magnetic field of the inductor, in joule.
12q02C=12LI02⇒I0=q0LC=V0CL\dfrac{1}{2}\dfrac{q_0^2}{C} = \dfrac{1}{2}LI_0^2 \qquad \Rightarrow \qquad I_0 = \dfrac{q_0}{\sqrt{LC}} = V_0\sqrt{\dfrac{C}{L}}
  • Conservation of energy in an ideal (resistanceless) LC circuit: q0=CV0q_0 = CV_0 is the initial charge on the capacitor, I0I_0 the maximum current in the coil.
f=12πLC,ω=1LCf = \dfrac{1}{2\pi\sqrt{LC}}, \qquad \omega = \dfrac{1}{\sqrt{LC}}
  • Natural (free) frequency of LC oscillations; LL in henry, CC in farad, ff in hertz.
C=14π2f2LC = \dfrac{1}{4\pi^2 f^2 L}
  • Rearranged form, used when the capacitance needed for a given oscillation frequency is asked.

Electric Resonance

FormulaSymbols, units and conditions
XL=XC⇒ωrL=1ωrCX_L = X_C \qquad \Rightarrow \qquad \omega_r L = \dfrac{1}{\omega_r C}
  • Condition for resonance, series or parallel. Both LL and CC must be present.
ωr=1LC,fr=12πLC\omega_r = \dfrac{1}{\sqrt{LC}}, \qquad f_r = \dfrac{1}{2\pi\sqrt{LC}}
  • Resonant frequency. Independent of RR.
Zmin=R,i0,max=e0R,irms=ermsRZ_{min} = R, \qquad i_{0,max} = \dfrac{e_0}{R}, \qquad i_{rms} = \dfrac{e_{rms}}{R}
  • Series resonance: impedance is least, circuit is purely resistive, current is maximum, cos⁡ϕ=1\cos\phi = 1.
i=e0(ωC−1ωL)cos⁡ωti = e_0\left(\omega C - \dfrac{1}{\omega L}\right)\cos\omega t
  • Total current in a parallel LC circuit; it is minimum (and ZZ maximum) when ωC=1/ωL\omega C = 1/\omega L.

Sharpness of Resonance — Q Factor

FormulaSymbols, units and conditions
i0=e0R2+(ωL−1ωC)2i_0 = \dfrac{e_0}{\sqrt{R^2 + \left(\omega L - \dfrac{1}{\omega C}\right)^2}}
  • Current amplitude at any angular frequency ω\omega in a series LCR circuit.
ω1=ωr+Δω,ω2=ωr−Δω,Bandwidth=ω1−ω2=2Δω\omega_1 = \omega_r + \Delta\omega, \qquad \omega_2 = \omega_r - \Delta\omega, \qquad \text{Bandwidth} = \omega_1 - \omega_2 = 2\Delta\omega
  • The half-power frequencies, at which the current amplitude falls to 1/21/\sqrt2 of its maximum value and the power dissipated becomes half.
Q=ωrω1−ω2=ωr2Δω=Resonant frequencyBandwidthQ = \dfrac{\omega_r}{\omega_1 - \omega_2} = \dfrac{\omega_r}{2\Delta\omega} = \dfrac{\text{Resonant frequency}}{\text{Bandwidth}}
  • Quality factor — a dimensionless measure of the sharpness of resonance. Larger QQ means smaller bandwidth, sharper peak, more selective circuit.

Choke Coil

FormulaSymbols, units and conditions
XL=ωL,Pav=erms irmscos⁡ϕX_L = \omega L, \qquad P_{av} = e_{rms}\,i_{rms}\cos\phi—
cos⁡ϕ=RR2+ω2L2\cos\phi = \dfrac{R}{\sqrt{R^2 + \omega^2L^2}}
  • Power factor of a choke coil of inductance LL and small resistance RR. Since LL is very large and RR very small, cos⁡ϕ≈0\cos\phi \approx 0 and the power loss is nearly zero.
Short NotesThe chapter subtopic by subtopic, in the textbook's own order.10 subtopics
  1. 1Introduction
  2. 2AC Generator1 board question
  3. 3Average and RMS Values1 board question
  4. 4Phasors
  5. 5Different types of AC circuits7 board questions
  6. 6Power in AC circuit4 board questions
  7. 7LC Oscillations
  8. 8Electric Resonance8 board questions
  9. 9Sharpness of Resonance: Q Factor
  10. 10Choke Coil
The whole chapter in brief

13.1 Introduction

  • DC (direct current): the polarity of the supply voltage is fixed — the positive and negative ends of the supply do not change.
  • AC (alternating current): the polarity of the voltage keeps reversing periodically.
  • Electricity is supplied mostly as AC because it is cheap to generate and convenient to transport.
  • Appliances such as TV, computer and radio use a rectifier to convert AC to DC; fans, fridges, air conditioners and heaters run directly on AC.
  • Remember this: an AC shock is attractive while a DC shock is repulsive, so 220 V AC is more dangerous than 220 V DC. Also 220 V AC has a peak value of ±2×220≈±311\pm\sqrt2\times220 \approx \pm311 V, whereas DC has a fixed value of 220 V only.

