Board Previous Year Questions

How to Use This Section

CBSE Board-pattern questions on Alternating Current by mark value, with model answers in examiner-rewarded phrasing — definitions first, phasor diagrams described, derivations stepwise, units carried. The Board regulars: rms derivation, phase relations for pure L and C with diagrams, the LCR impedance derivation, resonance and its applications, power factor/wattless current, and the transformer long answer.

1-Mark Questions (Definitions & Direct)

Q1. Define the rms value of an alternating current. Answer: It is the square root of the mean of $i^2$ over a cycle — equivalently, the steady (DC) current that would produce the same average heating in a given resistor: $I = i_m/\sqrt 2$.

Q2. What is the phase relation between voltage and current in a pure inductor? Answer: The current lags the voltage by $\pi/2$ (a quarter cycle).

Q3. Define capacitive reactance and give its SI unit. Answer: $X_C = 1/\omega C$, the effective opposition of a capacitor to AC; SI unit: ohm.

Q4. What is wattless current? Answer: The current in a purely inductive or capacitive circuit ($\phi = 90$ degrees, $\cos\phi = 0$): current flows but no average power is dissipated.

Q5. Write the resonance condition for a series LCR circuit and the resonant frequency. Answer: $X_L = X_C$, giving $\omega_0 = 1/\sqrt{LC}$ (i.e. $\nu_0 = \frac{1}{2\pi\sqrt{LC}}$).

Q6. On what principle does a transformer work? Answer: Mutual induction — an alternating flux in the shared core induces emfs in both windings in proportion to their turns.

2-Mark Questions (Short Answer)

Q7. Show that the average power consumed by an ideal inductor over a full cycle is zero. Answer: With $v = v_m\sin\omega t$ and $i = i_m\sin(\omega t - \pi/2)$, the instantaneous power is $p = vi = -\frac{v_mi_m}{2}\sin 2\omega t$. Since $\langle\sin 2\omega t\rangle = 0$ over a cycle, $\bar P = 0$: energy is alternately stored in and returned by the magnetic field.

Q8. The peak value of the 220 V mains is 311 V. Explain the two numbers. Answer: Specified AC voltages are rms values: $V = 220$ V. The sinusoidal peak is $v_m = \sqrt 2 V = 311$ V. The rms value is quoted because it gives DC-equivalent heating ($P = V^2/R$).

Q9. Distinguish between resistance and reactance. Answer: Resistance opposes current and dissipates energy ($\phi = 0$); it is frequency-independent. Reactance ($X_L = \omega L$ or $X_C = 1/\omega C$) opposes AC without average dissipation (voltage and current 90 degrees apart) and depends on frequency.

Q10. Why can a capacitor conduct AC but not steady DC? Answer: On DC ($\omega = 0$), $X_C = 1/\omega C \to \infty$: once charged, current stops. On AC the capacitor is alternately charged and discharged each half cycle, so charge flows continuously in the circuit, limited by the finite $X_C$.

Q11. A series LCR circuit is at resonance. What are (i) its impedance, (ii) the phase angle, (iii) the power factor? Answer: (i) $Z = R$ (minimum); (ii) $\phi = 0$; (iii) $\cos\phi = 1$ — maximum power transfer.

3-Mark Questions (Derivations & Numericals)

Q12. Derive the impedance of a series LCR circuit by the phasor method. Answer: The common current is $i = i_m\sin(\omega t + \phi)$. Voltage phasors: $V_R$ ($= i_mR$) along I; $V_L$ ($= i_mX_L$) ahead of I by 90 degrees; $V_C$ ($= i_mX_C$) behind by 90 degrees. $V_L$ and $V_C$ oppose, combining to $|i_mX_C - i_mX_L|$. Pythagoras on the phasor sum: $v_m^2 = (i_mR)^2 + (i_mX_C - i_mX_L)^2$, so $i_m = v_m/Z$ with $Z = \sqrt{R^2 + (X_C - X_L)^2}$ and $\tan\phi = (X_C - X_L)/R$.

Q13. A 15 $\mu$F capacitor is connected to a 220 V, 50 Hz source. Find the reactance and the rms and peak currents. Answer: $X_C = \frac{1}{2\pi(50)(15\times10^{-6})} = 212\ \Omega$; $I = 220/212 = 1.04$ A; $i_m = \sqrt2 I = 1.47$ A (current leads by 90 degrees).

Q14. State the condition for resonance and show the current is maximum there. Answer: Resonance: $X_L = X_C$, i.e. $\omega_0 = 1/\sqrt{LC}$. Then $Z = \sqrt{R^2 + 0} = R$, its minimum; hence $i_m = v_m/Z = v_m/R$, the maximum possible. $V_L$ and $V_C$, equal and opposite, cancel; the whole source voltage sits across R. Resonance requires both L and C — no RL/RC resonance.

Q15. R = 3 $\Omega$, $X_L = 8\ \Omega$, $X_C = 4\ \Omega$, source 283 V peak at 50 Hz. Find Z, the rms current and the power dissipated. Answer: $Z = \sqrt{9 + 16} = 5\ \Omega$; $I = \frac{283/\sqrt2}{5} = 40$ A; $P = I^2R = 4800$ W (power factor 0.6, current lagging by 53.1 degrees).

5-Mark Questions (Long Answer)

Q16. (a) With a phasor diagram, obtain the average power of a series LCR circuit and define the power factor. (b) Discuss the cases: pure R, pure L or C, and resonance. (c) Why is low power factor undesirable in power supply? Answer:

1. (a) With $v = v_m\sin\omega t$, $i = i_m\sin(\omega t + \phi)$: $p = vi = \frac{v_mi_m}{2}[\cos\phi - \cos(2\omega t + \phi)]$. The oscillating term averages to zero: $P = VI\cos\phi$. The factor $\cos\phi = R/Z$ is the power factor.
2. (b) Pure R: $\phi = 0$, $P = VI$ (max). Pure L or C: $\phi = 90$ degrees, $P = 0$ — wattless current. Resonance: $\phi = 0$, $P = I^2R$ maximum.
3. (c) For required power P at voltage V, $I = P/(V\cos\phi)$: small $\cos\phi$ forces large current and $I^2R$ line losses grow as its square. Suppliers therefore require pf near 1 (improved by adding capacitors across inductive loads).

Q17. (a) Describe the construction and working of a transformer and derive $v_s/v_p = N_s/N_p$. (b) List the sources of energy loss and their remedies. (c) A step-down transformer converts 2200 V to 220 V for a 4.4 kW load. Find the turns ratio and the primary and secondary currents (ideal). Answer:

1. (a) Primary ($N_p$) and secondary ($N_s$) windings, insulated, on a common soft-iron core. AC in the primary creates an alternating core flux $\phi$ linking both coils. Faraday: $v_s = -N_s\,d\phi/dt$ and $v_p = -N_p\,d\phi/dt$ (back emf equals applied voltage for negligible primary resistance). Dividing: $v_s/v_p = N_s/N_p$. With no losses, $i_pv_p = i_sv_s$.
2. (b) Flux leakage (wind coils one over the other); winding resistance (thick wire); eddy currents (laminated core); hysteresis (low-hysteresis core material).
3. (c) $N_s/N_p = 220/2200 = 1/10$; $i_s = 4400/220 = 20$ A; $i_p = 4400/2200 = 2$ A.