Hello dear learner. Welcome to HSLC GURU. In this lesson we present the complete English-medium question-and-answer set, key concepts, derivations, formula sheet and additional important questions of the seventh chapter — Alternating Current — of the ASSEB (Assam State School Education Board) Class 12 Physics syllabus. Alternating current (AC) is the form of electricity that runs almost every modern household appliance, the long-distance power-transmission grid, radio receivers, induction heaters and the local 50 Hz mains supply that lights your room. Mastering this chapter is therefore essential for both your board examination and any future engineering / medical entrance test.
Summary
An alternating current is a current whose magnitude changes with time and whose direction reverses periodically. The simplest and most common form is the sinusoidal AC, generated by an AC generator whose armature rotates uniformly in a uniform magnetic field. If the source EMF is $v = v_m\sin\omega t$, the steady-state current through a circuit element generally has the form $i = i_m\sin(\omega t – \phi)$, where $\phi$ is the phase angle by which the current lags (or leads) the voltage.
Because the instantaneous values change continuously, AC quantities are usually quoted by their root-mean-square (rms) values: $i_{\text{rms}} = i_m/\sqrt{2}$ and $V_{\text{rms}} = V_m/\sqrt{2}$. The rms current is also called the effective or virtual current — it is the value of a steady DC current that would dissipate the same heat in the same resistor in the same time.
The chapter studies AC across a pure resistor (current and voltage in phase), a pure inductor (voltage leads current by $\pi/2$ with reactance $X_L = \omega L$), a pure capacitor (current leads voltage by $\pi/2$ with reactance $X_C = 1/\omega C$) and the series LCR circuit (impedance $Z=\sqrt{R^2+(X_L-X_C)^2}$). At resonance ($X_L=X_C$) the impedance becomes purely resistive, the current is maximum, and the resonant angular frequency is $\omega_0 = 1/\sqrt{LC}$. The sharpness of resonance is described by the quality factor $Q = \omega_0L/R$. Average power in any AC circuit is $P_{\text{avg}}=V_{\text{rms}}I_{\text{rms}}\cos\phi$, where $\cos\phi$ is the power factor; in pure inductors or capacitors $\cos\phi=0$ giving the so-called wattless current. Finally, the transformer — a static device based on mutual induction — steps voltage up or down according to the turn ratio $V_s/V_p = N_s/N_p = I_p/I_s$ and is the workhorse of AC power transmission.
Key Formulas
| Quantity | Formula | SI Unit |
|---|---|---|
| Instantaneous EMF | $v = v_m\sin\omega t$ | volt (V) |
| Instantaneous current | $i = i_m\sin(\omega t – \phi)$ | ampere (A) |
| Peak / amplitude | $v_m,\; i_m$ | V, A |
| RMS voltage | $V_{\text{rms}} = V_m/\sqrt{2}$ | volt (V) |
| RMS current | $I_{\text{rms}} = I_m/\sqrt{2}$ | ampere (A) |
| Inductive reactance | $X_L = \omega L = 2\pi f L$ | ohm ($\Omega$) |
| Capacitive reactance | $X_C = \dfrac{1}{\omega C} = \dfrac{1}{2\pi f C}$ | ohm ($\Omega$) |
| Impedance (series LCR) | $Z = \sqrt{R^2+(X_L-X_C)^2}$ | ohm ($\Omega$) |
| Phase angle | $\tan\phi = \dfrac{X_L-X_C}{R}$ | radian |
| Resonant angular frequency | $\omega_0 = \dfrac{1}{\sqrt{LC}}$ | rad/s |
| Resonant frequency | $f_0 = \dfrac{1}{2\pi\sqrt{LC}}$ | hertz (Hz) |
| Quality factor | $Q = \dfrac{\omega_0 L}{R} = \dfrac{1}{\omega_0 RC} = \dfrac{1}{R}\sqrt{\dfrac{L}{C}}$ | dimensionless |
| Average power | $P_{\text{avg}} = V_{\text{rms}}I_{\text{rms}}\cos\phi$ | watt (W) |
| Power factor | $\cos\phi = \dfrac{R}{Z}$ | dimensionless |
| Transformer ratio | $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}$ | dimensionless |
| LC oscillation energy | $U = \tfrac{1}{2}LI^2 + \tfrac{q^2}{2C}$ | joule (J) |
7.1 AC Voltage Applied to a Resistor
If an AC source $v = v_m\sin\omega t$ is applied across a pure resistor $R$, by Kirchhoff’s loop rule the instantaneous current is
$$i = \dfrac{v}{R} = \dfrac{v_m}{R}\sin\omega t = i_m\sin\omega t$$
Hence the current and voltage are in phase; both reach maxima, zeros and minima together. Average power dissipated:
$$P_{\text{avg}} = \dfrac{1}{2}v_m i_m = V_{\text{rms}}I_{\text{rms}} = I_{\text{rms}}^2 R$$
7.2 RMS or Effective Value
For a sinusoidal current $i = i_m\sin\omega t$ the mean value over a full cycle is zero, so we instead average $i^2$ and take the square root:
$$I_{\text{rms}} = \sqrt{\dfrac{1}{T}\int_0^T i^2\,dt} = \dfrac{i_m}{\sqrt{2}} \approx 0.707\,i_m$$
Similarly $V_{\text{rms}} = V_m/\sqrt{2}$. The 220 V mains marked on Indian appliances is an rms value; the actual peak is $220\sqrt{2}\approx 311$ V.
7.3 AC Voltage Applied to an Inductor
For a pure inductor $L$, applying $v=v_m\sin\omega t$ and the inductor equation $v = L\,di/dt$ gives
$$i = \dfrac{v_m}{\omega L}\sin\!\left(\omega t – \dfrac{\pi}{2}\right) = i_m\sin\!\left(\omega t-\dfrac{\pi}{2}\right)$$
where the inductive reactance $X_L = \omega L$ replaces resistance. The current lags the voltage by 90°. Because the average of $\sin\omega t\cos\omega t$ over a full cycle is zero, an ideal inductor consumes no average power.
7.4 AC Voltage Applied to a Capacitor
For a capacitor $C$ with $q = Cv$, the current is $i = dq/dt = C\,dv/dt$. Substituting $v=v_m\sin\omega t$:
$$i = \omega C\,v_m\cos\omega t = \dfrac{v_m}{X_C}\sin\!\left(\omega t + \dfrac{\pi}{2}\right)$$
where capacitive reactance $X_C = 1/(\omega C)$. The current leads the voltage by 90°. Again the average power is zero. Note that DC ($\omega\to 0$) gives $X_C\to\infty$: the capacitor blocks DC.