13.2 AC Generator

  • An AC generator produces a time-dependent emf e=e0sin⁡ωte = e_0\sin\omega t, where e0e_0 is the peak emf and ω\omega the angular frequency of rotation of the coil.
  • The current in the connected circuit has the same time dependence: i=i0sin⁡(ωt+α)i = i_0\sin(\omega t + \alpha), with α\alpha the phase difference between current and emf.
  • Graph of ee against ωt\omega t is a sine curve: the direction of the emf reverses after every half revolution of the coil. Such an emf is called alternating emf and the corresponding current alternating current.
  • Peak value: the maximum value of the alternating current (or emf) in either direction.
  • The average value of alternating current or emf over a complete cycle is zero.

13.3 Average and RMS Values

  • Average value of AC is defined over one half cycle, because over a full cycle the average of sin⁡ωt\sin\omega t — and hence of the AC — is zero and carries no meaning.
  • Average value of sin⁡θ\sin\theta over 00 to π\pi is 2/π=0.6372/\pi = 0.637, so iav=0.637 i0i_{av} = 0.637\,i_0 and eav=0.637 e0e_{av} = 0.637\,e_0.
  • A moving-coil ammeter or voltmeter responds to the average value, so it cannot measure AC. What is needed is a property independent of the direction of flow — the heating effect, which depends on the square of the current.
  • RMS (effective, virtual) value: the value of that steady current which produces the same heat in the same resistance in the same time as the given AC does.
  • irms=i0/2=0.707 i0i_{rms} = i_0/\sqrt2 = 0.707\,i_0 and erms=e0/2=0.707 e0e_{rms} = e_0/\sqrt2 = 0.707\,e_0.
  • Heat produced by a sinusoidal AC over a complete cycle is H=irms2RtH = i_{rms}^2Rt — the same as that produced by a DC of magnitude irmsi_{rms} for the same time.
  • All AC ammeters, voltmeters and quoted supply ratings (220 V, 110 V) are rms values.

13.4 Phasors

  • A rotating vector that represents a sinusoidally varying quantity is called a phasor; the diagram representing it is a phasor diagram.
  • The length of the arrow represents the peak value of the quantity (i0i_0 or e0e_0); the phasor is inclined at ωt\omega t (or ωt+α\omega t + \alpha) to the horizontal axis and rotates anticlockwise.
  • The projection of the phasor on a fixed axis gives the instantaneous value: on the Y-axis for the sine form, on the X-axis for the cosine form.
  • Representing emf and current as rotating vectors lets us add them by the laws of vector addition, which is why phasors simplify AC circuit analysis.

13.5 Different Types of AC Circuits

Resistor alone

  • e=e0sin⁡ωte = e_0\sin\omega t gives i=(e0/R)sin⁡ωt=i0sin⁡ωti = (e_0/R)\sin\omega t = i_0\sin\omega t, with i0=e0/Ri_0 = e_0/R.
  • Current and emf are in phase — both reach zero, minimum and maximum at the same instants.
  • RR behaves identically for AC and DC, and can reduce both equally effectively.
  • Phasor diagram: the phasors for ii and ee point in the same direction, making an angle ωt\omega t with OX; phase angle between them is zero.

Inductor alone

  • Applied emf opposes the self-induced emf e′=−L di/dte' = -L\,di/dt. With no resistance, Kirchhoff's voltage law gives di=(e0/L)sin⁡ωt dtdi = (e_0/L)\sin\omega t\,dt; integrating (the constant of integration is zero, since ii must oscillate about zero) gives i=i0sin⁡(ωt−π/2)i = i_0\sin(\omega t - \pi/2) with i0=e0/ωLi_0 = e_0/\omega L.
  • Current lags the emf by π/2\pi/2 rad (90°); equivalently the voltage across LL leads the current by 90°.
  • Inductive reactance: XL=ωL=2πfLX_L = \omega L = 2\pi fL, the opposing nature of an inductor to the flow of AC. SI unit ohm; dimensions the same as those of resistance.
  • XL∝fX_L \propto f and XL∝LX_L \propto L. For DC, f=0f = 0, so XL=0X_L = 0 — a pure inductor cannot reduce DC. It passes DC and blocks high-frequency AC.
  • Phasor diagram: the i0i_0 phasor is turned 90° clockwise from the e0e_0 phasor.

Capacitor alone

  • q=Ceq = Ce, so i=dq/dt=C d(e0sin⁡ωt)/dt=ωCe0cos⁡ωt=i0sin⁡(ωt+π/2)i = dq/dt = C\,d(e_0\sin\omega t)/dt = \omega Ce_0\cos\omega t = i_0\sin(\omega t + \pi/2) with i0=e0/(1/ωC)i_0 = e_0/(1/\omega C).
  • Current leads the emf by π/2\pi/2 rad (90°).
  • Capacitive reactance: XC=1/ωC=1/2πfCX_C = 1/\omega C = 1/2\pi fC. It is defined as the ratio of the rms voltage across the capacitor to the corresponding rms current. SI unit ohm.
  • XC∝1/fX_C \propto 1/f and XC∝1/CX_C \propto 1/C. For DC, f=0f = 0, so XC=∞X_C = \infty — a capacitor blocks DC and acts as an open circuit, while it passes high-frequency AC.
  • Phasor diagram: the i0i_0 phasor is turned 90° anticlockwise from the e0e_0 phasor.
ResistanceReactance
Equally effective for AC and DCCurrent is reduced but no energy is consumed; heat is generated only by the resistive component
Value is independent of the frequency of the ACXL=2πfLX_L = 2\pi fL is directly proportional and XC=1/2πfCX_C = 1/2\pi fC inversely proportional to the frequency
Current opposed by a resistor is in phase with the voltageCurrent opposed by a pure inductor lags, and by an ideal capacitor leads, the voltage by π/2\pi/2

