1.The average value of alternating current over one complete cycle is exactly ____________.
Ans: Zero ($0$). The positive half perfectly cancels the negative half.
2.The RMS value of alternating current is also known as the ____________ value.
Ans: Virtual (or Effective) value.
3.Relation between $I_{rms}$ and $I_0$ is:
Ans: (B) $I_{rms} = I_0 / \sqrt{2}$
4.Domestic power supply "$220\text{ V}$" specifically represents:
Ans: (C) RMS voltage
5.Define RMS value in terms of heating effect.
Ans: RMS value of AC is that steady (DC) current which produces the exact same amount of heat in a given resistor in a given time as is produced by the AC in the same resistor in the same time.
6.Find RMS if peak $I_0 = 14.14\text{ A}$.
Ans: $I_{rms} = I_0 / \sqrt{2} = 14.14 / 1.414 = 10\text{ A}$.
7.Calculate peak voltage of $220\text{ V}$ RMS supply.
Ans: $V_0 = \sqrt{2} \times V_{rms} = 1.414 \times 220 \approx 311\text{ V}$.
8.Match AC terms to formulas:
Ans: (a) $\rightarrow$ (iii) $2 I_0 / \pi$
(b) $\rightarrow$ (i) Zero
(c) $\rightarrow$ (ii) $I_0 / \sqrt{2}$
9.In a purely resistive AC circuit, voltage and current are in ____________.
Ans: Phase
10.In a purely inductive AC circuit, the voltage ____________ the current by $\pi/2$.
Ans: Leads
11.In a purely capacitive AC circuit, the current ____________ the voltage by $90^\circ$.
Ans: Leads
12.Phase difference in purely inductive circuit?
Ans: (C) $\pi/2$
Ans: A phasor is a rotating vector that represents a sinusoidally varying quantity (like AC voltage or current), rotating with an angular velocity equal to the angular frequency ($\omega$) of the AC.
14.Write AC current equation if $V = V_0 \sin(\omega t)$ in pure C.
Ans: Current leads by $90^\circ$ ($\pi/2$). Equation: $I = I_0 \sin(\omega t + \pi/2)$.
15.Based on phasor diagram, which phasor rotates ahead in a pure inductor?
Ans: The Voltage phasor ($V_0$) rotates ahead of the Current phasor ($I_0$) by $90^\circ$.
16.Calculate frequency from $V = 100 \sin(314 t)$.
Ans: Standard form: $V = V_0 \sin(2\pi f t)$. Here, $2\pi f = 314 \implies 2(3.14)f = 314 \implies 6.28f = 314 \implies f = 314 / 6.28 = 50\text{ Hz}$.
17.Match circuit elements to phase relationships:
Ans: (a) $\rightarrow$ (iii) Voltage and Current are in phase
(b) $\rightarrow$ (ii) Voltage leads Current by $90^\circ$
(c) $\rightarrow$ (i) Current leads Voltage by $90^\circ$
18.SI unit of Reactance ($X_L$ and $X_C$) is...
Ans: Ohm ($\Omega$).
19.Inductive reactance is directly proportional, whereas capacitive reactance is ____________ proportional to frequency.
Ans: Inversely (since $X_C = 1 / 2\pi f C$).
20.Formula for impedance ($Z$) of series LCR is:
Ans: (C) $Z = \sqrt{R^2 + (X_L - X_C)^2}$
21.Define Impedance ($Z$).
Ans: Impedance is the total effective opposition offered by a circuit (containing R, L, and C) to the flow of alternating current.
22.Calculate $X_L$ for $L=0.1\text{ H}, f=50\text{ Hz}$.
Ans: $X_L = 2\pi f L = 2 \times 3.14 \times 50 \times 0.1 = 314 \times 0.1 = 31.4\text{ }\Omega$.
23.Calculate $X_C$ for $C=10\text{ }\mu\text{F}, \omega=100\text{ rad/s}$.
Ans: $X_C = \frac{1}{\omega C} = \frac{1}{100 \times 10 \times 10^{-6}} = \frac{1}{10^{-3}} = 1000\text{ }\Omega$.
24.Find $Z$ if $R=3\text{ }\Omega, X_L=8\text{ }\Omega, X_C=4\text{ }\Omega$.
Ans: $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{3^2 + (8 - 4)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\text{ }\Omega$.
25.In series LCR, what quantity remains the same for all components?
Ans: Alternating Current ($I$). (Because they are in a series combination).
26.Match quantities to formulas:
Ans: (a) $\rightarrow$ (ii) $\omega L$
(b) $\rightarrow$ (i) $1 / (\omega C)$
(c) $\rightarrow$ (iii) $2\pi f$
27.Resonance occurs when $X_L$ becomes exactly ____________ to $X_C$.
Ans: Equal
28.At resonance, total impedance is at its ____________ value.
Ans: Minimum (since $Z = R$)
29.Formula for resonant frequency is:
Ans: (B) $1 / (2\pi\sqrt{LC})$
30.Current at resonance in series LCR is:
Ans: (B) Maximum (because impedance is minimum).