7.5 AC Voltage Applied to a Series LCR Circuit
When $R$, $L$ and $C$ are connected in series across $v=v_m\sin\omega t$, the same current $i = i_m\sin(\omega t-\phi)$ flows through all three. Using a phasor diagram with the current phasor along the +x axis: $V_R$ is along $+x$, $V_L$ is along $+y$, $V_C$ is along $-y$. The resultant source voltage has magnitude
$$V_m = \sqrt{V_R^2 + (V_L – V_C)^2} = i_m\sqrt{R^2+(X_L-X_C)^2}$$
Hence the impedance $Z = \sqrt{R^2+(X_L-X_C)^2}$ and the phase
$$\tan\phi = \dfrac{X_L-X_C}{R}$$
If $X_L>X_C$ the circuit is inductive (current lags); if $X_L
7.6 Resonance
The series LCR circuit resonates when $X_L = X_C$, i.e.
$$\omega_0 L = \dfrac{1}{\omega_0 C}\quad\Rightarrow\quad \omega_0 = \dfrac{1}{\sqrt{LC}},\qquad f_0 = \dfrac{1}{2\pi\sqrt{LC}}$$
At resonance $Z = R$ (minimum), current $I_m = V_m/R$ (maximum), and $\phi = 0$ (V and I in phase). Resonance is the principle behind radio tuning: only the station whose carrier frequency equals $f_0$ of the LC tank gives a large response.
Sharpness and Q-factor
The bandwidth $2\Delta\omega$ is the width between the half-power points where current falls to $I_m/\sqrt{2}$. The dimensionless quality factor measures how sharp the resonance peak is:
$$Q = \dfrac{\omega_0}{2\Delta\omega} = \dfrac{\omega_0 L}{R} = \dfrac{1}{\omega_0 R C} = \dfrac{1}{R}\sqrt{\dfrac{L}{C}}$$
Higher $Q$ means a sharper, more selective resonance — desirable for a radio receiver.
7.7 Power in an AC Circuit — Power Factor
Instantaneous power $p = vi$. Averaging over one cycle:
$$P_{\text{avg}} = \dfrac{1}{2}v_m i_m\cos\phi = V_{\text{rms}}I_{\text{rms}}\cos\phi$$
The factor $\cos\phi = R/Z$ is called the power factor. Special cases:
- Pure $R$: $\phi=0$, $\cos\phi=1$ — maximum power transfer.
- Pure $L$ or pure $C$: $\phi = \pm\pi/2$, $\cos\phi=0$ — no average power. The current is then called wattless current or idle current.
- Series LCR at resonance: $X_L = X_C \Rightarrow \cos\phi=1$.
7.8 LC Oscillations
A charged capacitor connected across an inductor produces electrical oscillations analogous to a mechanical spring–mass system. With negligible resistance, charge varies as $q = q_0\cos\omega_0 t$ where $\omega_0 = 1/\sqrt{LC}$. Energy oscillates between the electric field of the capacitor and the magnetic field of the inductor while remaining constant overall:
$$\dfrac{q^2}{2C} + \dfrac{1}{2}LI^2 = \dfrac{q_0^2}{2C} = \text{const}$$
7.9 Transformer
A transformer is a static device that transfers electrical energy between two coupled coils through mutual induction. A laminated soft-iron core links a primary coil of $N_p$ turns to a secondary coil of $N_s$ turns. For an ideal lossless transformer driven by AC,
$$\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}$$
If $N_s>N_p$ it is a step-up transformer (voltage rises, current falls). If $N_s
Textbook Exercise — Question Answer
Q1. A 100 Ω resistor is connected to a 220 V, 50 Hz AC supply.
(a) Find the rms current in the circuit.
(b) What is the net power consumed over a full cycle?
Answer: Given $R = 100\,\Omega$, $V_{\text{rms}} = 220$ V.
$$\text{(a)}\; I_{\text{rms}} = \dfrac{V_{\text{rms}}}{R} = \dfrac{220}{100} = 2.2\;\text{A}$$
$$\text{(b)}\; P = V_{\text{rms}}I_{\text{rms}} = 220\times 2.2 = 484\;\text{W}$$
Q2. (a) The peak voltage of an AC supply is 300 V. What is the rms voltage? (b) The rms value of current in an AC circuit is 10 A. What is the peak current?
Answer:
$$\text{(a)}\; V_{\text{rms}} = \dfrac{V_m}{\sqrt 2} = \dfrac{300}{1.414} \approx 212.1\;\text{V}$$
$$\text{(b)}\; I_m = \sqrt 2\, I_{\text{rms}} = 1.414\times 10 \approx 14.14\;\text{A}$$
Q3. A 44 mH inductor is connected to a 220 V, 50 Hz AC supply. Determine the rms current.
Answer: $X_L = \omega L = 2\pi(50)(44\times 10^{-3}) = 13.82\,\Omega$.
$$I_{\text{rms}} = \dfrac{V_{\text{rms}}}{X_L} = \dfrac{220}{13.82} \approx 15.92\;\text{A}$$
Q4. A 60 µF capacitor is connected to a 110 V, 60 Hz AC supply. Find the rms current.
Answer: $X_C = 1/(\omega C) = 1/[2\pi(60)(60\times 10^{-6})] = 44.21\,\Omega$.
$$I_{\text{rms}} = \dfrac{110}{44.21}\approx 2.49\;\text{A}$$
Q5. In Q3 and Q4, what is the net power absorbed by each circuit over a complete cycle? Explain.
Answer: For an ideal inductor and an ideal capacitor the phase difference between voltage and current is $\pi/2$. Therefore $\cos\phi = 0$ and
$$P_{\text{avg}} = V_{\text{rms}}I_{\text{rms}}\cos(\pi/2) = 0$$
The energy supplied during one quarter cycle is stored in the magnetic field (inductor) or electric field (capacitor) and returned to the source during the next quarter cycle. No net energy is dissipated.
Q6. Obtain the resonant frequency $\omega_0$ of a series LCR circuit with $L = 2.0$ H, $C = 32\,\mu$F and $R = 10\,\Omega$. What is the Q-factor?
Answer:
$$\omega_0 = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{2.0\times 32\times 10^{-6}}} = 125\;\text{rad/s}$$
$$Q = \dfrac{\omega_0 L}{R} = \dfrac{125\times 2.0}{10} = 25$$
Q7. A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?
Answer:
$$\omega_0 = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{27\times 10^{-3}\times 30\times 10^{-6}}} \approx 1.11\times 10^{3}\;\text{rad/s}$$
Q8. Suppose the initial charge on the capacitor in Q7 is 6 mC. What is the total energy initially stored?
Answer:
$$U = \dfrac{q_0^2}{2C} = \dfrac{(6\times 10^{-3})^2}{2\times 30\times 10^{-6}} = \dfrac{36\times 10^{-6}}{60\times 10^{-6}} = 0.6\;\text{J}$$
If $R=0$ this energy is conserved as oscillations between the electric and magnetic fields.
Q9. A series LCR circuit with $R = 20\,\Omega$, $L = 1.5$ H and $C = 35\,\mu$F is connected to a variable-frequency 200 V AC source. When the frequency equals the natural frequency of the circuit, what is the average power absorbed?
