Vardaan Learning Institute

Ans: In parallel, Voltage is reference $V = V_0 \sin\omega t$. Current in R: $I_R = \frac{V_0}{R} \sin\omega t$. Current in C: $I_C = V_0(\omega C) \sin(\omega t + 90^\circ) = V_0 \omega C \cos\omega t$.
Total current $I(t) = I_R + I_C = V_0 [\frac{1}{R} \sin\omega t + \omega C \cos\omega t]$.
Let $I(t) = I_0 \sin(\omega t + \phi)$, where $I_0 = V_0 \sqrt{\frac{1}{R^2} + (\omega C)^2}$.
Power factor $\cos\phi = \frac{I_{R, rms}}{I_{rms}} = \frac{1/R}{\sqrt{1/R^2 + (\omega C)^2}} = \frac{1}{\sqrt{1 + (R\omega C)^2}}$.

Ans: Admittance $Y = Y_C + Y_{RL} = j\omega C + \frac{1}{R + j\omega L}$. Rationalize: $Y = j\omega C + \frac{R - j\omega L}{R^2 + \omega^2 L^2} = \frac{R}{R^2 + \omega^2 L^2} + j \left( \omega C - \frac{\omega L}{R^2 + \omega^2 L^2} \right)$.
For minimum current (anti-resonance), imaginary part of admittance must be zero.
$\omega C = \frac{\omega L}{R^2 + \omega^2 L^2} \implies R^2 + \omega^2 L^2 = L/C \implies \omega^2 = \frac{1}{LC} - \frac{R^2}{L^2}$.
Resonant frequency $\omega_r = \sqrt{\frac{1}{LC} - \frac{R^2}{L^2}}$.

Ans: The general condition is $Z_1 Z_4 = Z_2 Z_3$. Expressing impedances in polar form: $Z = |Z| \angle \theta$.
$(|Z_1| \angle \theta_1) \cdot (|Z_4| \angle \theta_4) = (|Z_2| \angle \theta_2) \cdot (|Z_3| \angle \theta_3)$.
This yields two independent conditions:
1) Magnitude condition: $|Z_1| |Z_4| = |Z_2| |Z_3|$
2) Phase angle condition: $\theta_1 + \theta_4 = \theta_2 + \theta_3$. Both must be satisfied for a perfect null.

Ans: $V_{net} = \sqrt{V_R^2 + (V_L - V_C)^2} = \sqrt{120^2 + (250 - 90)^2} = \sqrt{120^2 + 160^2} = \sqrt{14400 + 25600} = \sqrt{40000} = 200\text{ V}$. The data matches the applied source voltage $200\text{V}$, so it is consistent.
Phase angle: $\tan\phi = \frac{V_L - V_C}{V_R} = \frac{160}{120} = \frac{4}{3}$. $\phi = \tan^{-1}(4/3) \approx 53.1^\circ$ (Voltage leads current).

Ans: Skin effect is the tendency of high-frequency alternating current (AC) to distribute itself such that the current density is largest near the surface of the conductor and decreases exponentially toward the center. This severely reduces the effective cross-sectional area available for conduction, hence significantly increasing the effective AC resistance compared to DC resistance.

Ans: Total $R = 5 + 10 = 15\text{ }\Omega$. Total $L = 0.03 + 0.06 = 0.09\text{ H}$.
$X_L = 2\pi f L = 2(3.14)(50)(0.09) = 314 \times 0.09 = 28.26\text{ }\Omega$.
$Z = \sqrt{R^2 + X_L^2} = \sqrt{15^2 + 28.26^2} = \sqrt{225 + 798.6} = \sqrt{1023.6} \approx 32\text{ }\Omega$.
Current $I = 240 / 32 = 7.5\text{ A}$. Power Factor $\cos\phi = R/Z = 15/32 \approx 0.468$ lagging.

Ans: $\tan\phi = \frac{V_L - V_C}{V_R} = \frac{2 V_R - 0.5 V_R}{V_R} = \frac{1.5 V_R}{V_R} = 1.5 = 3/2$.
From right triangle, if Opp $= 3, \text{Adj} = 2$, then Hyp $= \sqrt{3^2 + 2^2} = \sqrt{13}$.
Power factor $\cos\phi = \text{Adj}/\text{Hyp} = 2 / \sqrt{13} \approx 2 / 3.605 \approx 0.554$.