Series LCR circuit

  • The same current i=i0sin⁡ωti = i_0\sin\omega t flows through all three; the voltages are eR=iRe_R = iR (in phase with ii), eL=iXLe_L = iX_L (leads ii by 90°) and eC=iXCe_C = iX_C (lags ii by 90°).
  • eLe_L and eCe_C are 180° out of phase, so the net reactive voltage is (eL−eC)(e_L - e_C).
  • Phasor addition gives e=eR2+(eL−eC)2=iR2+(XL−XC)2=iZe = \sqrt{e_R^2 + (e_L-e_C)^2} = i\sqrt{R^2+(X_L-X_C)^2} = iZ.
  • Impedance Z=R2+(XL−XC)2Z = \sqrt{R^2+(X_L-X_C)^2}, in ohm; it is the ratio of rms voltage to rms current. Its reciprocal is admittance, unit ohm−1^{-1} or siemens.
  • tan⁡ϕ=(XL−XC)/R\tan\phi = (X_L-X_C)/R, where ϕ\phi is the angle by which the voltage leads the current.
  • Impedance triangle: base RR, perpendicular (XL−XC)(X_L-X_C), hypotenuse ZZ, angle at the base ϕ\phi.
CaseSign of tan⁡ϕ\tan\phiNature of circuit
XL=XCX_L = X_CzeroVoltage and current in phase; non-inductive (purely resistive)
XL>XCX_L > X_CpositiveVoltage leads the current; inductance dominated
XL<XCX_L < X_CnegativeVoltage lags the current; capacitance dominated

13.6 Power in an AC Circuit

  • Instantaneous power P=eiP = ei; because both change every instant, the useful quantity is the average over one full cycle.
  • Pure resistor: P=e0i0sin⁡2ωtP = e_0i_0\sin^2\omega t, and since ∫0Tsin⁡2ωt dt=T/2\int_0^T\sin^2\omega t\,dt = T/2, Pav=e0i0/2=ermsirmsP_{av} = e_0i_0/2 = e_{rms}i_{rms}. This product is also called the apparent power.
  • Pure inductor: i=i0sin⁡(ωt−π/2)i = i_0\sin(\omega t - \pi/2), so P=−e0i0sin⁡ωtcos⁡ωtP = -e_0i_0\sin\omega t\cos\omega t, and ∫0Tsin⁡ωtcos⁡ωt dt=0\int_0^T\sin\omega t\cos\omega t\,dt = 0, giving Pav=0P_{av} = 0.
  • Ideal capacitor: i=i0cos⁡ωti = i_0\cos\omega t, so P=e0i0sin⁡ωtcos⁡ωtP = e_0i_0\sin\omega t\cos\omega t, and again Pav=0P_{av} = 0.
  • Series LCR: Pav=ermsirmscos⁡ϕP_{av} = e_{rms}i_{rms}\cos\phi — the true power.
  • Power factor cos⁡ϕ=Pav/(ermsirms)=\cos\phi = P_{av}/(e_{rms}i_{rms}) = true power / apparent power =R/Z= R/Z.
  • cos⁡ϕ=1\cos\phi = 1 (ϕ=0\phi = 0) for a non-inductive circuit; cos⁡ϕ=0\cos\phi = 0 (ϕ=90∘\phi = 90^\circ) for a pure inductor or ideal capacitor.
  • Wattless (idle) current: the current through a pure inductor or an ideal capacitor, which consumes no power for its maintenance in the circuit. Power is dissipated only in the resistance.
  • A low power factor in a transmission line means a large current is needed to deliver the same power, and the i2Ri^2R loss in the line therefore becomes large.

13.7 LC Oscillations

  • A charged capacitor connected across a resistanceless inductor sets up electrical oscillations of constant amplitude and frequency, called LC oscillations.
  • Energy surges back and forth between the electric field of the capacitor, UE=q2/2CU_E = q^2/2C, and the magnetic field of the inductor, UB=12LI2U_B = \dfrac{1}{2}LI^2.
  • Stage by stage: capacitor fully charged with q0q_0 (all energy electrostatic, I=0I = 0); capacitor discharges and II grows (energy shared); capacitor fully discharged and I=I0I = I_0 (all energy magnetic, 12LI02\dfrac{1}{2}LI_0^2); flux decreases and the induced emf drives the current in the same direction, charging the capacitor with reversed polarity; the process then repeats in the opposite direction and the circuit returns to its initial state.
  • Frequency: f=12πLCf = \dfrac{1}{2\pi\sqrt{LC}}. Maximum current from energy conservation: 12q02C=12LI02\dfrac{1}{2}\dfrac{q_0^2}{C} = \dfrac{1}{2}LI_0^2.
  • If there is no loss of energy the amplitude stays constant and the oscillations are undamped.
  • In practice LC oscillations are damped, for two reasons: (a) every inductor has some resistance, so energy is lost as heat and the amplitude dies out; (b) even with zero resistance, energy is radiated away as electromagnetic waves — the basis of radio and TV transmitters.