Answer: At resonance $X_L = X_C$, so $Z = R$, $\cos\phi=1$.
$$P = \dfrac{V_{\text{rms}}^2}{R} = \dfrac{200^2}{20} = 2000\;\text{W} = 2\;\text{kW}$$
Q10. A radio can tune frequencies in the MW band 800 – 1200 kHz. The effective inductance in its receiver circuit is 200 μH. What range of capacitance is required?
Answer: Using $C = 1/(\omega^2 L) = 1/(4\pi^2 f^2 L)$.
$$C_{\max}=\dfrac{1}{4\pi^2(800\times 10^3)^2(200\times 10^{-6})} \approx 1.98\times 10^{-10}\;\text{F} \approx 198\;\text{pF}$$
$$C_{\min}=\dfrac{1}{4\pi^2(1200\times 10^3)^2(200\times 10^{-6})} \approx 88\;\text{pF}$$
So $C$ must vary roughly from 88 pF to 198 pF.
Q11. Figure shows a series LCR circuit connected to a variable-frequency 230 V source. $L = 5.0$ H, $C = 80\,\mu$F, $R = 40\,\Omega$.
(a) Determine the source frequency that drives the circuit in resonance.
(b) Obtain the impedance and the amplitude of current at resonance.
(c) Determine the rms potential drops across the three elements at resonance. Show that the potential drop across the LC combination is zero at the resonating frequency.
Answer: Given $V_{\text{rms}}=230$ V so $V_m = 230\sqrt 2$ V.
$$\text{(a)}\;\omega_0=\dfrac{1}{\sqrt{LC}}=\dfrac{1}{\sqrt{5.0\times 80\times 10^{-6}}}=50\;\text{rad/s},\; f_0=\dfrac{\omega_0}{2\pi}\approx 7.96\;\text{Hz}$$
$$\text{(b)}\;Z=R=40\,\Omega,\; I_m=\dfrac{V_m}{R}=\dfrac{230\sqrt 2}{40}\approx 8.13\;\text{A}$$
(c) At resonance $I_{\text{rms}} = 230/40 = 5.75$ A.
$$V_R = I_{\text{rms}}R = 5.75\times 40 = 230\;\text{V}$$
$$V_L = I_{\text{rms}}\omega_0 L = 5.75\times 50\times 5 = 1437.5\;\text{V}$$
$$V_C = \dfrac{I_{\text{rms}}}{\omega_0 C} = \dfrac{5.75}{50\times 80\times 10^{-6}} = 1437.5\;\text{V}$$
Since $V_L$ and $V_C$ are 180° out of phase, their phasor sum $V_L – V_C = 0$. Hence the rms drop across the LC combination is zero — even though each element individually has a very large 1437.5 V drop. This is the famous voltage-magnification effect at resonance.
Q12. An LC circuit contains a 20 mH inductor and a 50 μF capacitor with an initial charge of 10 mC. The resistance is negligible. The circuit is closed at $t=0$.
(a) What is the total energy stored initially? Is it conserved?
(b) What is the natural frequency?
(c) At what time is the energy stored (i) entirely electrical, (ii) entirely magnetic?
(d) At what times is the total energy shared equally between the inductor and the capacitor?
Answer:
$$\text{(a)}\;U = \dfrac{q_0^2}{2C} = \dfrac{(10\times 10^{-3})^2}{2\times 50\times 10^{-6}} = 1.0\;\text{J}$$
Yes, $U$ is conserved because the resistance is zero — energy oscillates between $C$ and $L$.
$$\text{(b)}\;\omega_0 = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{20\times 10^{-3}\times 50\times 10^{-6}}} = 10^{3}\;\text{rad/s},\; f_0 \approx 159\;\text{Hz}$$
Period $T = 1/f_0 \approx 6.28$ ms.
(c) (i) Entirely electrical at $t = 0,\; T/2,\; T,\;3T/2,\dots$ — i.e. multiples of $T/2$ when $\cos\omega_0 t = \pm 1$. (ii) Entirely magnetic at odd multiples of $T/4$ — $T/4,\; 3T/4,\dots$ — when the capacitor is momentarily uncharged.
(d) Energy is equally shared when $q^2/(2C) = U/2$, i.e. $\cos^2\omega_0 t = 1/2$, giving $\omega_0 t = (2n+1)\pi/4$. So $t = T/8,\; 3T/8,\; 5T/8,\dots$.
Q13. A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V, 50 Hz AC supply.
(a) What is the maximum current in the coil?
(b) What is the time lag between the voltage maximum and the current maximum?
Answer: $X_L = 2\pi(50)(0.5) = 157.1\,\Omega$, $Z = \sqrt{R^2+X_L^2} = \sqrt{100^2+157.1^2}=186.2\,\Omega$.
$$I_m = \dfrac{V_m}{Z} = \dfrac{240\sqrt 2}{186.2} \approx 1.82\;\text{A}$$
$$\tan\phi = \dfrac{X_L}{R} = 1.571 \Rightarrow \phi = 57.5° = 1.004\;\text{rad}$$
$$\Delta t = \dfrac{\phi}{\omega} = \dfrac{1.004}{2\pi(50)} \approx 3.2\times 10^{-3}\;\text{s} = 3.2\;\text{ms}$$
So the current peaks 3.2 ms after the voltage peaks.
Q14. Repeat Q13 with the same coil now driven by a 240 V, 10 kHz supply. Comment on the result.
Answer: $X_L = 2\pi(10^4)(0.5) = 3.14\times 10^4\,\Omega$, much greater than $R$. So $Z \approx X_L$.
$$I_m = \dfrac{240\sqrt 2}{3.14\times 10^4} \approx 1.08\times 10^{-2}\;\text{A}$$
The phase $\phi\to\pi/2$ and the time lag $\Delta t = (\pi/2)/(2\pi\times 10^4) = 25\,\mu$s. At high frequencies an inductor effectively chokes the current — this is the principle of the choke coil and of low-pass filters.
Q15. A 100 µF capacitor in series with a 40 Ω resistor is connected to a 110 V, 60 Hz supply.
(a) What is the maximum current in the circuit?
(b) What is the time lag between the current maximum and the voltage maximum?
Answer: $X_C = 1/[2\pi(60)(100\times 10^{-6})] = 26.53\,\Omega$, $Z = \sqrt{40^2+26.53^2} = 48.0\,\Omega$.
$$I_m = \dfrac{110\sqrt 2}{48.0} \approx 3.24\;\text{A}$$
$$\tan\phi = \dfrac{X_C}{R} = 0.663 \Rightarrow \phi = 33.5° = 0.585\;\text{rad}$$
$$\Delta t = \dfrac{\phi}{\omega} = \dfrac{0.585}{2\pi(60)} \approx 1.55\;\text{ms}$$
The current here leads the voltage, so the current peak occurs 1.55 ms before the voltage peak.
Q16. Repeat Q15 with the same circuit driven by a 12 kHz supply. Comment.
