Vardaan Learning Institute

Ans: $I_{mean} = \frac{\int_0^{T/2} I_0 \sin(\omega t) dt}{\int_0^{T/2} dt} = \frac{I_0}{T/2} \left[ \frac{-\cos(\omega t)}{\omega} \right]_0^{T/2}$.
Since $\omega = \frac{2\pi}{T}$, upper limit is $\cos(\frac{2\pi}{T} \frac{T}{2}) = \cos(\pi) = -1$. Lower limit is $\cos(0) = 1$.
$I_{mean} = \frac{2I_0}{T\omega} [-(-1) - (-1)] = \frac{2I_0}{T(2\pi/T)} [2] = \frac{4I_0}{2\pi} = \frac{2I_0}{\pi} \approx 0.637 I_0$.

Ans: The current is a superposition of a DC component ($I_{DC} = 3$) and an AC component ($I_{AC} = 4\sin\omega t$).
The RMS value of such a combined signal is $I_{rms} = \sqrt{I_{DC}^2 + I_{rms, AC}^2}$.
$I_{rms, AC} = \frac{4}{\sqrt{2}}$.
$I_{rms} = \sqrt{3^2 + \left(\frac{4}{\sqrt{2}}\right)^2} = \sqrt{9 + \frac{16}{2}} = \sqrt{9 + 8} = \sqrt{17}\text{ A} \approx 4.12\text{ A}$.

Ans: A hot-wire instrument measures the heating effect, which depends on the square of the RMS current ($H \propto I_{rms}^2$). Since the expansion (and thus deflection $\theta$) is directly proportional to $I^2$, the scale is non-linear ($\theta \propto I^2$). It is crowded near zero and spreads out at higher values.

Ans: Let $I = I_0 \sin(\omega t)$. We need time $t$ when $I = I_{rms} = I_0 / \sqrt{2}$.
$\frac{I_0}{\sqrt{2}} = I_0 \sin(\omega t) \implies \sin(\omega t) = \frac{1}{\sqrt{2}} \implies \omega t = \frac{\pi}{4}$.
Since $\omega = 2\pi f$, we have $2\pi f t = \frac{\pi}{4} \implies t = \frac{1}{8f}$.
For $f = 50\text{ Hz}$, $t = \frac{1}{8 \times 50} = \frac{1}{400}\text{ s} = 2.5\text{ ms}$.

Ans: RMS means Root of Mean of Square. The square of the current is $(I_0)^2 = I_0^2$ for the positive half, and $(-I_0)^2 = I_0^2$ for the negative half. Thus, the squared current is a constant $I_0^2$ at all times.
Mean of square is $I_0^2$. Root of mean is $\sqrt{I_0^2} = I_0$. So, $I_{rms} = I_0$.

Ans: For full-wave rectification, all negative half cycles are flipped to positive. The area under the curve is identical to two positive half-cycles of normal AC. Therefore, $V_{mean} = \frac{2V_0}{\pi}$. This is exactly double the mean value of a half-wave rectified AC ($V_0/\pi$).

Ans: No. A DC moving coil ammeter reads the *average* current. For a full cycle of AC, the average is perfectly zero. Multiplying a zero reading by any correction factor will still yield zero, failing to measure the RMS value.

Ans: $V = 100\sqrt{2} \sin(100\pi \times \frac{1}{600}) = 100\sqrt{2} \sin(\frac{\pi}{6})$.
$\sin(30^\circ) = 1/2$.
$V = 100\sqrt{2} \times \frac{1}{2} = 50\sqrt{2}\text{ V} \approx 70.7\text{ V}$.

Ans: Current is the *rate of flow* of charge ($I = dq/dt$). Voltage across the capacitor is proportional to the *accumulated* charge ($V = q/C$). When the AC cycle starts from zero, the capacitor is empty, so charge rushes in rapidly (Current is maximum). As charge accumulates, the voltage builds up. By the time voltage hits its maximum, the capacitor is fully charged and current momentarily stops (Current is zero). Thus, the current wave leads the voltage wave by a quarter cycle ($90^\circ$).