Ans: $Z_L = j\omega L$ and $Z_C = \frac{1}{j\omega C} = -\frac{j}{\omega C}$.
Parallel: $\frac{1}{Z_{eq}} = \frac{1}{Z_L} + \frac{1}{Z_C} = \frac{1}{j\omega L} + j\omega C = \frac{1 - \omega^2 LC}{j\omega L}$.
$Z_{eq} = \frac{j\omega L}{1 - \omega^2 LC} = j \left( \frac{\omega L}{1 - \omega^2 LC} \right)$. (It is purely reactive).

Ans: When an AC circuit is switched on, the complete mathematical solution contains both a steady-state sinusoidal term and a transient exponential decay term ($e^{-t/\tau}$). The time constant $\tau$ ($L/R$ or $RC$) controls the rate of this decay. After roughly $5\tau$, the transient term effectively vanishes, leaving only the pure steady-state AC response.

Ans: From Q2, Admittance $Y = \frac{R}{R^2 + \omega^2 L^2} + j(...)$. At resonance (unity power factor), the imaginary part is zero. $Y = \frac{R}{R^2 + \omega^2 L^2}$.
We found condition $\omega_r^2 L^2 = L/C - R^2$. Substitute this into the denominator:
$Y = \frac{R}{R^2 + (L/C - R^2)} = \frac{R}{L/C} = \frac{CR}{L}$.
Dynamic impedance $Z_d = 1/Y = \frac{L}{CR}$.

Ans: $I^2 = I_1^2 \sin^2\omega t + I_2^2 \cos^2\omega t + 2I_1 I_2 \sin\omega t \cos\omega t$.
Take time average over one cycle: $\langle \sin^2\omega t \rangle = 1/2$, $\langle \cos^2\omega t \rangle = 1/2$, and $\langle \sin 2\omega t \rangle = 0$.
Mean square value $= \frac{I_1^2}{2} + \frac{I_2^2}{2} + 0$.
RMS value $= \sqrt{\text{Mean square}} = \sqrt{\frac{I_1^2 + I_2^2}{2}}$.

Ans: Mean square $I^2_{mean} = \frac{1}{T} \int_0^T \left( \frac{I_0}{T} t \right)^2 dt = \frac{I_0^2}{T^3} \int_0^T t^2 dt = \frac{I_0^2}{T^3} \left[ \frac{T^3}{3} \right] = \frac{I_0^2}{3}$.
$I_{rms} = \sqrt{I_0^2 / 3} = \frac{I_0}{\sqrt{3}}$.

Ans: $V_{avg} = \frac{1}{T} \int_0^T V_0 \sin^2(\omega t) dt = \frac{V_0}{T} \int_0^T \frac{1 - \cos(2\omega t)}{2} dt$.
Since integral of $\cos(2\omega t)$ over $T$ is $0$, $V_{avg} = \frac{V_0}{2T} \int_0^T dt = \frac{V_0}{2T} [T] = \frac{V_0}{2}$.

Ans: The wave is $V = V_0 \sin\omega t$ for $0$ to $T/2$, and $V = 0$ for $T/2$ to $T$.
Mean square $= \frac{1}{T} \int_0^{T/2} V_0^2 \sin^2\omega t dt + 0$.
$= \frac{V_0^2}{T} \int_0^{T/2} \frac{1 - \cos(2\omega t)}{2} dt = \frac{V_0^2}{2T} [T/2 - 0] = \frac{V_0^2}{4}$.
RMS value $= \sqrt{V_0^2/4} = \frac{V_0}{2}$.

Ans: $V_L = L \frac{dI}{dt} = 2.0 \cdot \frac{d}{dt} (10\sin(50\pi t)) = 2.0 \cdot 10(50\pi)\cos(50\pi t) = 1000\pi \cos(50\pi t)$.
Instantaneous voltage $V(t) = 1000\pi \cos(50\pi t)$. Maximum value $V_0 = 1000\pi \approx 3140\text{ V}$.

Ans: $H = \int_0^\infty I^2 R dt = \int_0^\infty (I_0^2 e^{-2t/\tau}) R dt = I_0^2 R \left[ \frac{e^{-2t/\tau}}{-2/\tau} \right]_0^\infty$.
$H = I_0^2 R (\frac{-\tau}{2}) (0 - 1) = \frac{1}{2} I_0^2 R \tau$.

Ans: $P_{avg} = \frac{1}{T} \int_0^T V_0 \sin(\omega t) \cdot I_0 \sin(2\omega t) dt$.
Since $\sin(\omega t)$ and $\sin(2\omega t)$ are orthogonal functions over period $T$ (harmonics), their integral is strictly zero. Average power is $0$.