Answer: $X_C = 1/[2\pi(12000)(100\times 10^{-6})] = 0.133\,\Omega$, far smaller than $R$. Hence $Z\approx R$, $I_m \approx 110\sqrt 2/40 \approx 3.89$ A and $\phi\approx 0$. At very high frequency the capacitor behaves almost like a short circuit (zero reactance) — this is why coupling capacitors pass AC signals freely while blocking DC.
Q17. Keeping the source frequency equal to the resonant frequency of the series LCR circuit (Q11), if the three elements $L$, $C$ and $R$ are arranged in parallel, show that the total current in the parallel combination is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Q11 for this frequency.
Answer: For the parallel LCR circuit, the admittance is $Y = 1/R + j(\omega C – 1/\omega L)$. At $\omega_0 = 1/\sqrt{LC}$ the imaginary part vanishes, so $Y = 1/R$ — the smallest possible (most resistive) admittance. Therefore the source current $I = V/Z = V/R$ is minimum at resonance — opposite to the series case where it is maximum. Branch currents: $I_R = 230/40 = 5.75$ A, $I_L = V/X_L = 230/(50\times 5) = 0.92$ A, $I_C = V\omega_0 C = 230\times 50\times 80\times 10^{-6} = 0.92$ A. The two reactive branch currents are equal in magnitude but 180° out of phase, cancelling in the source line. This is the basis of the parallel “tank” circuit used in oscillators.
Q18. A circuit containing a 80 mH inductor and a 250 μF capacitor in series is connected to a 240 V, 100 rad s⁻¹ supply. The resistance of the circuit is negligible.
(a) Obtain the rms value of current.
(b) What is the total average power absorbed by the circuit?
Answer: $X_L = (100)(0.08) = 8\,\Omega$, $X_C = 1/[(100)(250\times 10^{-6})] = 40\,\Omega$.
$$Z = |X_L – X_C| = 32\;\Omega$$
$$I_{\text{rms}} = \dfrac{240}{32} = 7.5\;\text{A}$$
(b) Since $R=0$, $\cos\phi=0$. Hence $P_{\text{avg}}=0$ — the circuit is purely reactive and the energy oscillates without dissipation.
Q19. Suppose the circuit in Q18 has $R = 15\,\Omega$. Obtain the average power transferred to each element of the circuit and the total power absorbed.
Answer: $Z = \sqrt{15^2 + 32^2} = \sqrt{225+1024} = 35.34\,\Omega$.
$$I_{\text{rms}} = \dfrac{240}{35.34} \approx 6.79\;\text{A}$$
Power dissipated only in the resistor: $P_R = I_{\text{rms}}^2 R = (6.79)^2(15) = 691.7\;\text{W}$. Inductor and capacitor each have $P_L = P_C = 0$. Total absorbed = 691.7 W (all in $R$).
Q20. A series LCR circuit with $L = 0.12$ H, $C = 480$ nF, $R = 23\,\Omega$ is connected to a 230 V variable-frequency supply.
(a) What is the source frequency for which the current amplitude is maximum? Obtain this maximum value.
(b) What frequency for maximum absorbed power and the value of this maximum power?
(c) For what frequencies is the power transferred to the circuit half the resonant power? What is the current amplitude at these frequencies?
(d) What is the Q-factor of the given circuit?
Answer:
$$\text{(a)}\;\omega_0 = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{0.12\times 480\times 10^{-9}}} = 4.166\times 10^{3}\;\text{rad/s}$$
$$f_0 \approx 663\;\text{Hz},\; I_m^{\text{(max)}} = \dfrac{V_m}{R} = \dfrac{230\sqrt 2}{23} \approx 14.14\;\text{A}$$
(b) Maximum absorbed power also occurs at $f_0$: $P_{\max} = V_{\text{rms}}^2/R = 230^2/23 = 2300\;\text{W}$.
(c) Half-power frequencies satisfy $\omega = \omega_0\pm \Delta\omega$ where $\Delta\omega = R/(2L) = 23/(0.24) = 95.83$ rad/s. So $\omega_{1,2} = 4166\pm 96\approx 4070,\;4262$ rad/s, i.e. about 648 Hz and 678 Hz. At these frequencies $I = I_m^{\max}/\sqrt 2 \approx 10.0$ A.
$$\text{(d)}\;Q = \dfrac{\omega_0 L}{R} = \dfrac{4166\times 0.12}{23} \approx 21.7$$
Q21. Obtain the resonant frequency and Q-factor of a series LCR circuit with $L = 3.0$ H, $C = 27\,\mu$F, $R = 7.4\,\Omega$. It is desired to improve the sharpness of the resonance of the circuit by reducing its “full width at half maximum” by a factor of 2. Suggest a suitable way.
Answer:
$$\omega_0 = \dfrac{1}{\sqrt{LC}}=\dfrac{1}{\sqrt{3.0\times 27\times 10^{-6}}}\approx 111\;\text{rad/s}$$
$$Q = \dfrac{\omega_0 L}{R} = \dfrac{111\times 3.0}{7.4} = 45$$
Bandwidth $2\Delta\omega = R/L$. To halve the bandwidth (double $Q$ to 90), reduce $R$ to half: $R’ = 3.7\,\Omega$.
Q22. Answer the following:
(a) In any AC circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements? Is the same true for rms voltages?
(b) A capacitor is used in the primary circuit of an induction coil.
(c) An applied voltage signal consists of a superposition of a DC voltage and an AC voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the DC signal will appear across $C$ and the AC signal across $L$.
(d) A choke coil in series with a lamp is connected to a DC line. The lamp glows brightly. Insertion of an iron core in the choke causes no change in the lamp’s brightness. Predict the corresponding observations if the connection is to an AC line.
(e) Why is choke coil needed in the use of fluorescent tubes with AC mains? Why cannot we use an ordinary resistor?
Answer:
(a) Yes for instantaneous, by Kirchhoff’s voltage law applied to instants. No for rms, because the voltage phasors across $R$, $L$ and $C$ are not in phase, so their rms magnitudes add as vectors, not algebraically.
(b) When the primary current is suddenly interrupted in an induction coil, a high inductive EMF is set up. A capacitor in parallel with the breaker absorbs this energy, suppresses arcing, and recharges to drive a sharper interruption — making the induced EMF in the secondary larger and the operation safer.
(c) For the DC component, $X_C = \infty$ and $X_L = 0$; the capacitor blocks DC, so the entire DC voltage appears across $C$. For the high-frequency AC component, $X_L \gg X_C$; the AC therefore drops mostly across the inductor.
(d) On DC, the choke offers only its small ohmic resistance; the lamp glows brightly. Inserting an iron core does not change anything because steady current produces no induced EMF. On AC, however, the choke’s reactance $X_L = \omega L$ is significant; the lamp glows dimmer. Inserting the iron core greatly increases $L$, increases $X_L$, reduces current, and the lamp dims further. This is exactly the principle used in fluorescent-tube ballasts.