Ans: $V = V_0 \sin\omega t$. For inductor, $V = L \frac{dI}{dt} \implies \frac{dI}{dt} = \frac{V_0}{L} \sin\omega t$.
$I = \int \frac{V_0}{L} \sin\omega t dt = -\frac{V_0}{\omega L} \cos\omega t$.
Since $-\cos\omega t = \sin(\omega t - \pi/2)$, $I = I_0 \sin(\omega t - \pi/2)$. The "$-\pi/2$" term proves the current mathematically lags the voltage.

Ans: $X_L = \omega L = 2\pi f L = 2\pi(50)(0.5) = 50\pi \approx 157\text{ }\Omega$.
RMS Current $I_{rms} = \frac{V_{rms}}{X_L} = \frac{200}{157} \approx 1.274\text{ A}$.
Peak current $I_0 = I_{rms} \times \sqrt{2} = 1.274 \times 1.414 \approx 1.80\text{ A}$.

Ans: $X_C = \frac{1}{2\pi f C} = \frac{1}{2 \times 3.14 \times 60 \times 100 \times 10^{-6}} = \frac{10^4}{37.68} \approx 26.5\text{ }\Omega$.
$I_{rms} = \frac{40}{26.5} \approx 1.51\text{ A}$.
Peak current $I_0 = I_{rms} \times \sqrt{2} = 1.51 \times 1.414 \approx 2.13\text{ A}$.

Ans: A soft iron core greatly increases the magnetic permeability ($\mu$) of the coil, which makes its Inductance ($L$) high. If withdrawn, $L$ drops significantly. Since $X_L = \omega L$, withdrawing the core causes the inductive reactance to **decrease** sharply.

Ans: RMS voltage $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{311}{1.414} \approx 220\text{ V}$.
RMS Current $I_{rms} = \frac{V_{rms}}{R} = \frac{220}{50} = 4.4\text{ A}$.

Ans: Phase lag $\phi = \pi/2$ radians (which is exactly $1/4$ of a full cycle).
Time period $T = 1/f = 1/50 = 0.02\text{ s} = 20\text{ ms}$.
Time lag $\Delta t = \frac{\phi}{2\pi} T = \frac{\pi/2}{2\pi} T = \frac{T}{4} = \frac{20}{4} = 5\text{ ms}$.

Ans: Capacitive reactance $X_C = \frac{1}{2\pi f C}$.
For steady DC, the frequency is exactly zero ($f = 0$).
Substituting $f = 0$ yields $X_C = \frac{1}{0} \to \infty$. Infinite reactance means it offers infinite opposition to current, thus perfectly blocking steady DC.

Ans: Let current $I$ be the reference horizontal phasor. Voltage across resistor $V_R = IR$ is in phase with $I$. Voltage across inductor $V_L = IX_L$ leads by $90^\circ$ (points up). Voltage across capacitor $V_C = IX_C$ lags by $90^\circ$ (points down).
The net reactive voltage is a single vertical phasor $(V_L - V_C)$.
By Pythagoras theorem, net source voltage $V = \sqrt{V_R^2 + (V_L - V_C)^2}$.
Substitute $V = IZ$: $I Z = \sqrt{(IR)^2 + (I X_L - I X_C)^2} \implies Z = \sqrt{R^2 + (X_L - X_C)^2}$.

Ans: Inductive reactance $X_L = 2\pi f L$. Since $X_L \propto f$, when frequency doubles from $50\text{Hz}$ to $100\text{Hz}$, the new reactance becomes $X_L' = 2 \times 40 = 80\text{ }\Omega$. Ohmic resistance $R$ remains $30\text{ }\Omega$.
New Impedance $Z = \sqrt{R^2 + (X_L')^2} = \sqrt{30^2 + 80^2} = \sqrt{900 + 6400} = \sqrt{7300} \approx 85.44\text{ }\Omega$.

Ans: $X_C = \frac{1}{2\pi f C} = \frac{1}{2 \times 3.14 \times 50 \times 50 \times 10^{-6}} = \frac{10^6}{15700} \approx 63.7\text{ }\Omega$.
Impedance $Z = \sqrt{10^2 + 63.7^2} = \sqrt{100 + 4057} = \sqrt{4157} \approx 64.5\text{ }\Omega$.
RMS current $I_{rms} = \frac{220}{64.5} \approx 3.41\text{ A}$.
Peak current $I_0 = I_{rms} \times \sqrt{2} = 3.41 \times 1.414 \approx 4.82\text{ A}$.