Ans: Instantaneous $U_E(t) = \frac{1}{2}CV^2 = \frac{1}{2} C V_0^2 \cos^2(\omega t)$.
Time average $\langle U_E \rangle = \frac{1}{2} C V_0^2 \langle \cos^2(\omega t) \rangle = \frac{1}{2} C V_0^2 \times \frac{1}{2} = \frac{1}{4} C V_0^2$.

Ans: Conduction current $I_c = dq/dt$. The electric field $E = q/\epsilon_0 A$.
Displacement current $I_d = \epsilon_0 A \frac{dE}{dt} = \epsilon_0 A \frac{d}{dt}(q/\epsilon_0 A) = dq/dt = I_c$.
Since $I_d(t) = I_c(t)$ identically at every instant, they are perfectly in phase (Phase difference = $0^\circ$).

Ans: Form Factor = RMS value / Mean value.
$F_F = (I_0/\sqrt{2}) / (2I_0/\pi) = \frac{\pi}{2\sqrt{2}} \approx 1.11$.
Peak Factor (Crest Factor) = Peak value / RMS value.
$C_F = I_0 / (I_0/\sqrt{2}) = \sqrt{2} \approx 1.414$.

Ans: At half power, $I = I_0/\sqrt{2}$, implying $Z = \sqrt{2}R$.
$\implies R^2 + (\omega L - 1/\omega C)^2 = 2R^2 \implies \omega L - 1/\omega C = \pm R$.
For narrow bandwidth, let $\omega = \omega_0 \pm \Delta\omega/2$. Substituting and using binomial approximation ($\omega_0 L = 1/\omega_0 C$), we get $\Delta\omega \cdot L \approx R$.
Thus, $Q = \frac{\omega_0 L}{R} = \frac{\omega_0}{\Delta\omega}$.

Ans: At resonance, $I = I_0 \sin\omega_0 t$. $q = \int I dt = -\frac{I_0}{\omega_0} \cos\omega_0 t$.
$U_L = \frac{1}{2} L I_0^2 \sin^2\omega_0 t$. $U_C = \frac{q^2}{2C} = \frac{I_0^2 \cos^2\omega_0 t}{2C\omega_0^2}$. Since $\omega_0^2 = 1/LC$, $U_C = \frac{1}{2} L I_0^2 \cos^2\omega_0 t$.
Total $U = \frac{1}{2} L I_0^2 (\sin^2\omega_0 t + \cos^2\omega_0 t) = \frac{1}{2} L I_0^2 = L (\frac{I_0}{\sqrt{2}})^2 = L I_{rms}^2 = \text{Constant}$.

Ans: From Section A Q2, $\omega_r = \sqrt{\frac{1}{LC} - \frac{R^2}{L^2}}$.
It strictly equals the series resonant frequency ($1/\sqrt{LC}$) only if the resistance of the inductor branch is exactly zero ($R = 0$).

Ans: $Z_d = L/CR$ is purely resistive. In radio receivers, parallel LC circuits are used as load resistors in amplifier stages. A very high $Z_d$ (high $L$, low $C$ and $R$) provides maximum voltage gain exclusively at the resonant frequency, ensuring extremely sharp selectivity and amplifying only the desired station signal.

Ans: If $LC=$ constant, the resonant frequency $\omega_0 = 1/\sqrt{LC}$ is constant. Since the applied frequency $\omega$ is constant and equals $\omega_0$, we always have $X_L = X_C$. Thus, $Z = \sqrt{R^2 + (X_L-X_C)^2} = R$. The impedance $Z$ is a constant real number. Its locus in the complex plane is simply a fixed point $(R, 0)$ on the real axis.

Ans: At half-power frequencies, the total impedance magnitude is $|Z| = \sqrt{2} R$.
This requires $R^2 + X_{net}^2 = 2R^2 \implies X_{net}^2 = R^2 \implies X_{net} = \pm R$.
Phase angle $\tan\phi = \frac{X_{net}}{R} = \frac{\pm R}{R} = \pm 1$.
Therefore, $\phi = \pm 45^\circ$ (or $\pm \pi/4$ radians).

Ans: Voltage Magnification $= \frac{V_L}{V_{applied}} = \frac{I X_L}{I Z}$.
At resonance, $Z = R$. Therefore, ratio $= \frac{X_L}{R} = \frac{\omega_0 L}{R}$.
This is precisely the mathematical definition of the Q-factor.