(e) A fluorescent tube needs a large voltage to start the discharge but has very low operating resistance afterwards. A choke coil drops the necessary voltage by inductive reactance, dissipating negligible energy because of $\cos\phi \to 0$. An ordinary resistor would dissipate the same drop as $I^2R$ heat — wasteful and causing the resistor to run hot.
Q23. A power transmission line feeds input power at 2300 V to a step-down transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary in order to get output power at 230 V?
Answer: Using $V_s/V_p = N_s/N_p$:
$$N_s = N_p\dfrac{V_s}{V_p} = 4000\times \dfrac{230}{2300} = 400\;\text{turns}$$
Q24. At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is $100\,\text{m}^3\,\text{s}^{-1}$. If the turbine generator efficiency is 60%, estimate the electric power available from the plant. (Take $g = 9.8\;\text{m/s}^2$.)
Answer: Mass flow rate $\dot m = \rho Q = 1000\times 100 = 10^5$ kg/s.
$$P_{\text{in}} = \dot m\,gh = 10^5\times 9.8\times 300 = 2.94\times 10^{8}\;\text{W}$$
$$P_{\text{electrical}} = 0.60\times 2.94\times 10^{8} \approx 1.76\times 10^{8}\;\text{W} \approx 176\;\text{MW}$$
Q25. A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two-wire line carrying power is $0.5\,\Omega$ per km. The town gets power from the line through a 4000–220 V step-down transformer at a substation in the town.
(a) Estimate the line power loss in the form of heat.
(b) How much power must the plant supply, assuming there is negligible power loss due to leakage?
(c) Characterise the step-up transformer at the plant.
Answer: The line carries 800 kW at the high-side voltage 4000 V (substation primary).
$$I = \dfrac{P}{V} = \dfrac{8\times 10^5}{4000} = 200\;\text{A}$$
$$R_{\text{line}} = 0.5\times 2\times 15 = 15\;\Omega$$
$$\text{(a)}\;P_{\text{loss}} = I^2 R = 200^2\times 15 = 6\times 10^{5}\;\text{W} = 600\;\text{kW}$$
$$\text{(b)}\;P_{\text{plant}} = 800 + 600 = 1400\;\text{kW}$$
(c) Voltage drop along the line $= IR = 200\times 15 = 3000$ V. So the plant transformer secondary must be at $4000 + 3000 = 7000$ V. Plant transformer rating: 440 V – 7000 V step-up.
Q26. Do the same exercise as in Q25 but with the replacement of the 4000–220 V step-down transformer by an 40000–220 V step-down transformer. (Maintain the same power demand of 800 kW.) Comment on the result.
Answer: Now line voltage is 40 000 V.
$$I = \dfrac{8\times 10^{5}}{40\,000} = 20\;\text{A}$$
$$P_{\text{loss}} = 20^2\times 15 = 6000\;\text{W} = 6\;\text{kW}$$
So the loss falls from 600 kW to just 6 kW. Plant must supply $800+6=806$ kW. Plant transformer rating: $440$ V – $40\,300$ V step-up. Conclusion: By stepping up to a higher transmission voltage we slash the resistive line loss by a factor of $(V_1/V_2)^2 = 100$. This is precisely why power is transmitted at hundreds of kilovolts.
Additional Important Questions
A. Very Short Answer (1 mark)
Q1. Define rms value of an alternating current.
Answer: It is that steady DC current which, when passed through a given resistor for a given time, produces the same heat as the AC current does in the same resistor in the same time. For sinusoidal AC, $I_{\text{rms}} = I_m/\sqrt 2$.
Q2. Why is the average value of AC over a complete cycle zero?
Answer: Because for a sinusoid the positive half-cycle is exactly cancelled by the equal and opposite negative half-cycle, so $\int_0^T \sin\omega t\,dt = 0$.
Q3. What is the SI unit of inductive reactance?
Answer: Ohm ($\Omega$), the same as resistance, since $X_L$ relates voltage and current.
Q4. What is meant by wattless current?
Answer: An AC current whose phase difference with the voltage is $\pm 90°$ so that $\cos\phi = 0$ and the average power $P = V_{\text{rms}}I_{\text{rms}}\cos\phi = 0$. Found in pure inductors and pure capacitors.
Q5. State the principle of a transformer.
Answer: Mutual induction — a changing current in the primary coil produces a changing magnetic flux that induces an EMF in the secondary coil.
Q6. What is the phase difference between voltage and current in an AC circuit containing a pure capacitor?
Answer: Current leads voltage by $\pi/2$ (90°).
Q7. Why is the core of a transformer laminated?
Answer: Lamination breaks up the cross-section of the core into thin insulated sheets, restricting the area of any closed loop available for eddy currents. This drastically reduces eddy-current losses and the heating of the core.
Q8. The peak value of EMF in an AC circuit is 311 V. What is the rms value?
Answer: $V_{\text{rms}} = 311/\sqrt 2 \approx 220$ V (the standard Indian mains).
Q9. What is the impedance of a series LCR circuit at resonance?
Answer: Just the resistance $R$, because $X_L = X_C$ cancel.
Q10. Define quality factor of a series LCR circuit.
Answer: $Q = \omega_0 L/R$. It is the ratio of resonant frequency to the bandwidth and measures the sharpness of the resonance peak.
B. Short Answer (2-3 marks)
Q11. Show that the average power dissipated in a pure inductor over one full cycle is zero.
Answer: Let $v=v_m\sin\omega t$, then $i = i_m\sin(\omega t-\pi/2) = -i_m\cos\omega t$. Instantaneous power
$$p = vi = -v_m i_m\sin\omega t\cos\omega t = -\dfrac{v_m i_m}{2}\sin 2\omega t$$
The integral of $\sin 2\omega t$ over one period is zero, hence $\langle p\rangle = 0$. The inductor alternately stores and releases energy with no net dissipation.
Q12. Define inductive reactance and capacitive reactance and state how each depends on frequency.
Answer: Inductive reactance $X_L = \omega L$ — directly proportional to frequency. Capacitive reactance $X_C = 1/(\omega C)$ — inversely proportional to frequency. Hence inductors block high frequencies (used as chokes) and capacitors block low frequencies / DC (used in coupling).
Q13. Distinguish between resistance, reactance and impedance.
| Resistance ($R$) | Reactance ($X$) | Impedance ($Z$) |
|---|---|---|
| Opposes flow of current in any element (DC or AC). | Opposition due to inductor or capacitor in AC only. | Net opposition in an AC circuit containing $R$ and $X$. |
| Independent of frequency. | Depends on frequency: $X_L$ rises, $X_C$ falls. | Combines both: $Z = \sqrt{R^2+(X_L-X_C)^2}$. |
| Causes heat dissipation. | Causes phase shift but no power loss. | Determines current and average power. |
Q14. Derive an expression for the impedance of a series LCR circuit using the phasor method.