Ans: $\tan\phi = \frac{X_C}{R} = \frac{63.7}{10} = 6.37$.
Phase angle $\phi = \tan^{-1}(6.37) \approx 81^\circ$. Since it is an RC circuit, the current **leads** the voltage by approximately $81^\circ$.

Ans: Bulb safe current $I_{rms} = P/V = 10/60 = 1/6\text{ A}$. Bulb resistance $R = V/I = 60 / (1/6) = 360\text{ }\Omega$.
Required circuit impedance $Z = V_{source} / I_{rms} = 100 / (1/6) = 600\text{ }\Omega$.
$Z^2 = R^2 + X_L^2 \implies 600^2 = 360^2 + X_L^2 \implies 360000 = 129600 + X_L^2 \implies X_L^2 = 230400 \implies X_L = 480\text{ }\Omega$.
$X_L = 2\pi f L \implies 480 = 2 \times 3.14 \times 50 \times L = 314 L \implies L = 480 / 314 \approx 1.53\text{ H}$.

Ans: Both can drop voltage to control current. A rheostat uses ohmic resistance, dissipating substantial power as waste heat ($P = I^2R$). A choke coil uses inductive reactance; since voltage and current are $90^\circ$ out of phase ($\cos 90^\circ = 0$), the average power dissipated by an ideal choke is zero. It controls current without massive energy loss.

Ans: When $X_L = X_C$, the reactive components perfectly cancel each other out ($X_L - X_C = 0$).
Effective impedance $Z = \sqrt{R^2 + 0^2} = R$. This is the minimum possible impedance.
This specific electrical state is called **Resonance**.

Ans: $X_L = 2\pi f L = 2 \times 3.14 \times 100 \times 0.1 = 62.8\text{ }\Omega$.
$X_C = \frac{1}{2\pi f C} = \frac{1}{2 \times 3.14 \times 100 \times 20 \times 10^{-6}} = \frac{10^6}{12560} \approx 79.6\text{ }\Omega$.
$Z = \sqrt{10^2 + (62.8 - 79.6)^2} = \sqrt{100 + (-16.8)^2} = \sqrt{100 + 282.2} = \sqrt{382.2} \approx 19.5\text{ }\Omega$.

Ans: At resonance, imaginary part of impedance is zero: $X_L - X_C = 0 \implies X_L = X_C$.
Substitute formulas: $2\pi f_r L = \frac{1}{2\pi f_r C}$.
Rearrange for $f_r$: $(2\pi f_r)^2 = \frac{1}{LC} \implies 2\pi f_r = \frac{1}{\sqrt{LC}}$.
Final expression: $f_r = \frac{1}{2\pi\sqrt{LC}}$.

Ans: Q-factor indicates the sharpness of the resonant peak, defined physically as $2\pi \times (\text{Max Energy Stored} / \text{Energy Dissipated per cycle})$. Mathematically, $Q = \frac{\omega_r L}{R}$. Since $Q \propto 1/R$, a circuit with lower ohmic resistance $R$ dissipates less energy, resulting in a much higher Q-factor, meaning a sharper, narrower resonance curve (high selectivity).

Ans: $\omega_r = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{1.5 \times 35 \times 10^{-6}}} = \frac{1}{\sqrt{52.5 \times 10^{-6}}} \approx \frac{1}{0.007245} \approx 138\text{ rad/s}$.
Q-factor $Q = \frac{\omega_r L}{R} = \frac{138 \times 1.5}{20} = \frac{207}{20} = 10.35$.

Ans: Resonant frequency depends on the product $LC$ ($f_r \propto 1/\sqrt{LC}$). To keep $f_r$ constant, the product $LC$ must remain constant. If $C_{new} = 4C$, then $L_{new}$ must be $L/4$ so that $(L/4)(4C) = LC$. Inductance must be reduced to one-fourth of its original value.