13.8 Electric Resonance

  • Resonance occurs when a system with a natural frequency of oscillation is driven by a source whose frequency equals that natural frequency; the amplitude of oscillation then becomes large.
  • Series resonance circuit: LL, CC and RR in series, admitting maximum current for a given AC frequency.
  • As ff rises, XL=ωLX_L = \omega L increases and XC=1/ωCX_C = 1/\omega C decreases; at ωr\omega_r they are equal, so Z=R2+0=RZ = \sqrt{R^2+0} = R, the least possible value.
  • ωrL=1/ωrC\omega_r L = 1/\omega_r C gives ωr=1/LC\omega_r = 1/\sqrt{LC} and fr=12πLCf_r = \dfrac{1}{2\pi\sqrt{LC}}.
  • Series resonance curve: rms current plotted against frequency, peaking sharply at frf_r. The peak height is e0/Re_0/R, so a smaller RR gives a taller, sharper peak.
  • Parallel resonance circuit: coil LL and condenser CC in parallel across the source. The total current is i=e0(ωC−1ωL)cos⁡ωti = e_0\left(\omega C - \dfrac{1}{\omega L}\right)\cos\omega t, minimum when ωC=1/ωL\omega C = 1/\omega L, i.e. at the same fr=12πLCf_r = \dfrac{1}{2\pi\sqrt{LC}}.
  • Do you know: for resonance to occur, both LL and CC must be present, because only then do the mutually 180°-out-of-phase voltages across them cancel and the whole source voltage appear across RR. There is no resonance in an LR or a CR circuit.
Series resonanceParallel resonance
Occurs when XL=XCX_L = X_COccurs when XL=XCX_L = X_C
fr=12πLCf_r = \dfrac{1}{2\pi\sqrt{LC}}fr=12πLCf_r = \dfrac{1}{2\pi\sqrt{LC}}
Impedance is minimum and equals RRImpedance is maximum
Current is maximumCurrent is minimum
Accepts one frequency frf_r and rejects others — acceptor circuitRejects frf_r and passes the others — rejector circuit
Circuit is purely resistive; ee and ii in phaseUsed in wireless transmission, radio communication and filter circuits

13.9 Sharpness of Resonance — Q Factor

  • Current amplitude in a series LCR circuit is i0=e0/R2+(ωL−1ωC)2i_0 = e_0\Big/\sqrt{R^2+\left(\omega L - \dfrac{1}{\omega C}\right)^2}; it is maximum when ωrL−1ωrC=0\omega_r L - \dfrac{1}{\omega_r C} = 0.
  • Half-power frequencies: the two values of ω\omega at which the current amplitude falls to 1/21/\sqrt2 of its maximum, so that the power dissipated is halved. They lie symmetrically about ωr\omega_r: ω1=ωr+Δω\omega_1 = \omega_r + \Delta\omega and ω2=ωr−Δω\omega_2 = \omega_r - \Delta\omega.
  • Bandwidth =ω1−ω2=2Δω= \omega_1 - \omega_2 = 2\Delta\omega.
  • Q factor of a series resonant circuit is defined as the ratio of the resonant frequency to the difference between the two frequencies, taken one on either side of resonance, at which the current amplitude is 1/21/\sqrt2 times its value at resonance.
  • Q=ωrω1−ω2=ωr2ΔωQ = \dfrac{\omega_r}{\omega_1-\omega_2} = \dfrac{\omega_r}{2\Delta\omega}; it is dimensionless.
  • Larger QQ means smaller bandwidth, a sharper peak and a more selective circuit.
  • On the resonance curve the low-frequency side is dominated by the capacitor's reactance and the high-frequency side by the inductor's reactance; resonance occurs in the middle.
  • Tuning a radio or TV is an LCR resonance: the antenna picks up many frequencies, but maximum current flows for the one equal to fr=12πLCf_r = \dfrac{1}{2\pi\sqrt{LC}}. A large QQ keeps other stations weak, and varying the adjustable capacitor CC tunes in the desired station.

13.10 Choke Coil

  • Reducing AC with an ordinary resistance wastes energy as Joule heat (i2Rti^2Rt). A choke coil reduces the current without much loss of energy.
  • Construction: thick insulated copper wire wound closely in a large number of turns over a soft-iron laminated core.
  • It offers a large reactance XL=ωLX_L = \omega L to AC, so the current is reduced; the laminated core reduces eddy-current loss.
  • Power dissipated P=ermsirmscos⁡ϕP = e_{rms}i_{rms}\cos\phi with cos⁡ϕ=RR2+ω2L2\cos\phi = \dfrac{R}{\sqrt{R^2+\omega^2L^2}}.
  • LL is very large and RR very small, so cos⁡ϕ\cos\phi is nearly zero and the power loss is very small.
  • The only real loss is hysteresis loss in the core, reduced by using soft iron.