Answer: Take the current phasor $\vec I$ along the +x axis. Then $\vec V_R = IR\hat x$ (in phase), $\vec V_L = IX_L\hat y$ (90° ahead), $\vec V_C = -IX_C\hat y$ (90° behind). The source-voltage phasor is the resultant:
$$\vec V = IR\hat x + I(X_L-X_C)\hat y$$
Magnitude $V = I\sqrt{R^2+(X_L-X_C)^2}$. Hence $Z=V/I=\sqrt{R^2+(X_L-X_C)^2}$ and $\tan\phi=(X_L-X_C)/R$.
Q15. Show that the resonant frequency of a series LCR circuit is $f_0=1/(2\pi\sqrt{LC})$ and explain why current is maximum at resonance.
Answer: Resonance ⇔ $X_L = X_C$ ⇔ $\omega L = 1/(\omega C)$ ⇔ $\omega^2 = 1/LC$ ⇔ $\omega_0 = 1/\sqrt{LC}$ and $f_0 = \omega_0/2\pi$. At this frequency $Z = R$, the smallest possible impedance, so $I_m = V_m/R$ is maximum.
Q16. Explain why power factor should be high in a power-transmission line.
Answer: The current required to deliver a given useful power $P$ is $I = P/(V\cos\phi)$. A low $\cos\phi$ means a much larger current for the same useful power, leading to higher $I^2R$ losses in the transmission cables and overheating of generators. Industries are therefore required to maintain a power factor close to 1, often by adding capacitor banks (“power-factor correction”).
Q17. Why is AC preferred over DC for long-distance power transmission?
Answer: AC voltage can easily be stepped up to very high values using transformers. At high voltage the transmission current — and hence $I^2R$ losses — drop drastically. DC voltages cannot be transformed using simple static devices, making efficient long-haul transmission expensive (HVDC needs costly converter stations).
Q18. List the various energy losses in a transformer and how each is minimised.
- Copper loss ($I^2R$ in windings) — minimised by using thick low-resistance copper wires.
- Eddy-current loss in the core — minimised by laminating the core.
- Hysteresis loss — minimised by using soft iron (small hysteresis loop) such as silicon steel or permalloy.
- Flux leakage — minimised by winding primary and secondary on the same core, often interleaved.
- Humming / mechanical vibration — minimised by tightly clamping core laminations.
Q19. A series LCR circuit has $R = 5\,\Omega$, $L = 25$ mH, $C = 100\,\mu$F. It is driven at 50 Hz, 200 V rms. Find $Z$, $I_{\text{rms}}$, $\cos\phi$ and the average power.
Answer: $X_L = 2\pi(50)(0.025) = 7.85\,\Omega$, $X_C = 1/[2\pi(50)(10^{-4})] = 31.83\,\Omega$.
$$Z = \sqrt{5^2 + (7.85-31.83)^2} = \sqrt{25 + 575.0} = 24.5\;\Omega$$
$$I_{\text{rms}} = 200/24.5 = 8.16\;\text{A},\; \cos\phi = R/Z = 5/24.5 = 0.204$$
$$P = V_{\text{rms}}I_{\text{rms}}\cos\phi = 200\times 8.16\times 0.204 \approx 333\;\text{W}$$
Q20. A choke coil and a 60 W lamp are connected in series across a 220 V, 50 Hz mains. The lamp glows at full brilliance. Find the inductance of the choke. (Resistance of lamp filament 100 Ω.)
Answer: Lamp current $I = \sqrt{P/R} = \sqrt{60/100}=0.775$ A. Required impedance $Z = V/I = 220/0.775 = 284\,\Omega$. Coil reactance $X_L = \sqrt{Z^2-R^2} = \sqrt{284^2-100^2}=265.8\,\Omega$. Hence $L = X_L/\omega = 265.8/(100\pi) = 0.846$ H.
C. Long Answer (5 marks)
Q21. Derive an expression for the average power dissipated in a series LCR circuit. Hence define power factor and discuss the cases of pure resistor, pure inductor, pure capacitor, series LCR at resonance and wattless current.
Answer: Let $v = v_m\sin\omega t$ and $i = i_m\sin(\omega t – \phi)$. Instantaneous power
$$p = vi = v_m i_m\sin\omega t\sin(\omega t-\phi)$$
$$= \dfrac{v_m i_m}{2}[\cos\phi – \cos(2\omega t – \phi)]$$
Averaging over one cycle, the second term vanishes, so
$$P_{\text{avg}} = \dfrac{v_m i_m}{2}\cos\phi = V_{\text{rms}}I_{\text{rms}}\cos\phi$$
The factor $\cos\phi = R/Z$ is the power factor. Cases:
- Pure $R$: $\phi = 0$, $\cos\phi = 1$, $P = V_{\text{rms}}I_{\text{rms}}$.
- Pure $L$: $\phi = +\pi/2$, $\cos\phi = 0$, $P = 0$ (wattless).
- Pure $C$: $\phi = -\pi/2$, $\cos\phi = 0$, $P = 0$ (wattless).
- Series LCR at resonance: $X_L = X_C$, $\phi = 0$, $\cos\phi = 1$, $P_{\max} = V_{\text{rms}}^2/R$.
- The component of current $I_{\text{rms}}\sin\phi$ that is 90° out of phase with the voltage is called wattless current.
Q22. Describe the construction and working of a transformer. Explain step-up and step-down transformers, derive the turns ratio and discuss energy losses with the methods used to reduce them.
Answer: A transformer has two coils — primary ($N_p$ turns) and secondary ($N_s$ turns) — wound on a common laminated soft-iron core. AC is applied to the primary; the changing flux $\Phi$ links both coils. By Faraday’s law the EMF induced in each coil equals $-N\,d\Phi/dt$, so
$$\dfrac{V_s}{V_p}=\dfrac{N_s}{N_p}$$
For an ideal lossless transformer, $V_p I_p = V_s I_s$, hence
$$\dfrac{I_p}{I_s}=\dfrac{N_s}{N_p}$$
Step-up: $N_s>N_p$, so $V_s>V_p$ and $I_s
Q23. Discuss LC oscillations qualitatively and quantitatively. Compare with the simple harmonic motion of a spring–mass system.
Answer: Consider a charged capacitor $q_0$ connected at $t=0$ to a pure inductor. Kirchhoff’s voltage law gives
$$L\dfrac{d^2 q}{dt^2} + \dfrac{q}{C} = 0\quad\Rightarrow\quad \dfrac{d^2 q}{dt^2}+\omega_0^2 q = 0,\;\omega_0^2 = \dfrac{1}{LC}$$
Solution: $q = q_0\cos\omega_0 t$, $i = -q_0\omega_0\sin\omega_0 t$. The capacitor energy and inductor energy interconvert sinusoidally with total energy conserved. The analogy with SHM is exact:
| Mechanical | Electrical (LC) |
|---|---|
| Mass $m$ | Inductance $L$ |
| Spring constant $k$ | $1/C$ |
| Displacement $x$ | Charge $q$ |
| Velocity $\dot x$ | Current $i$ |
| $\tfrac{1}{2}kx^2$ | $q^2/2C$ |
| $\tfrac{1}{2}m\dot x^2$ | $\tfrac{1}{2}Li^2$ |
| $\omega = \sqrt{k/m}$ | $\omega_0 = 1/\sqrt{LC}$ |
Real LC circuits include resistance $R$, which damps the oscillations exponentially.