Ans: We know $Q = \frac{\omega_r}{\Delta\omega}$.
Bandwidth $\Delta\omega = \frac{\omega_r}{Q} = \frac{138}{10.35} \approx 13.3\text{ rad/s}$.
(Alternatively, $\Delta\omega = R/L = 20 / 1.5 = 13.33\text{ rad/s}$).

Ans: At resonance (or any state in a series circuit), $V_L$ leads the current $I$ by $90^\circ$ and $V_C$ lags the current $I$ by $90^\circ$. Therefore, the absolute phase difference between $V_L$ and $V_C$ is always exactly $180^\circ$ ($\pi\text{ radians}$). They are directly out of phase.

Ans: Use $f = \frac{1}{2\pi\sqrt{LC}} \implies C = \frac{1}{4\pi^2 f^2 L}$.
For $f_{max} = 1200\text{ kHz}$: $C_{min} = \frac{1}{4(3.14)^2 (1200 \times 10^3)^2 (200 \times 10^{-6})} \approx 88\text{ pF}$.
For $f_{min} = 800\text{ kHz}$: $C_{max} = \frac{1}{4(3.14)^2 (800 \times 10^3)^2 (200 \times 10^{-6})} \approx 198\text{ pF}$.
Range of variable capacitor is $88\text{ pF}$ to $198\text{ pF}$.

Ans: At resonance, total impedance is minimum ($Z=R$), so the circuit draws a very large current ($I=V/R$). Because $X_L$ and $X_C$ can be individually very large compared to $R$, the voltage drops across them ($V_L = I X_L$, $V_C = I X_C$) can mathematically exceed the total applied source voltage $V$. They cancel each other perfectly ($V_L - V_C = 0$), so Kirchhoff's rules are not violated.

Ans: Power factor is the ratio of true power (Watts) to apparent power (Volt-Amperes), given by $\cos\phi = R/Z$.
Derivation: $P_{inst} = V I = (V_0 \sin\omega t) (I_0 \sin(\omega t - \phi)) = V_0 I_0 [\sin^2\omega t \cos\phi - \sin\omega t \cos\omega t \sin\phi]$.
Integrate over a cycle ($T$): average of $\sin^2\omega t$ is $1/2$, average of $\sin\omega t \cos\omega t$ is $0$.
$P_{avg} = \frac{V_0 I_0}{2} \cos\phi = (\frac{V_0}{\sqrt{2}})(\frac{I_0}{\sqrt{2}}) \cos\phi = V_{rms} I_{rms} \cos\phi$.

Ans: $X_L = 2\pi f L = 2 \times 3.14 \times 50 \times 0.2 = 62.8\text{ }\Omega$.
Impedance $Z = \sqrt{R^2 + X_L^2} = \sqrt{20^2 + 62.8^2} = \sqrt{400 + 3943.8} = \sqrt{4343.8} \approx 65.9\text{ }\Omega$.
Power Factor $\cos\phi = R / Z = 20 / 65.9 \approx 0.30$.

Ans: Wattless current is the perpendicular component of AC current ($I_{rms} \sin\phi$) that is $90^\circ$ out of phase with the voltage, contributing exactly zero to the average power dissipation.
Wattless component $= I_{rms} \sin\phi = 10 \sin(60^\circ) = 10 \times (\sqrt{3}/2) \approx 10 \times 0.866 = 8.66\text{ A}$.

Ans: Inserting a soft iron core into the choke heavily increases its inductance $L$, and thus its inductive reactance $X_L$. The total circuit impedance $Z = \sqrt{R_{bulb}^2 + X_L^2}$ increases significantly. Consequently, the current $I = V/Z$ drawn from the source drops, causing the bulb's brightness (which depends on $I^2R$) to **decrease**.

Ans: Active current $I \cos\phi = P/V = 1000/220 = 4.54\text{ A}$. Total current $I = 4.54/0.8 = 5.68\text{ A}$.
Reactive inductive current $I_L = I \sin\phi = 5.68 \times 0.6 = 3.41\text{ A}$.
For unity power factor, capacitor must draw an equal but leading reactive current: $I_C = 3.41\text{ A}$.
$I_C = V / X_C = V \cdot 2\pi f C \implies 3.41 = 220 \times 2 \times 3.14 \times 50 \times C \implies 3.41 = 69080 \times C$.
$C = 3.41 / 69080 \approx 4.9 \times 10^{-5}\text{ F} = 49\text{ }\mu\text{F}$.