Ans: Quality factor (Q-factor) is the ratio of resonant frequency to the bandwidth of the resonant curve. It measures the sharpness of resonance. Formula: $Q = \frac{1}{R}\sqrt{\frac{L}{C}}$ or $Q = \frac{\omega_r L}{R}$.
32.What is sharpness of resonance?
Ans: Sharpness refers to how quickly the current drops as the frequency moves away from the resonant frequency. A sharper curve (high Q) means better frequency selectivity.
33.Calculate $\omega_r$ if $L = 1\text{ H}, C = 1\text{ }\mu\text{F}$.
Ans: $\omega_r = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{1 \times 10^{-6}}} = \frac{1}{10^{-3}} = 1000\text{ rad/s}$.
34.Find Q-factor if $X_L = 1000\text{ }\Omega, R = 10\text{ }\Omega$.
Ans: $Q = \frac{X_L}{R} = \frac{1000}{10} = 100$.
35.Match resonance conditions:
Ans: (a) $\rightarrow$ (ii) $Z = R$
(b) $\rightarrow$ (iii) Equal in magnitude, opposite in phase
(c) $\rightarrow$ (i) $0^\circ$
36.Power Factor is the cosine of the ____________ angle.
Ans: Phase ($\phi$)
37.Wattless current flows when power is strictly ____________.
Ans: Zero
38.Power factor of pure resistive circuit is:
Ans: (C) $1$ (Unity, since $\phi=0^\circ$ and $\cos 0^\circ = 1$).
39.Power factor of pure inductive/capacitive circuit is:
Ans: (A) Zero (since $\phi=90^\circ$ and $\cos 90^\circ = 0$).
40.Formula for average power in LCR.
Ans: $P_{avg} = V_{rms} I_{rms} \cos\phi$.
41.Formula for Power Factor in terms of R and Z.
Ans: $\cos\phi = \frac{R}{Z}$.
42.Calculate power factor if $R=4\text{ }\Omega, Z=5\text{ }\Omega$.
Ans: $\cos\phi = \frac{R}{Z} = \frac{4}{5} = 0.8$.
43.Ratio of True Power to Apparent Power gives:
Ans: Power Factor ($\cos\phi = \frac{\text{True Power}}{\text{Apparent Power}}$).
44.Match circuit types to power factors:
Ans: (a) $\rightarrow$ (ii) $\cos\phi = 1$
(b) $\rightarrow$ (iii) $\cos\phi = 0$
(c) $\rightarrow$ (i) $\cos\phi = R / Z$
45.Energy oscillates between electric field of capacitor and ____________ field of inductor.
Ans: Magnetic
46.Total energy in ideal LC circuit remains ____________.
Ans: Constant (Conserved).
47.Natural angular frequency ($\omega$) is:
Ans: (B) $1 / \sqrt{LC}$
48.Electrostatic energy formula (max charge $Q_0$).
Ans: $U_E = \frac{Q_0^2}{2C}$.
49.Magnetic energy formula (max current $I_0$).
Ans: $U_B = \frac{1}{2} L I_0^2$.
50.Calculate total energy if $Q_0 = 4\text{ mC}, C = 2\text{ mF}$.
Ans: Total energy is the maximum electric energy.
$U = \frac{Q_0^2}{2C} = \frac{(4 \times 10^{-3})^2}{2 \times 2 \times 10^{-3}} = \frac{16 \times 10^{-6}}{4 \times 10^{-3}} = 4 \times 10^{-3}\text{ Joules} = 4\text{ mJ}$.
51.Match LC parameters to Mechanical analogies:
Ans: (a) $\rightarrow$ (ii) Mass ($m$ / Inertia)
(b) $\rightarrow$ (iii) Spring constant ($k$)
(c) $\rightarrow$ (i) Displacement ($x$)
52.Transformer works purely on the principle of ____________ induction.
Ans: Mutual
53.Step-up transformer increases voltage but decreases AC ____________.
Ans: Current
54.Core is laminated to minimize losses due to:
Ans: (B) Eddy currents
55.Turns ratio ($N_s/N_p$) for step-down transformer is:
Ans: (B) Less than 1
56.Can transformer step up DC? Reason?
Ans: No. A steady DC produces a constant magnetic flux, which means there is zero change in flux ($\frac{d\Phi}{dt} = 0$), so no EMF will be induced in the secondary coil.
57.Primary transformer equation.
Ans: $\frac{V_s}{V_p} = \frac{N_s}{N_p}$
58.$N_p=100, N_s=500, V_p=220\text{V}$. Find $V_s$.
Ans: $V_s = V_p \times \frac{N_s}{N_p} = 220 \times \frac{500}{100} = 220 \times 5 = 1100\text{ V}$.
59.$V_p=100\text{V}, I_p=2\text{A}, V_s=200\text{V}$. Find $I_s$.
Ans: $V_p I_p = V_s I_s \implies 100 \times 2 = 200 \times I_s \implies 200 = 200 I_s \implies I_s = 1\text{ A}$.
60.Match transformer losses to remedies:
Ans: (a) $\rightarrow$ (ii) Laminated iron core
(b) $\rightarrow$ (iii) Using soft iron core
(c) $\rightarrow$ (i) Using thick copper wires