Standard One-Mark and Two-Mark Answers

  • Brightness of a lamp in series with a capacitor when CC is increased: XC=1/2πfCX_C = 1/2\pi fC decreases, so ZZ decreases, current increases and the lamp glows brighter.
  • Why the total impedance falls when a capacitor is added in series with LL and RR: XCX_C acts opposite to XLX_L, so the net reactance (XL−XC)(X_L - X_C) is smaller than XLX_L alone and hence ZZ is smaller.
  • Why a capacitor behaves like a pure conductor at very high frequency: XC=1/2πfC→0X_C = 1/2\pi fC \to 0 as ff becomes very large, so it offers almost no opposition.
  • Wattless current: the current in a pure inductor or ideal capacitor, where ϕ=90∘\phi = 90^\circ and Pav=ermsirmscos⁡90∘=0P_{av} = e_{rms}i_{rms}\cos 90^\circ = 0, so it consumes no power.
  • Natural frequency of an LC circuit is f=12πLCf = \dfrac{1}{2\pi\sqrt{LC}}, and at this frequency XL=XCX_L = X_C, so the net reactance of the circuit is zero.
  • In an LR circuit with XL=RX_L = R, power factor P1=R/R2+R2=1/2P_1 = R/\sqrt{R^2+R^2} = 1/\sqrt2. Adding a capacitor with XC=XLX_C = X_L makes the circuit purely resistive, so P2=1P_2 = 1 and P1/P2=1/2=0.707P_1/P_2 = 1/\sqrt2 = 0.707.

Traps That Cost Marks

  • Writing the average value of AC over a full cycle as 0.637 i00.637\,i_0. Over a full cycle it is zero; 0.637 i00.637\,i_0 is the half-cycle average.
  • Adding the voltages across RR, LL and CC arithmetically. They must be added as phasors: e=eR2+(eL−eC)2e = \sqrt{e_R^2+(e_L-e_C)^2}. The reading across LL or CC alone can even exceed the source voltage.
  • Confusing XLX_L and XCX_C behaviour with frequency. XLX_L rises with ff (inductor blocks high-frequency AC, passes DC); XCX_C falls with ff (capacitor blocks DC, passes high-frequency AC).
  • Forgetting the 2π2\pi when converting: XL=2πfLX_L = 2\pi fL, not fLfL; XC=1/2πfCX_C = 1/2\pi fC, not 1/fC1/fC.
  • Using the peak value where the question gives an rms value. A supply described as "220 V, 50 Hz" is rms; convert with e0=2 ermse_0 = \sqrt2\,e_{rms} before using i0=e0/Zi_0 = e_0/Z.
  • Treating the impedance at resonance as zero. It is minimum and equals RR, not zero.
  • Interchanging the two resonance circuits: series is minimum ZZ, maximum current, acceptor; parallel is maximum ZZ, minimum current, rejector.
  • Saying resonance can happen in an LR or CR circuit. Both LL and CC are essential.
  • Forgetting to convert mH, μ\muF and pF to henry and farad before substituting. 1 mH =10−3= 10^{-3} H, 1 μ\muF =10−6= 10^{-6} F, 1 pF =10−12= 10^{-12} F.
  • Quoting the power in a pure inductor or capacitor as ermsirmse_{rms}i_{rms}. It is zero — the power factor is zero.
  • Dropping cos⁡ϕ\cos\phi from Pav=ermsirmscos⁡ϕP_{av} = e_{rms}i_{rms}\cos\phi in an LCR numerical, which turns true power into apparent power.
  • Leaving ϕ\phi in degrees when a time lag is asked. Δt=ϕ/ω\Delta t = \phi/\omega needs ϕ\phi in radian.
Solved Numerical ExamplesWorked numericals to attempt first, each with a hint before the full solution.Coming soon

These are being added chapter by chapter. They are solved numericals from the prescribed book and the reference guides — not previous board questions, which live under Practice — so that you can work through the method before meeting it in a real paper.

Important DerivationsEvery derive / obtain / prove question in this chapter. 2 have come up in two or more papers.5
  1. 1
    Impedance of a series LCR circuit connected to an AC source (phasor method) Repeats
    Derive an expression for the impedance of an LCR circuit connected to an AC power supply. Draw the phasor diagram.