Ans: No. The phase angle $\phi$ between voltage and current in any passive LCR circuit is strictly restricted between $-90^\circ$ (pure capacitor) and $+90^\circ$ (pure inductor). The cosine of any angle in the range $[-\pi/2, +\pi/2]$ is always positive or zero. Hence, $\cos\phi$ can never be negative.

Ans: Here, $X_L = X_C = 200\text{ }\Omega$. The reactive components perfectly cancel out. The circuit is in a state of **resonance**.
Impedance $Z = \sqrt{R^2 + 0} = R = 400\text{ }\Omega$.
Power factor $\cos\phi = R/Z = 400/400 = 1$ (Unity).

Ans: $V = V_0 \sin\omega t$. Current $I = I_0 \cos\omega t$ (since it leads by $90^\circ$).
Instantaneous power $P = VI = V_0 I_0 \sin\omega t \cos\omega t = \frac{V_0 I_0}{2} \sin(2\omega t)$.
Average power $P_{avg} = \frac{1}{T} \int_0^T \frac{V_0 I_0}{2} \sin(2\omega t) dt$.
Since the integral of a full sine wave ($\sin(2\omega t)$ completes two cycles in time $T$) over a complete period is exactly zero, $P_{avg} = 0$.

Ans: A charged capacitor stores energy in an electric field. When connected to an inductor, it discharges, building a current. This current creates a magnetic field in the inductor, transferring the energy. Once the capacitor is empty, the collapsing magnetic field induces an EMF that keeps the current flowing, recharging the capacitor with opposite polarity. This rhythmic sloshing of energy between fields constitutes LC oscillations.

Ans: Conservation of energy: Max Electric Energy = Max Magnetic Energy.
$\frac{1}{2} C V_0^2 = \frac{1}{2} L I_0^2 \implies I_0 = V_0 \sqrt{\frac{C}{L}}$.
$I_0 = 60 \times \sqrt{\frac{1.5 \times 10^{-6}}{15 \times 10^{-3}}} = 60 \times \sqrt{10^{-4}} = 60 \times 0.01 = 0.6\text{ A}$.

Ans: Let $q = Q_0 \cos\omega t$ and $i = -Q_0 \omega \sin\omega t$.
$U = U_E + U_B = \frac{q^2}{2C} + \frac{1}{2}Li^2 = \frac{Q_0^2 \cos^2\omega t}{2C} + \frac{1}{2}L (Q_0^2 \omega^2 \sin^2\omega t)$.
Since $\omega^2 = 1/LC \implies L\omega^2 = 1/C$.
$U = \frac{Q_0^2}{2C} \cos^2\omega t + \frac{Q_0^2}{2C} \sin^2\omega t = \frac{Q_0^2}{2C} (\cos^2\omega t + \sin^2\omega t) = \frac{Q_0^2}{2C}$. Since $Q_0$ and $C$ are constants, total energy $U$ is conserved.

Ans: $f = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2 \times 3.14 \times \sqrt{15 \times 10^{-3} \times 1.5 \times 10^{-6}}} = \frac{1}{6.28 \times \sqrt{22.5 \times 10^{-9}}} = \frac{1}{6.28 \times 1.5 \times 10^{-4}} = \frac{10^4}{9.42} \approx 1061\text{ Hz}$ (or $1.06\text{ kHz}$).

Ans: Equation: $\frac{d^2q}{dt^2} + \frac{1}{LC} q = 0$ is analogous to $\frac{d^2x}{dt^2} + \frac{k}{m} x = 0$.
Charge ($q$) $\leftrightarrow$ Displacement ($x$).
Current ($dq/dt$) $\leftrightarrow$ Velocity ($v$).
Inductance ($L$) $\leftrightarrow$ Mass ($m$, inertia).
Reciprocal of Capacitance ($1/C$) $\leftrightarrow$ Spring constant ($k$, stiffness).