MCQ — Multiple Choice Questions
1. The rms value of an AC of peak value 4 A is —
(a) 4 A (b) $4\sqrt 2$ A (c) $2\sqrt 2$ A (d) 2 A
Answer: (c) $2\sqrt 2$ A.
2. In a pure inductor, current —
(a) leads the voltage by 90° (b) lags the voltage by 90° (c) is in phase (d) lags by 45°
Answer: (b).
3. Capacitive reactance is inversely proportional to —
(a) frequency only (b) capacitance only (c) frequency and capacitance (d) impedance
Answer: (c).
4. At resonance the impedance of a series LCR circuit equals —
(a) $X_L$ (b) $X_C$ (c) $R$ (d) zero
Answer: (c) $R$.
5. Power factor of a pure inductor is —
(a) 1 (b) 0 (c) 0.5 (d) $1/\sqrt 2$
Answer: (b) 0.
6. A transformer cannot work on —
(a) AC (b) DC (c) high-frequency AC (d) low-frequency AC
Answer: (b) DC.
7. The Q-factor of a series LCR circuit is —
(a) $\omega_0 R/L$ (b) $\omega_0 L/R$ (c) $RC$ (d) $1/RC$
Answer: (b).
8. The natural frequency of an LC circuit increases when —
(a) $L$ alone increases (b) $C$ alone increases (c) both increase (d) both decrease
Answer: (d) both decrease (since $f_0 = 1/2\pi\sqrt{LC}$).
9. A step-down transformer has $N_p = 1000$, $N_s = 100$. If primary voltage is 220 V, secondary voltage is —
(a) 22 V (b) 2200 V (c) 220 V (d) 11 V
Answer: (a) 22 V.
10. Eddy currents in a transformer core are minimised by —
(a) using a thicker core (b) using a copper core (c) laminating the core (d) increasing frequency
Answer: (c) laminating the core.
Fill in the Blanks
- The rms voltage of the Indian mains supply is _____. Answer: 220 V.
- An ideal capacitor consumes _____ average power in an AC circuit. Answer: zero.
- At resonance the phase difference between current and voltage is _____. Answer: zero.
- Inductive reactance is _____ proportional to frequency. Answer: directly.
- The factor $\cos\phi$ in the power expression is called the _____. Answer: power factor.
- The unit of impedance is _____. Answer: ohm ($\Omega$).
- The transformer works on the principle of _____ induction. Answer: mutual.
- The resonant frequency of an LC circuit is given by $f_0 = $ _____. Answer: $1/(2\pi\sqrt{LC})$.
True / False
- An ideal inductor in an AC circuit dissipates power. False.
- The current in a pure capacitor leads the voltage by 90°. True.
- A transformer can step up DC voltage. False.
- The impedance of a series LCR circuit is minimum at resonance. True.
- The power factor of a pure resistor is zero. False (it is 1).
- The rms value of $i_m\sin\omega t$ is $i_m/\sqrt 2$. True.
- The Q-factor is dimensionless. True.
- An ideal step-up transformer increases both voltage and current. False (current decreases).
Match the Following
| Column A | Column B | Match |
|---|---|---|
| 1. Inductive reactance | (p) $1/(\omega C)$ | 1 – q |
| 2. Capacitive reactance | (q) $\omega L$ | 2 – p |
| 3. Resonant frequency | (r) $V_{\text{rms}}I_{\text{rms}}\cos\phi$ | 3 – t |
| 4. Power factor | (s) $\sqrt{R^2+(X_L-X_C)^2}$ | 4 – u |
| 5. Average power | (t) $1/(2\pi\sqrt{LC})$ | 5 – r |
| 6. Impedance | (u) $R/Z$ | 6 – s |
HOTS — Higher-Order Thinking Questions
HOTS-1. Why is a fluorescent tube fitted with a choke and not with a resistor?
Answer: A fluorescent tube needs a high voltage to ionise the gas at start-up but only a small voltage to sustain the discharge. The starter creates this high voltage and a series choke then drops the difference between the supply voltage and the operating voltage of the tube. Because the phase angle between current and voltage in a choke is very close to 90°, the choke dissipates almost no power; a resistor used in its place would dissipate the same voltage drop as $I^2R$ heat, wasting power and overheating.
HOTS-2. The voltage across the inductor in a series LCR circuit at resonance can be many times larger than the source voltage. Is this a violation of energy conservation?
Answer: No. Although $V_L$ can be $Q$ times the source voltage at resonance, the voltage across the capacitor at the same instant is exactly equal in magnitude but 180° out of phase, so the algebraic sum across the LC pair is zero. The energy oscillates between $L$ and $C$; only $R$ dissipates energy supplied by the source. Hence energy conservation is not violated — only the reactive voltages can momentarily exceed the source value.
HOTS-3. Why does a DC ammeter not read the rms value of an AC current passed through it?
Answer: A moving-coil DC ammeter responds to the time-average of the current. For a sinusoidal AC the average over a full cycle is zero, so the pointer would not deflect (or would only flicker about zero). To read AC, instruments based on heating effect (hot-wire ammeter) or rectifier-bridge meters are used; these respond to the rms or average rectified value.
HOTS-4. A series LCR circuit is connected across an AC source with frequency below resonance. Will the current lead or lag the voltage? Justify.
Answer: Below resonance, $\omega < \omega_0$ ⇒ $X_L = \omega L < 1/(\omega C) = X_C$. So $X_C > X_L$ and $\tan\phi = (X_L-X_C)/R < 0$. A negative $\phi$ means the current leads the voltage — the circuit is effectively capacitive at sub-resonant frequencies.
HOTS-5. Why is the secondary coil of a transformer not used to step up DC voltage?
Answer: A transformer relies on a changing flux to induce EMF in the secondary. DC produces a constant flux ($d\Phi/dt = 0$); hence no EMF is induced in the secondary. Worse, a steady DC in the primary would heat the windings (no inductive choking) and could damage the transformer.
HOTS-6. Two students argue: Student A says that an inductor blocks high-frequency current. Student B says it blocks low-frequency current. Who is right?
Answer: Student A is right. The reactance of an inductor is $X_L = \omega L$. As frequency rises, $X_L$ rises and the current falls — so the inductor preferentially blocks high frequencies. That is why inductors are used as RF chokes in audio circuits and as ballasts in fluorescent lamps. Student B’s claim describes capacitors, not inductors.
HOTS-7. Why does increasing the resistance in a series LCR circuit reduce the sharpness of resonance?