    Asked in: July 2018; February 2024 (Q.21, 3 marks); July 2024 (Q.17, 3 marks); June 2025 (Q.28, 4 marks)

    Show the derivation
    Step 1
    A circuit in which inductance L, capacitance C and resistance R are connected in series and the circuit admits maximum current corresponding to a given frequency of AC, is called a series resonance circuit.
    Step 2
    The impedance (Z) of an LCR circuit is given by,
    Z=R2+(ωL−1ωC)2Z=\sqrt{R^2+\left(\omega L-\dfrac{1}{\omega C}\right)^2}
    Step 3
    At very low frequencies, inductive reactance XL=ωLX_L=\omega L is negligible but capacitive reactance XC=1ωCX_C=\dfrac{1}{\omega C} is very high.
    Step 4
    As we increase the applied frequency then XLX_L increases and XCX_C decreases.
    Step 5
    At some angular frequency (ωr\omega_r), XL=XCX_L=X_C
    i.e., ωrL=1ωrC\omega_rL=\dfrac{1}{\omega_rC}
    ∴(ωr)2=1LC  or  (2πfr)2=1LC\therefore(\omega_r)^2=\dfrac{1}{LC}\ \text{ or }\ (2\pi f_r)^2=\dfrac{1}{LC}
    ∴2πfr=1LC\therefore2\pi f_r=\dfrac{1}{\sqrt{LC}}
    ∴fr=12πLC\therefore f_r=\dfrac{1}{2\pi\sqrt{LC}}
    Where frf_r is called the resonant frequency.
    Step 6
    At this particular frequency frf_r, since XL=XCX_L=X_C we get Z=R2+0=RZ=\sqrt{R^2+0}=R. This is the least value of Z.
    Step 7
    Thus, when the impedance of an LCR circuit is minimum, circuit is said to be purely resistive, current and voltage are in phase and hence the current i0=e0Z=e0Ri_0=\dfrac{e_0}{Z}=\dfrac{e_0}{R} is maximum. This condition of the LCR circuit is called resonance condition and this frequency is called series resonant frequency.
  2. 2
    Resonant frequency of a series resonant circuit Repeats
    Derive an expression for the resonant frequency of a series resonant circuit.

    Asked in: July 2018; February 2026 (Q.16, 3 marks)

    Show the derivation
    Step 1
    A circuit in which inductance L, capacitance C and resistance R are connected in series and the circuit admits maximum current corresponding to a given frequency of AC, is called a series resonance circuit.
    Step 2
    The impedance (Z) of an LCR circuit is given by,
    Z=R2+(ωL−1ωC)2Z=\sqrt{R^2+\left(\omega L-\dfrac{1}{\omega C}\right)^2}
    Step 3
    At very low frequencies, inductive reactance XL=ωLX_L=\omega L is negligible but capacitive reactance XC=1ωCX_C=\dfrac{1}{\omega C} is very high.
    Step 4
    As we increase the applied frequency then XLX_L increases and XCX_C decreases.
    Step 5
    At some angular frequency (ωr\omega_r), XL=XCX_L=X_C
    i.e., ωrL=1ωrC\omega_rL=\dfrac{1}{\omega_rC}
    ∴(ωr)2=1LC  or  (2πfr)2=1LC\therefore(\omega_r)^2=\dfrac{1}{LC}\ \text{ or }\ (2\pi f_r)^2=\dfrac{1}{LC}
    ∴2πfr=1LC\therefore2\pi f_r=\dfrac{1}{\sqrt{LC}}
    ∴fr=12πLC\therefore f_r=\dfrac{1}{2\pi\sqrt{LC}}
    Where frf_r is called the resonant frequency.
    Step 6
    At this particular frequency frf_r, since XL=XCX_L=X_C we get Z=R2+0=RZ=\sqrt{R^2+0}=R. This is the least value of Z.
    Step 7
    Thus, when the impedance of an LCR circuit is minimum, circuit is said to be purely resistive, current and voltage are in phase and hence the current i0=e0Z=e0Ri_0=\dfrac{e_0}{Z}=\dfrac{e_0}{R} is maximum. This condition of the LCR circuit is called resonance condition and this frequency is called series resonant frequency.
  3. 3
    Average power dissipated in a purely resistive AC circuit; power factor
    Obtain an expression for the average power dissipated in a purely resistive A.C. circuit. Define the power factor of the circuit and state its value for a purely resistive A.C. circuit.