Ans: If $R > 0$, energy is continually dissipated as Joule heat ($I^2R$). The amplitude of the oscillations (max charge and max current) will exponentially decay over time until the oscillations die out completely. This is called **Damped Oscillation**.
[Graph: A sinusoidal wave whose peak envelope exponentially shrinks toward the horizontal time axis].

Ans: Total energy $U_0 = Q_0^2/2C$. We want electric energy $U_E = U_0/2$.
$q^2/2C = \frac{1}{2} (Q_0^2/2C) \implies q^2 = Q_0^2 / 2 \implies q = Q_0 / \sqrt{2}$.
Since $q = Q_0 \cos(\omega t) \implies \cos(\omega t) = 1/\sqrt{2} \implies \omega t = \pi/4$.
Since $\omega = 2\pi/T$, we have $(2\pi/T) t = \pi/4 \implies t = T/8$.

Ans: 1. **Ohmic heating:** Practical wires and inductors have resistance, converting electrical energy into heat.
2. **Electromagnetic radiation:** Oscillating charges act as an antenna and radiate energy away into space as electromagnetic waves.

Ans: Principle: Mutual Induction. When AC flows in the primary, it creates an alternating magnetic flux in the core. This changing flux links with the secondary coil, inducing an EMF (Faraday's Law).
Derivation: $V_p = -N_p \frac{d\Phi}{dt}$ and $V_s = -N_s \frac{d\Phi}{dt}$ (assuming ideal flux linkage). Dividing them yields the transformer ratio: $\frac{V_s}{V_p} = \frac{N_s}{N_p}$.

Ans: $\frac{V_s}{V_p} = \frac{N_s}{N_p} \implies \frac{220}{2200} = \frac{N_s}{4000} \implies N_s = \frac{1}{10} \times 4000 = 400\text{ turns}$.
Efficiency $\eta = \frac{P_{out}}{P_{in}} \implies 0.9 = \frac{8\text{ kW}}{P_{in}} \implies P_{in} = \frac{8}{0.9} \approx 8.89\text{ kW}$.

Ans: Secondary current $I_s = \frac{P_{out}}{V_s} = \frac{8000\text{ W}}{220\text{ V}} \approx 36.36\text{ A}$.
Primary current $I_p = \frac{P_{in}}{V_p} = \frac{8888.89\text{ W}}{2200\text{ V}} \approx 4.04\text{ A}$.

Ans: The changing magnetic flux induces swirling "eddy currents" within the bulk iron core itself, causing massive $I^2R$ heating losses. Laminating the core (slicing it into thin sheets separated by insulating varnish) breaks the conductive paths, greatly increasing the electrical resistance to these eddy currents and minimizing the heat loss.

Ans: (a) Copper loss: Joule heating ($I^2R$) in the primary/secondary windings.
(b) Iron loss: Eddy current heating + Hysteresis (energy lost realigning magnetic domains).
(c) Flux leakage: Not all magnetic flux generated by primary links to secondary.
(d) Humming (Magnetostriction): Core expands/contracts slightly due to changing magnetic fields, creating audible sound and vibrating away energy.

Ans: $\frac{V_s}{V_p} = \frac{N_s}{N_p} \implies V_s = 120 \times \frac{10}{500} = 120 \times \frac{1}{50} = 2.4\text{ V}$.
Since $V_s < V_p$, it is a **Step-down** transformer.

Ans: Faraday's law of induction requires a *changing* magnetic flux ($d\Phi/dt \neq 0$) to induce an EMF in the secondary coil. A steady DC current creates a constant, unchanging magnetic field ($d\Phi/dt = 0$). Therefore, induced EMF in the secondary is zero.

Ans: Transformation ratio $k = \frac{V_s}{V_p} = \frac{20}{200} = \frac{1}{10} = 0.1$.
Input Power $P_{in} = V_p I_p = 200 \times 5 = 1000\text{ W}$.
Output Power $P_{out} = V_s I_s = 20 \times 40 = 800\text{ W}$.
Efficiency $\eta = \frac{P_{out}}{P_{in}} \times 100\% = \frac{800}{1000} \times 100\% = 80\%$.