Vardaan Learning Institute

Ans: Mean value of AC over a half-cycle is that steady DC current which would send the exact same amount of charge through a circuit in the time of half a cycle ($T/2$) as is sent by the AC in the same time.
Relation: $I_{mean} = \frac{2I_0}{\pi} \approx 0.637 I_0$.

Ans: Heat produced in time $dt$ is $dH = i^2 R dt = (I_0 \sin\omega t)^2 R dt$.
Total heat in one cycle ($T$): $H = \int_0^T I_0^2 R \sin^2(\omega t) dt = I_0^2 R \int_0^T \left[\frac{1 - \cos(2\omega t)}{2}\right] dt$.
Integrating gives $H = \frac{I_0^2 R T}{2}$. By definition of RMS, $H = I_{rms}^2 R T$.
Equating: $I_{rms}^2 R T = \frac{I_0^2 R T}{2} \implies I_{rms}^2 = \frac{I_0^2}{2} \implies I_{rms} = \frac{I_0}{\sqrt{2}}$.

Ans: "$220\text{V}$ AC" refers to the RMS value. Its peak value is much higher: $V_0 = \sqrt{2} \times V_{rms} = 1.414 \times 220 \approx 311\text{ V}$. A $220\text{V}$ DC supply stays constantly at $220\text{V}$. Since the AC voltage reaches $311\text{V}$ twice every cycle, the electric shock from it is significantly more severe.

Ans: Comparing with standard $I = I_0 \sin(2\pi f t)$:
Peak current $I_0 = 50\text{ A}$. $RMS$ current $I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{50}{\sqrt{2}} = 25\sqrt{2}\text{ A} \approx 35.35\text{ A}$.
Angular frequency $\omega = 100\pi \implies 2\pi f = 100\pi \implies f = 50\text{ Hz}$.

Ans: A moving coil galvanometer measures the *average* value of current. Since the average value of a sinusoidal alternating current over one complete cycle is exactly zero, the pointer simply vibrates very slightly around the zero mark (due to inertia) and reads a net zero.

Ans: $V_{avg} = \frac{\int_0^T V dt}{\int_0^T dt} = \frac{\int_0^T V_0 \sin(\omega t) dt}{T} = \frac{V_0}{T} \left[ \frac{-\cos(\omega t)}{\omega} \right]_0^T$.
$= \frac{-V_0}{\omega T} [\cos(\omega T) - \cos(0)]$. Since $\omega T = 2\pi$, $\cos(2\pi) = 1$ and $\cos(0) = 1$.
$V_{avg} = \frac{-V_0}{\omega T} [1 - 1] = 0$.

Ans: The peak value of the current is the amplitude of the cosine wave, so $I_0 = 4\text{ A}$. (The phase angle $\pi/4$ does not affect the RMS value).
$I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}\text{ A} \approx 2.828\text{ A}$.

Ans: AC can be easily and efficiently stepped up to very high voltages using transformers. Transmitting power at high voltage drastically reduces the current ($P=VI$), which fundamentally minimizes $I^2R$ (Joule heating) power losses in the transmission cables. DC cannot be stepped up using simple transformers.

Ans: Applied Voltage: $V = V_0 \sin(\omega t)$.
Resulting Current: $I = I_0 \sin(\omega t + \pi/2)$.
This shows that the current leads the voltage by a phase angle of $90^\circ$ ($\pi/2\text{ rad}$).

Ans: The phase difference is exactly $0^\circ$ (Voltage and Current are perfectly in phase).
Equations: $V = V_0 \sin(\omega t)$ and $I = I_0 \sin(\omega t)$.

Ans: In a pure inductor, the phase difference between voltage and current is $\phi = 90^\circ$. Average power $P_{avg} = V_{rms} I_{rms} \cos\phi = V_{rms} I_{rms} \cos(90^\circ)$. Since $\cos(90^\circ) = 0$, $P_{avg} = 0$. Energy is just alternately stored in the magnetic field and returned to the source.

Ans: Peak voltage $V_0 = 200\text{ V}$. Peak current $I_0 = \frac{V_0}{R} = \frac{200}{100} = 2\text{ A}$.
RMS current $I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}\text{ A} \approx 1.414\text{ A}$.

Ans: Applied $V = V_0 \sin(\omega t)$. Back EMF in inductor is $-L(di/dt)$. By Kirchhoff's loop rule, $V - L(di/dt) = 0 \implies di = \frac{V}{L} dt = \frac{V_0}{L} \sin(\omega t) dt$.
Integrate: $i = \int \frac{V_0}{L} \sin(\omega t) dt = \frac{-V_0}{\omega L} \cos(\omega t)$.
Since $-\cos(\omega t) = \sin(\omega t - \pi/2)$, we get $i = I_0 \sin(\omega t - \pi/2)$, where $I_0 = V_0/(\omega L)$.