    Asked in: October 2015

    Show the derivation
    Step 1
    Expression for average power in purely resistive circuit:
    a. Let, e=e0sin⁡ωte=e_0\sin\omega t be applied e.m.f across a resistor of resistance 'R' as shown in figure. At certain instant, current i=i0sin⁡ωti=i_0\sin\omega t is flowing through the resistor. In this case both 'e' and 'i' are in phase.
    b. Instantaneous power in circuit is given by,
    P=ei=(e0sin⁡ωt)(i0sin⁡ωt)P=ei=(e_0\sin\omega t)(i_0\sin\omega t)
    ∴P=e0i0sin⁡2ωt....(1)\therefore P=e_0i_0\sin^2\omega t\qquad....(1)
    c. Average power for a complete cycle can be obtained by integrating equation (1).
    ∴Pav=work done by emf on the charges in one cycletime for one cycle\therefore P_{av}=\dfrac{\text{work done by emf on the charges in one cycle}}{\text{time for one cycle}}
    =∫0TPdtT=∫0Te0i0sin⁡2ωt dtT=\dfrac{\displaystyle\int_0^T Pdt}{T}=\dfrac{\displaystyle\int_0^T e_0i_0\sin^2\omega t\,dt}{T}
    ∴Pav=e0i0T(∫0Tsin⁡2ωt dt)....(2)\therefore P_{av}=\dfrac{e_0i_0}{T}\left(\int_0^T\sin^2\omega t\,dt\right)\qquad....(2)
    ∴Pav=e0i0T(T2)(∵∫0Tsin⁡2ωt dt=T2)\therefore P_{av}=\dfrac{e_0i_0}{T}\left(\dfrac{T}{2}\right)\qquad\left(\because\int_0^T\sin^2\omega t\,dt=\dfrac{T}{2}\right)
    =(12)e0i0=\left(\dfrac{1}{2}\right)e_0i_0
    ∴Pav=e02×i02\therefore P_{av}=\dfrac{e_0}{\sqrt{2}}\times\dfrac{i_0}{\sqrt{2}}
    ∴Pav=erms×irms\therefore P_{av}=e_{rms}\times i_{rms}
    P is also called as apparent power.
    Step 2
    Power factor for a circuit is given by,
    cos⁡ϕ=RZ=ResistanceImpendance\cos\phi=\dfrac{R}{Z}=\dfrac{\text{Resistance}}{\text{Impendance}}
    In purely resistive circuit,
    ∴cos⁡ϕ=RR2=RR=1\therefore\cos\phi=\dfrac{R}{\sqrt{R^2}}=\dfrac{R}{R}=1
    ∴ϕ=0°\therefore\phi=0°
  4. 4
    Average power dissipated in a series LCR AC circuit; power factor
    Obtain an expression for the average power dissipated in a series LCR A.C. circuit. Hence obtain an expression for the power factor of the circuit.

    Asked in: July 2019

    Show the derivation
    Step 1
    Expression for average power dissipated in LCR circuit:
    a. Suppose that an alternating e.m.f e=e0sin⁡ωte=e_0\sin\omega t is applied across an AC circuit containing L, C and R as shown in figure.
    b. In such circuit, there is a phase difference ϕ\phi between applied emf and current. Therefore, the instantaneous current is given by, i=i0sin⁡(ωt±ϕ)i=i_0\sin(\omega t\pm\phi).
    c. Instantaneous power in the circuit is given by,
    P=ei=e0sin⁡ωt×i0sin⁡(ωt±ϕ)P=ei=e_0\sin\omega t\times i_0\sin(\omega t\pm\phi)
    =e0i0sin⁡ωt×sin⁡(ωt±ϕ)=e_0i_0\sin\omega t\times\sin(\omega t\pm\phi)
    =e0i0sin⁡ωt(sin⁡ωtcos⁡ϕ±cos⁡ωtsin⁡ϕ)=e_0i_0\sin\omega t(\sin\omega t\cos\phi\pm\cos\omega t\sin\phi)
    =e0i0(sin⁡2ωtcos⁡ϕ±sin⁡ωtcos⁡ωtsin⁡ϕ)=e_0i_0(\sin^2\omega t\cos\phi\pm\sin\omega t\cos\omega t\sin\phi)
    =e0i0sin⁡2ωtcos⁡ϕ=e_0i_0\sin^2\omega t\cos\phi
    ±12e0i0×2sin⁡ωtcos⁡ωtsin⁡ϕ\pm\dfrac{1}{2}e_0i_0\times2\sin\omega t\cos\omega t\sin\phi
    ∴P=e0i0sin⁡2ωtcos⁡ϕ±e0i02sin⁡2ωtsin⁡ϕ....(1)\therefore P=e_0i_0\sin^2\omega t\cos\phi\pm\dfrac{e_0i_0}{2}\sin2\omega t\sin\phi\qquad....(1)
    This expression shows that P varies with t.
    d. Average power dissipated in the circuit can be obtained by integrating equation (1),
    ∴Pav=Work done in one cycletime for one cycle=∫0TPdtT\therefore P_{av}=\dfrac{\text{Work done in one cycle}}{\text{time for one cycle}}=\dfrac{\displaystyle\int_0^T Pdt}{T}
    ∴Pav=e0i0T∫0T[sin⁡2(ωt)cos⁡(ϕ)±12sin⁡(2ωt)sin⁡(ϕ)]\therefore P_{av}=\dfrac{e_0i_0}{T}\int_0^T\left[\sin^2(\omega t)\cos(\phi)\pm\dfrac{1}{2}\sin(2\omega t)\sin(\phi)\right]
    ∴Pav=e0i0T[∫0Tsin⁡2ωtcos⁡ϕ dt±∫0Tsin⁡2ωtsin⁡ϕ2dt]....(2)\therefore P_{av}=\dfrac{e_0i_0}{T}\left[\int_0^T\sin^2\omega t\cos\phi\,dt\pm\int_0^T\dfrac{\sin2\omega t\sin\phi}{2}dt\right]\qquad....(2)
    ∵∫0Tsin⁡2ωt dt=T2\because\int_0^T\sin^2\omega t\,dt=\dfrac{T}{2}
    ∴∫0Tsin⁡2ωtcos⁡ϕ dt=T2cos⁡ϕ....(3)\therefore\int_0^T\sin^2\omega t\cos\phi\,dt=\dfrac{T}{2}\cos\phi\qquad....(3)
    e. Also, ∫0Tsin⁡2ωt dt=0\displaystyle\int_0^T\sin2\omega t\,dt=0
    ∴∫0Tsin⁡2ωtsin⁡ϕ dt2=0....(4)\therefore\int_0^T\dfrac{\sin2\omega t\sin\phi\,dt}{2}=0\qquad....(4)
    f. Substitute equations (3) and (4) in equation (2), we get,
    Pav=e0i0T×T2cos⁡ϕP_{av}=\dfrac{e_0i_0}{T}\times\dfrac{T}{2}\cos\phi
    =e0i02cos⁡ϕ=\dfrac{e_0i_0}{2}\cos\phi
    =e02⋅i02cos⁡ϕ=\dfrac{e_0}{\sqrt2}\cdot\dfrac{i_0}{\sqrt2}\cos\phi
    =erms×irms×cos⁡ϕ=e_{rms}\times i_{rms}\times\cos\phi
    ∴Pav=erms×irmscos⁡ϕ....(5)\therefore P_{av}=e_{rms}\times i_{rms}\cos\phi\qquad....(5)
    Equation (5) represents required average power or true power dissipated in the AC circuit.
    Step 2
    Power factor:
    a. Average power dissipated in LCR series circuit is given by,
    Pav=erms×irmscos⁡ϕP_{av}=e_{rms}\times i_{rms}\cos\phi
    b. This implies that, the average power dissipated in LCR series circuit depends on the rms values of current and emf as well as the phase difference ϕ\phi between them. The term cos⁡ϕ\cos\phi is called power factor.
    c. It is given by, cos⁡ϕ=Paverms×irms\cos\phi=\dfrac{P_{av}}{e_{rms}\times i_{rms}}
    d. The average power PavP_{av} is also termed as true power and product (erms×Irmse_{rms}\times I_{rms}) is called apparent power.
    ∴Power factor, cos⁡ϕ=True powerApparent power\therefore\text{Power factor, }\cos\phi=\dfrac{\text{True power}}{\text{Apparent power}}
    e. Consider impedance triangle for series LCR circuit as shown in figure,
    ∴cos⁡ϕ=RZ=RR2+(XL−XC)2=resistanceimpedance\therefore\cos\phi=\dfrac{R}{Z}=\dfrac{R}{\sqrt{R^2+(X_L-X_C)^2}}=\dfrac{\text{resistance}}{\text{impedance}}
  5. 5
    Resonant frequency of a parallel resonant circuit
    Assuming the expression for impedance in a parallel resonant circuit, state the conditions for parallel resonance. Define resonant frequency and obtain an expression for it.