Ans: Max energy stored $E_{stored} = \frac{1}{2} L I_0^2$.
Energy dissipated per cycle $E_{diss} = P_{avg} \times T = (I_{rms}^2 R) T = \frac{I_0^2 R T}{2}$.
Ratio $= 2\pi \frac{L I_0^2 / 2}{I_0^2 R T / 2} = 2\pi \frac{L}{RT}$. Since $T = 2\pi/\omega_0$, Ratio $= 2\pi \frac{L}{R (2\pi/\omega_0)} = \frac{\omega_0 L}{R} = Q$.

Ans: Impedance of L-r parallel group is $\frac{j\omega L \cdot r}{r + j\omega L}$. Rationalizing yields an equivalent series inductance $L_{eq} = \frac{L r^2}{r^2 + \omega^2 L^2}$, which is less than $L$. Since the effective inductance has decreased, the resonant frequency $\omega_0 = 1/\sqrt{L_{eq}C}$ will **increase** (shift to a higher frequency).

Ans: A parallel resistor provides an alternate path for current to bypass the capacitor, meaning energy is constantly drained from the oscillating tank. This increases the total energy dissipation per cycle, thus significantly **lowering** the overall Q-factor. It also alters the reactive balance, slightly shifting the resonant frequency.

Ans: Equation: $L \frac{d^2q}{dt^2} + R \frac{dq}{dt} + \frac{q}{C} = 0$.
Damped frequency $\omega_d = \sqrt{\omega_0^2 - (R/2L)^2} = \sqrt{\frac{1}{LC} - \frac{R^2}{4L^2}}$.
Critical damping occurs when the term under the square root is exactly zero, i.e., $R = 2\sqrt{L/C}$. The circuit returns to zero without oscillating.

Ans: Since current $i$ is same in both, self-induced EMFs are $-L_1 \frac{di}{dt}$ and $-L_2 \frac{di}{dt}$. Since they are aiding, the mutual EMFs enhance the primary EMFs: $-M \frac{di}{dt}$ in coil 1 due to 2, and $-M \frac{di}{dt}$ in coil 2 due to 1.
Total EMF $E = -L_1 \frac{di}{dt} - M \frac{di}{dt} - L_2 \frac{di}{dt} - M \frac{di}{dt} = -(L_1 + L_2 + 2M)\frac{di}{dt}$.
Hence, $L_{eq} = L_1 + L_2 + 2M$.

Ans: Let induced EMFs be $E_p, E_s$. $\frac{E_s}{E_p} = \frac{N_s}{N_p}$.
In primary, applied voltage $V_p = E_p + I_p R_p \implies E_p = V_p - I_p R_p$.
In secondary, terminal voltage $V_s = E_s - I_s R_s$.
Therefore, $\frac{V_s + I_s R_s}{V_p - I_p R_p} = \frac{N_s}{N_p}$.

Ans: Principle: Works on self-induction and partly on direct electrical conduction. It uses a single continuous winding for both primary and secondary.
Advantage: Significantly uses less copper wire and is more compact/efficient for small voltage ratios.
Disadvantage: There is no electrical isolation between the primary high-voltage and secondary circuits, posing a severe safety hazard.

Ans: Energy lost per cycle $= (\text{Area of B-H loop}) \times \text{Volume } V$.
Total power loss is energy lost per second. Since there are $f$ cycles per second, $P_h = (\text{Area of B-H loop}) \times V \times f$.

Ans: $q = Q_0 \cos\omega t$. At $t = T/6$, $\omega t = \frac{2\pi}{T} \frac{T}{6} = \frac{\pi}{3} = 60^\circ$.
$q = Q_0 \cos 60^\circ = Q_0/2$. Electrical energy $U_E = \frac{q^2}{2C} = \frac{Q_0^2/4}{2C} = \frac{1}{4} U_{max}$.
Since total energy is $U_{max}$, magnetic energy $U_B = U_{max} - 1/4 U_{max} = 3/4 U_{max}$.
Ratio $U_E / U_B = (1/4) / (3/4) = 1/3$.

Ans: Total initial energy $U_{total} = \frac{q_0^2}{2C} + \frac{1}{2} L i_0^2$.
Maximum current $I_{max}$ occurs when all this energy is fully converted into magnetic energy.
$\frac{1}{2} L I_{max}^2 = \frac{q_0^2}{2C} + \frac{1}{2} L i_0^2 \implies I_{max}^2 = \frac{q_0^2}{LC} + i_0^2$.
$I_{max} = \sqrt{\frac{q_0^2}{LC} + i_0^2}$.