Ans: Capacitive reactance is $X_C = \frac{1}{2\pi f C}$. For steady DC, frequency $f = 0$, making $X_C = \infty$ (infinite opposition), so it blocks DC. For AC, $f > 0$, making $X_C$ finite and relatively low, allowing the alternating charge and discharge current to "pass" through the circuit.

Ans: $V_{rms} = 110\text{ V}$. $X_C = \frac{1}{2\pi f C} = \frac{1}{2 \times 3.14 \times 60 \times 60 \times 10^{-6}} = \frac{10^6}{22608} \approx 44.2\text{ }\Omega$.
$I_{rms} = \frac{V_{rms}}{X_C} = \frac{110}{44.2} \approx 2.49\text{ A}$.

Ans: $X_L = 2\pi f L = 2 \times 3.14 \times 50 \times 25 \times 10^{-3} = 314 \times 25 \times 10^{-3} = 7.85\text{ }\Omega$.
$I_{rms} = \frac{V_{rms}}{X_L} = \frac{220}{7.85} \approx 28.0\text{ A}$.

Ans: $X_L = 2\pi f L$. Higher frequency means faster changing current, which induces a stronger opposing back EMF, hence higher reactance (straight line).
$X_C = \frac{1}{2\pi f C}$. Higher frequency means the capacitor charges/discharges faster without ever fully opposing the source, hence reactance drops (hyperbola).

Ans: **Impedance ($Z$)** is the total opposition offered by an AC circuit (containing R, L, C) to current flow. Unit: Ohm ($\Omega$).
**Admittance ($Y$)** is the reciprocal of Impedance ($Y = 1/Z$), measuring how easily a circuit allows AC to flow. Unit: Siemens ($\text{S}$) or $\Omega^{-1}$.

Ans: [Student draws a right-angled triangle]. Base is horizontal, labeled $R$. Perpendicular side goes upwards, labeled $(X_L - X_C)$. Hypotenuse is $Z = \sqrt{R^2 + (X_L - X_C)^2}$. The angle between base $R$ and hypotenuse $Z$ is $\phi$ (phase angle).

Ans: Susceptance ($B$) is the reciprocal of Reactance ($X$). It is the imaginary part of admittance. Formula: $B = 1/X$. SI unit: Siemens ($\text{S}$) or $\Omega^{-1}$.

Ans: In a series circuit, current $I$ is the reference. $V_L$ leads current $I$ by $90^\circ$. $V_C$ lags current $I$ by $90^\circ$. Therefore, the total phase difference between $V_L$ and $V_C$ is $90^\circ + 90^\circ = 180^\circ$ ($\pi\text{ rad}$). They are exactly out of phase.

Ans: $X_C = \frac{1}{2\pi f C} = \frac{1}{2 \times 3.14 \times 50 \times 15 \times 10^{-6}} = \frac{10^6}{4710} \approx 212.3\text{ }\Omega$.
Circuit is RC series ($X_L = 0$).
$Z = \sqrt{R^2 + X_C^2} = \sqrt{200^2 + 212.3^2} = \sqrt{40000 + 45071} = \sqrt{85071} \approx 291.7\text{ }\Omega$.

Ans: $I_{rms} = \frac{V_{rms}}{Z} = \frac{220}{291.7} \approx 0.754\text{ A}$.

Ans: At very low frequencies ($f \to 0$), $X_C \to \infty$, making the circuit behave like an open circuit (blocks current). At very high frequencies ($f \to \infty$), $X_L \to \infty$, also causing massive impedance and blocking current. Hence, current only passes well at intermediate (resonant) frequencies.

Ans: Condition: Inductive reactance equals capacitive reactance ($X_L = X_C$).
$\implies 2\pi f_r L = \frac{1}{2\pi f_r C}$.
Cross-multiplying: $(2\pi f_r)^2 LC = 1 \implies 2\pi f_r = \frac{1}{\sqrt{LC}} \implies f_r = \frac{1}{2\pi\sqrt{LC}}$.

Ans: Sharpness indicates high selectivity (ability to tune into a very specific narrow frequency band). A smaller resistance ($R$) produces a sharper, taller peak because less energy is dissipated, leading to a much higher Q-factor. A large $R$ severely damps the circuit, flattening the curve.

Ans: Q-factor is the voltage magnification at resonance, defined as the ratio of voltage across L (or C) to the applied voltage at resonance.
Formula: $Q = \frac{1}{R} \sqrt{\frac{L}{C}}$.

Ans: Inserting an iron core drastically increases the inductance ($L$) of the coil due to high permeability. Since resonant frequency $f_r = 1 / (2\pi\sqrt{LC})$, an increase in $L$ will cause the resonant frequency $f_r$ to **decrease**.