Answer: The bandwidth $2\Delta\omega = R/L$ varies directly with $R$. Increasing $R$ widens the bandwidth, so the resonance peak becomes broader and shorter; the Q-factor $Q = \omega_0 L/R$ falls. A “sharp” resonance — needed in a radio tuner to discriminate between adjacent stations — therefore requires a small ohmic resistance.
HOTS-8. Show that a series LCR circuit with $R = 0$ has zero average power and infinite Q-factor.
Answer: With $R=0$ the impedance reduces to $Z = |X_L-X_C|$, purely reactive. Phase angle $\phi=\pm\pi/2$, so $\cos\phi = 0$ ⇒ $P = V_{\text{rms}}I_{\text{rms}}\cos\phi = 0$. The Q-factor $Q = \omega_0 L/R\to\infty$ as $R\to 0$ — bandwidth is zero, the resonance is infinitely sharp. In practice the unavoidable resistance of the wire keeps $Q$ finite (typically 50 – 500 in tuned circuits).
Numerical Practice — Worked Examples
Example 1. A 50 Ω resistor and a 1 H inductor are connected in series across a 100 V, 50 Hz AC supply. Find the impedance, the rms current and the phase angle.
Solution: $X_L = 2\pi(50)(1) = 314\,\Omega$.
$$Z = \sqrt{50^2 + 314^2} = \sqrt{2500 + 98596} \approx 318\;\Omega$$
$$I_{\text{rms}} = \dfrac{100}{318} \approx 0.314\;\text{A}$$
$$\tan\phi = \dfrac{X_L}{R} = 6.28 \Rightarrow \phi \approx 80.96°$$
The current lags the source voltage by about 81°.
Example 2. A 200 V, 50 Hz source is applied across a 1 µF capacitor in series with a 318 Ω resistor. Find the rms current and the phase angle.
Solution: $X_C = 1/[2\pi(50)(10^{-6})] = 3183\,\Omega$.
$$Z = \sqrt{318^2 + 3183^2} \approx 3199\;\Omega$$
$$I_{\text{rms}} = \dfrac{200}{3199} \approx 0.0625\;\text{A} = 62.5\;\text{mA}$$
$$\tan\phi = \dfrac{X_C}{R} = 10.0 \Rightarrow \phi \approx 84.3° \;(\text{current leads})$$
Example 3. An LCR circuit has $L = 100$ mH, $C = 10\,\mu$F, $R = 10\,\Omega$. Find the resonant frequency and the bandwidth at half-power.
$$\omega_0 = \dfrac{1}{\sqrt{0.1\times 10^{-5}}} = 1000\;\text{rad/s},\; f_0 = 159.2\;\text{Hz}$$
$$2\Delta\omega = \dfrac{R}{L} = \dfrac{10}{0.1} = 100\;\text{rad/s},\; \Delta f \approx 16\;\text{Hz}$$
So the half-power frequencies lie at about $159.2 \pm 8.0$ Hz and $Q = \omega_0/2\Delta\omega = 1000/100 = 10$.
Example 4. A 100% efficient transformer has 200 turns in the primary and 800 turns in the secondary. The primary draws 5 A at 220 V. Find the secondary voltage and current.
$$V_s = V_p\dfrac{N_s}{N_p} = 220\times\dfrac{800}{200} = 880\;\text{V}$$
$$I_s = I_p\dfrac{N_p}{N_s} = 5\times\dfrac{200}{800} = 1.25\;\text{A}$$
Power balance: $V_p I_p = 220\times 5 = 1100$ W $= V_s I_s = 880\times 1.25 = 1100$ W. ✓
Example 5. Find the rms value of the current $i = 4\sin\omega t + 3\cos\omega t$ ampere.
Solution: Combine into a single sinusoid using $a\sin\theta + b\cos\theta = \sqrt{a^2+b^2}\sin(\theta+\delta)$. Amplitude $i_m = \sqrt{4^2+3^2} = 5$ A. Hence $I_{\text{rms}} = 5/\sqrt 2 \approx 3.54$ A.
Example 6. An AC voltage $v = 200\sin(100\pi t)$ V is connected across a 50 Ω resistor. Calculate (i) frequency (ii) rms voltage (iii) rms current (iv) average power.
$$\omega = 100\pi \Rightarrow f = 50\;\text{Hz}$$
$$V_{\text{rms}} = 200/\sqrt 2 = 141.4\;\text{V}$$
$$I_{\text{rms}} = 141.4/50 = 2.83\;\text{A}$$
$$P = V_{\text{rms}}I_{\text{rms}} = 141.4\times 2.83 = 400\;\text{W}$$
Glossary — Key Terms
| Term | Meaning |
|---|---|
| Alternating current (AC) | Current whose magnitude varies sinusoidally and direction reverses periodically. |
| Peak / Amplitude | Maximum instantaneous value of an AC quantity ($v_m$, $i_m$). |
| RMS value | Root-mean-square value; for a sinusoid, $X_{\text{rms}} = X_m/\sqrt 2$. |
| Phasor | Rotating vector that represents a sinusoidally varying quantity by its length (amplitude) and orientation (phase). |
| Inductive reactance | Opposition by an inductor to AC, $X_L = \omega L$. |
| Capacitive reactance | Opposition by a capacitor to AC, $X_C = 1/\omega C$. |
| Impedance | Net opposition of an AC circuit, $Z = \sqrt{R^2+(X_L-X_C)^2}$. |
| Phase angle | Angle by which current leads or lags voltage, $\tan\phi = (X_L-X_C)/R$. |
| Resonance | Condition $X_L = X_C$ when impedance is minimum and current is maximum. |
| Resonant frequency | $f_0 = 1/(2\pi\sqrt{LC})$. |
| Bandwidth | $2\Delta\omega = R/L$; range of frequencies between half-power points. |
| Quality factor (Q) | $\omega_0 L/R$; measures sharpness of resonance. |
| Power factor | $\cos\phi = R/Z$; ratio of actual power to apparent power. |
| Wattless current | Component of current 90° out of phase with voltage; carries no average power. |
| Choke coil | Pure-inductor coil used to control AC current with negligible energy loss. |
| Transformer | Static device that changes AC voltage by mutual induction. |
| Step-up / Step-down | Transformer that raises / lowers voltage, depending on $N_s/N_p$. |
| Eddy currents | Induced circulating currents in a bulk conductor that cause heat loss. |
| LC oscillation | Sinusoidal exchange of energy between inductor and capacitor. |
| Hydroelectric plant | Plant where falling water drives turbines coupled to AC generators. |
This concludes the chapter Alternating Current. Practise every numerical at least twice — first by reading our solution, then by closing the book and reproducing the steps yourself. Pay special attention to phasor diagrams, the resonance condition $\omega_0 = 1/\sqrt{LC}$, the wattless-current concept, and transformer numericals — these are perennial AHSEC favourites. Keep visiting HSLC GURU for more chapter-wise ASSEB Class 12 Physics solutions.