Ans: $\eta = \frac{\text{Output}}{\text{Output} + P_{iron} + I^2R_{copper}} = \frac{VI\cos\phi}{VI\cos\phi + P_i + I^2R}$.
Divide by $I$: $\eta = \frac{V\cos\phi}{V\cos\phi + P_i/I + IR}$.
For $\eta$ to be max, denominator must be min. Differentiate wrt $I$: $\frac{d}{dI}(V\cos\phi + P_i/I + IR) = 0 \implies -P_i/I^2 + R = 0 \implies P_i = I^2R$. (Iron loss = Copper loss).

Ans: Core saturation occurs when all magnetic domains are aligned; further increases in current yield negligible flux increase. To maintain a sinusoidal flux (required to balance applied sinusoidal voltage), the primary winding must draw hugely disproportionate, sharp spikes of magnetizing current at the peaks. The current waveform becomes severely distorted (peaky) and non-sinusoidal.

Ans: The circuit forms a driven LC harmonic oscillator. Equation: $L\frac{di}{dt} + \frac{q}{C} = V \implies L\frac{d^2q}{dt^2} + \frac{q - CV}{C} = 0$.
The charge oscillates around the new steady state value $Q_{steady} = CV$ with amplitude $CV$.
$q(t) = CV(1 - \cos\omega t)$.
Max charge $q_{max} = CV(1 - (-1)) = 2CV$. Max voltage $V_{max} = q_{max}/C = 2V$. The voltage spikes to exactly double the supply voltage.

Ans: $I_{DC} = 4$. $I_{AC, peak} = 3\sqrt{2} \implies I_{AC, rms} = 3\sqrt{2} / \sqrt{2} = 3$.
$I_{rms} = \sqrt{I_{DC}^2 + I_{AC, rms}^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$. 5

Ans: $V_{source} = \sqrt{V_R^2 + (V_L - V_C)^2} = \sqrt{30^2 + (60 - 20)^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50$. 50

Ans: $Q = \frac{1}{R} \sqrt{\frac{L}{C}} = \frac{1}{10} \sqrt{\frac{2}{32 \times 10^{-6}}} = 0.1 \times \sqrt{\frac{1}{16 \times 10^{-6}}} = 0.1 \times \frac{1}{4 \times 10^{-3}} = 0.1 \times 250 = 25$. 25

Ans: $10\sin(100\pi t) = 5 \implies \sin(100\pi t) = 0.5 \implies 100\pi t = \pi/6 \implies t = \frac{1}{600}\text{ s} = \frac{1000}{600}\text{ ms} = 5/3\text{ ms} \approx 1.667$. 1.67

Ans: $I_0 = Q_0 \omega = \frac{Q_0}{\sqrt{LC}} = \frac{4 \times 10^{-3}}{\sqrt{10 \times 10^{-3} \times 40 \times 10^{-6}}} = \frac{4 \times 10^{-3}}{\sqrt{400 \times 10^{-9}}} = \frac{4 \times 10^{-3}}{20 \times 10^{-4}} = \frac{40}{20} = 2$. 2

Ans: Phase angle is exactly $90^\circ$. $\cos(90^\circ) = 0$. 0

Ans: $N_s = N_p (V_s/V_p) = 2000 (220/2200) = 2000 \times 0.1 = 200$. 200

Ans: $\cos\phi = R / Z = 40 / 50 = 0.8$. 0.8

Ans: Peak $V_0 = 100\sqrt{2}$. $V_{rms} = V_0 / \sqrt{2} = 100$. 100

Ans: $X_L = X_C \implies$ Resonance $\implies Z = R$. Circuit is purely resistive. Phase angle $\phi = 0^\circ$. 0

Ans: $R = V^2 / P = (200)^2 / 100 = 40000 / 100 = 400$. 400

Ans: Output $= \eta \times \text{Input} = 0.90 \times 10 = 9$. 9

Ans: $Q = \omega_0 / \Delta\omega = 500 / 10 = 50$. 50

Ans: $I_{mean} = 2 I_0 / \pi = 2(15.7) / 3.14 = 31.4 / 3.14 = 10$. 10

Ans: $P = V I \cos\phi = 100 \times 2 \times \cos(60^\circ) = 200 \times 0.5 = 100$. 100

Ans: $V = \sqrt{V_R^2 + (V_L - V_C)^2} = \sqrt{50^2 + 0^2} = 50$. 50

Ans: $Z = \sqrt{R^2 + X_C^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$. 5

Ans: Product $L'C' = (2L)(C/2) = LC$. Since product is same, frequency is same. Ratio is 1. 1

Ans: $X_L = \omega L = 100 \times 0.5 = 50$. 50

Ans: $M = k\sqrt{L_1 L_2} = 1 \cdot \sqrt{2 \times 8} = \sqrt{16} = 4$. 4