Ans: Bandwidth is the difference between the upper and lower half-power frequencies ($\omega_2 - \omega_1$). It is the range of frequencies over which current is at least $1/\sqrt{2}$ of its max value.
Relation: $Q = \frac{\omega_r}{\text{Bandwidth}} \implies \text{Bandwidth} = \frac{\omega_r}{Q}$.

Ans: $\omega_r = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{2.0 \times 32 \times 10^{-6}}} = \frac{1}{\sqrt{64 \times 10^{-6}}} = \frac{1}{8 \times 10^{-3}} = \frac{1000}{8} = 125\text{ rad/s}$.

Ans: $Q = \frac{\omega_r L}{R} = \frac{125 \times 2.0}{10} = \frac{250}{10} = 25$. (Alternatively: $Q = \frac{1}{R}\sqrt{\frac{L}{C}} = \frac{1}{10}\sqrt{\frac{2}{32 \times 10^{-6}}} = 25$).

Ans: At the resonant frequency, the circuit's impedance is minimum ($Z=R$), allowing maximum current to flow. Thus, it strongly "accepts" the AC signal of that specific frequency while largely blocking all other frequencies, making it ideal for radio tuning.

Ans: Power factor is the ratio of true power to apparent power ($\cos\phi = R/Z$). Its maximum possible value is $1$ (Unity). This occurs in a purely resistive circuit, or in a series LCR circuit perfectly tuned to resonance where $X_L = X_C$ and thus $Z=R$.

Ans: Instantaneous power $P = VI = (V_0 \sin\omega t)(I_0 \sin(\omega t - \phi)) = V_0 I_0 \sin\omega t (\sin\omega t \cos\phi - \cos\omega t \sin\phi)$.
$P = V_0 I_0 (\sin^2\omega t \cos\phi - \frac{1}{2}\sin 2\omega t \sin\phi)$.
Average over one cycle: $\langle \sin^2\omega t \rangle = 1/2$ and $\langle \sin 2\omega t \rangle = 0$.
$P_{avg} = \frac{V_0 I_0}{2} \cos\phi = (\frac{V_0}{\sqrt{2}}) (\frac{I_0}{\sqrt{2}}) \cos\phi = V_{rms} I_{rms} \cos\phi$.

Ans: Wattless current is the component of alternating current ($I_{rms} \sin\phi$) that does not consume any active power ($P_{avg} = 0$). It exists strictly in purely inductive or purely capacitive ideal circuits where the phase difference $\phi = \pm 90^\circ$.

Ans: A rheostat relies on high ohmic resistance, which reduces current but dissipates massive amounts of electrical energy as wasted heat ($I^2R$). A choke coil has high inductance but very low resistance, reducing current mainly via high inductive reactance without dissipating significant active power ($\cos\phi \approx 0$).

Ans: $V_0 = 100\text{ V}$, $I_0 = 100\text{ mA} = 0.1\text{ A}$. Phase angle $\phi = \pi/3 = 60^\circ$.
$P_{avg} = \frac{V_0 I_0}{2} \cos\phi = \frac{100 \times 0.1}{2} \cos(60^\circ) = 5 \times 0.5 = 2.5\text{ Watts}$.

Ans: First find $Z = \sqrt{R^2 + X_L^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\text{ }\Omega$.
Power factor $\cos\phi = R / Z = 8 / 10 = 0.8$.

Ans: In a pure capacitor, $\phi = 90^\circ$. $P_{avg} = V_{rms} I_{rms} \cos(90^\circ)$. Since $\cos(90^\circ) = 0$, $P_{avg} = 0$. During half the cycle, energy is absorbed from the source to charge the plates, and during the next half, it is fully returned to the source as it discharges.

Ans: By connecting a suitable Capacitor bank in parallel with the inductive circuit. The capacitor draws a leading reactive current which cancels out the lagging reactive current drawn by the inductors (motors), bringing the total phase angle $\phi$ closer to zero, thus making $\cos\phi$ closer to $1$.

Ans: A charged capacitor stores electrostatic energy. When connected to an inductor, it discharges, creating a growing current that builds a magnetic field in the inductor (transferring energy). Once fully discharged, the collapsing magnetic field keeps the current flowing, recharging the capacitor with opposite polarity. This cycle repeats endlessly.

Ans: Equation (Kirchhoff's loop rule): $-L \frac{d^2q}{dt^2} - \frac{q}{C} = 0 \implies \frac{d^2q}{dt^2} + \frac{1}{LC} q = 0$.
This is a standard SHM equation where the natural angular frequency is $\omega_0 = \frac{1}{\sqrt{LC}}$.