    Asked in: March 2019

    Show the derivation
    Step 1
    A parallel resonance circuit is an AC circuit in which a parallel combination of an inductor and capacitor is connected to a source of an alternating e.m.f.
    Step 2
    Let the alternating emf supplied by the source be, e=e0sin⁡ωte=e_0\sin\omega t
    Step 3
    In case of an inductor, the current lags behind the applied emf by a phase angle of π/2\pi/2, then the instantaneous current through L is given by
    iL=e0XLsin⁡(ωt−π/2)i_L=\dfrac{e_0}{X_L}\sin(\omega t-\pi/2)
    Step 4
    Similarly in a capacitor, as current leads the emf by a phase angle of π/2\pi/2, we can write
    iC=e0XCsin⁡(ωt+π/2)i_C=\dfrac{e_0}{X_C}\sin(\omega t+\pi/2)
    Step 5
    Therefore, the total current i in the circuit at this instant is,
    i=iC+iLi=i_C+i_L
    =e0XLsin⁡(ωt−π/2)+e0XCsin⁡(ωt+π/2)=\dfrac{e_0}{X_L}\sin(\omega t-\pi/2)+\dfrac{e_0}{X_C}\sin(\omega t+\pi/2)
    =e0XL(−cos⁡ωt)+e0XCcos⁡ωt=\dfrac{e_0}{X_L}(-\cos\omega t)+\dfrac{e_0}{X_C}\cos\omega t
    =e0(1XC−1XL)cos⁡ωt=e_0\left(\dfrac{1}{X_C}-\dfrac{1}{X_L}\right)\cos\omega t
    i=e0(ωC−1ωL)cos⁡ωti=e_0\left(\omega C-\dfrac{1}{\omega L}\right)\cos\omega t
    Step 6
    Therefore, when (ωC−1ωL)=0\left(\omega C-\dfrac{1}{\omega L}\right)=0, i = minimum,
    ∴ωC=1ωL\therefore\omega C=\dfrac{1}{\omega L}
    ∴ω2=1LC\therefore\omega^2=\dfrac{1}{LC}
    ∴ω=1LC\therefore\omega=\dfrac{1}{\sqrt{LC}}
    ∴2πf=1LC\therefore2\pi f=\dfrac{1}{\sqrt{LC}}
    ∴fr=12πLC\therefore f_r=\dfrac{1}{2\pi\sqrt{LC}}
    Where frf_r is called the resonant frequency.
    Step 7
    Conditions for parallel resonance to occur,
    ωC=1ωL i.e., XC=XL.\omega C=\dfrac{1}{\omega L}\text{ i.e., }X_C=X_L.

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