Ans: Charge $q = Q_0 \cos\omega t$, current $i = -dq/dt = Q_0 \omega \sin\omega t$.
Total $U = U_E + U_B = \frac{q^2}{2C} + \frac{1}{2}Li^2$.
$U = \frac{Q_0^2 \cos^2\omega t}{2C} + \frac{1}{2}L (Q_0^2 \omega^2 \sin^2\omega t)$. Since $\omega^2 = 1/LC \implies L\omega^2 = 1/C$.
$U = \frac{Q_0^2 \cos^2\omega t}{2C} + \frac{Q_0^2 \sin^2\omega t}{2C} = \frac{Q_0^2}{2C} (\cos^2\omega t + \sin^2\omega t) = \frac{Q_0^2}{2C} = \text{Constant}$.

Ans: Comparing $\frac{d^2q}{dt^2} + \frac{1}{LC} q = 0$ with $\frac{d^2x}{dt^2} + \frac{k}{m} x = 0$.
Charge ($q$) is analogous to displacement ($x$).
Inductance ($L$) acts as electrical inertia, so it is analogous to Mass ($m$).
The inverse of capacitance ($1/C$) acts as the electrical stiffness, analogous to Spring constant ($k$).

Ans: Maximum electrostatic energy equals maximum magnetic energy.
$\frac{1}{2} C V_0^2 = \frac{1}{2} L I_0^2 \implies I_0 = V_0 \sqrt{\frac{C}{L}} = 50 \sqrt{\frac{1.5 \times 10^{-6}}{20 \times 10^{-3}}} = 50 \sqrt{0.075 \times 10^{-3}} = 50 \sqrt{75 \times 10^{-6}} \approx 50 \times 8.66 \times 10^{-3} \approx 0.433\text{ A}$.

Ans: $f = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2 \times 3.14 \times \sqrt{10^{-3} \times 10^{-9}}} = \frac{1}{6.28 \times \sqrt{10^{-12}}} = \frac{1}{6.28 \times 10^{-6}} = \frac{10^6}{6.28} \approx 1.59 \times 10^5\text{ Hz}$ (or $159\text{ kHz}$).

Ans: 1. **Ohmic Loss:** Every real inductor and connecting wire has some non-zero electrical resistance ($R$). This constantly dissipates energy as Joule heat ($I^2R$). 2. **Radiation Loss:** The oscillating charges act like an antenna, radiating away a portion of the energy as electromagnetic waves.

Ans: Zero. By the principle of conservation of energy, when the current is at its absolute maximum, all the system's energy is stored purely in the magnetic field ($1/2 L I_0^2$), meaning the electrostatic energy must be zero, hence charge is zero.

Ans: Principle: Mutual Induction (changing magnetic flux in a primary coil induces an EMF in a magnetically linked secondary coil). It absolutely cannot step up/down steady DC, because DC produces a constant, non-changing magnetic field, yielding zero induced EMF.

Ans: Step-up transformer: Secondary turns > Primary turns ($N_s/N_p > 1$). Output voltage is higher ($V_s > V_p$).
Step-down transformer: Secondary turns < Primary turns ($N_s/N_p < 1$). Output voltage is lower ($V_s < V_p$).

Ans: 1. Copper Loss ($I^2R$ heat): Minimized by using thick copper wires. 2. Eddy Current Loss (heating of iron core): Minimized by using a laminated (sliced and insulated) iron core. 3. Hysteresis Loss (magnetic reversal friction): Minimized by using Soft Iron, which has a narrow hysteresis loop.

Ans: Copper loss is the energy wasted as heat ($I^2R$) due to the inherent electrical resistance of the primary and secondary windings. For the coil carrying high current (the low-voltage side), very thick copper wires are used because thicker wires have lower resistance ($R \propto 1/A$).

Ans: $\frac{V_s}{V_p} = \frac{N_s}{N_p} \implies \frac{220}{2200} = \frac{N_s}{2000}$.
$\frac{1}{10} = \frac{N_s}{2000} \implies N_s = \frac{2000}{10} = 200\text{ turns}$.

Ans: Secondary current $I_s = \frac{V_s}{R} = \frac{220}{22} = 10\text{ A}$.
For ideal transformer, power input = power output. $V_p I_p = V_s I_s \implies 2200 \times I_p = 220 \times 10 = 2200$.
$I_p = 2200 / 2200 = 1\text{ A}$.

Ans: Transmitted power $P = VI$. For a fixed power demand, stepping up transmission voltage $V$ by a massive factor proportionally reduces the transmission current $I$. Since line power loss is given by $I^2 R_{line}$, a small current drastically minimizes these expensive $I^2R$ heat losses over long distances.

Ans: Soft iron is highly ferromagnetic, so it efficiently channels and links almost $100\%$ of the magnetic flux from the primary to the secondary (minimizing flux leakage). More importantly, it has a very narrow hysteresis loop, meaning very little energy is lost when the AC magnetic field rapidly reverses direction.