Alternating Current

AC VOLTAGE APPLIED TO A RESISTOR

**Definition**: An **alternating voltage (AC)** is a potential difference that varies sinusoidally with time, represented as:

$$v = v_m \sin(\omega t)$$

where $v_m$ is the **peak/amplitude voltage** and $\omega$ is the **angular frequency** (rad/s).

**Current in a Pure Resistor**:

When AC voltage is applied across a resistor, applying Kirchhoff's loop rule:

$$v = iR$$

$$v_m \sin(\omega t) = iR$$

$$i = i_m \sin(\omega t)$$

where $i_m = \frac{v_m}{R}$ (Ohm's law holds for AC as well).

**Key Point**: In a **pure resistor, voltage and current are in phase** — they reach maximum, minimum, and zero values simultaneously. Phase angle = 0°.

Power Dissipation in AC Resistor

**Instantaneous Power**:

$$p = i^2 R = i_m^2 R \sin^2(\omega t)$$

**Average Power over One Cycle**:

$$\bar{P} = i_m^2 R \langle \sin^2(\omega t) \rangle$$

Using the trigonometric identity $\sin^2(\omega t) = \frac{1}{2}(1 - \cos 2\omega t)$ and noting that $\langle \cos 2\omega t \rangle = 0$:

$$\langle \sin^2(\omega t) \rangle = \frac{1}{2}$$

$$\bar{P} = \frac{1}{2} i_m^2 R$$

Root Mean Square (RMS) Values

The **RMS current** is defined as:

$$I = \frac{i_m}{\sqrt{2}} = 0.707 \, i_m$$

The **RMS voltage** is defined as:

$$V = \frac{v_m}{\sqrt{2}} = 0.707 \, v_m$$

**Physical Meaning**: The RMS value is the **equivalent DC current/voltage that produces the same average power** as the AC current/voltage.

**Average Power in terms of RMS values**:

$$\bar{P} = I^2 R = \frac{V^2}{R} = VI$$

This is identical to DC power formulas, showing why RMS values are preferred.

**Real-life Example**: Household mains supply rated at 220 V is an RMS value. The peak voltage is:

$$v_m = \sqrt{2} \times 220 = 311 \text{ V}$$

**Example 7.1 Working**:

Power P = 100 W, Voltage V = 220 V (RMS)

Resistance: $R = \frac{V^2}{P} = \frac{(220)^2}{100} = 484 \, \Omega$

Peak voltage: $v_m = \sqrt{2} \times 220 = 311 \text{ V}$

RMS current: $I = \frac{P}{V} = \frac{100}{220} = 0.454 \text{ A}$

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PHASORS AND PHASOR DIAGRAMS

**Definition**: A **phasor** is a rotating vector that rotates about the origin with angular speed $\omega$. The vertical component of a phasor at any instant represents the instantaneous value of the sinusoidally varying quantity.

**Key Properties**:

**Magnitude** of phasor = amplitude (peak value) of the AC quantity

**Angle** from reference axis = phase angle at that instant

**Vertical projection** = instantaneous value of the quantity

**Advantages of Phasor Representation**:

Visually shows phase relationships between voltage and current

Simplifies analysis of AC circuits with multiple elements

Phase difference between two quantities = angle between their phasors

**For a Pure Resistor**: Voltage and current phasors are aligned in the **same direction** (phase angle = 0°), confirming they are in phase.

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AC VOLTAGE APPLIED TO AN INDUCTOR

**Definition**: A **pure inductor** is an ideal coil with inductance L but negligible resistance.

Voltage-Current Relationship

Applying Kirchhoff's loop rule to a circuit with inductor and AC source:

$$v - L\frac{di}{dt} = 0$$

$$L\frac{di}{dt} = v_m \sin(\omega t)$$

$$\frac{di}{dt} = \frac{v_m}{L} \sin(\omega t)$$

**Integrating to find current**:

$$i = \int \frac{v_m}{L} \sin(\omega t) \, dt = -\frac{v_m}{\omega L} \cos(\omega t)$$

Using the identity $-\cos(\omega t) = \sin(\omega t - \frac{\pi}{2})$:

$$i = i_m \sin(\omega t - \frac{\pi}{2})$$

where $i_m = \frac{v_m}{\omega L}$

**Key Observation**: **In an inductor, current lags voltage by $\pi/2$ or 90°** — current reaches its peak one-quarter period after voltage.

Inductive Reactance

**Definition**: **Inductive reactance** ($X_L$) is the effective "resistance" offered by an inductor to AC current:

$$X_L = \omega L$$

**Formula for Current Amplitude**:

$$i_m = \frac{v_m}{X_L}$$

**Properties**:

SI unit: Ohm (Ω)

Directly proportional to frequency and inductance

At DC (f = 0): $X_L = 0$ (inductor acts as short circuit)

At very high frequency: $X_L$ becomes very large (acts as open circuit)

Power in an Inductor

**Instantaneous Power**:

$$p = iv = i_m \sin(\omega t - \frac{\pi}{2}) \times v_m \sin(\omega t)$$

$$p = -i_m v_m \cos(\omega t) \sin(\omega t) = -\frac{i_m v_m}{2} \sin(2\omega t)$$

**Average Power**:

$$\bar{P}_L = \left\langle -\frac{i_m v_m}{2} \sin(2\omega t) \right\rangle = 0$$

**Important Conclusion**: **An ideal inductor consumes ZERO average power**. It stores energy during one quarter-cycle and returns it during the next quarter-cycle.

**Example 7.2 Working**:

L = 25.0 mH = 25 × 10⁻³ H, f = 50 Hz, V = 220 V (RMS)

$\omega = 2\pi f = 2\pi(50) = 314.16 \text{ rad/s}$

$X_L = \omega L = 314.16 \times 25 \times 10^{-3} = 7.85 \, \Omega$

$I = \frac{V}{X_L} = \frac{220}{7.85} = 28 \text{ A}$

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AC VOLTAGE APPLIED TO A CAPACITOR

**Definition**: A **pure capacitor** is an ideal capacitor with capacitance C but no resistance or inductance.

Voltage-Current Relationship

For a capacitor: $v = \frac{q}{C}$

Applying Kirchhoff's loop rule:

$$v_m \sin(\omega t) = \frac{q}{C}$$

$$q = Cv_m \sin(\omega t)$$

**Current** is the rate of charge flow:

$$i = \frac{dq}{dt} = Cv_m \omega \cos(\omega t)$$

Using the identity $\cos(\omega t) = \sin(\omega t + \frac{\pi}{2})$:

$$i = i_m \sin(\omega t + \frac{\pi}{2})$$

where $i_m = \omega C v_m$

**Key Observation**: **In a capacitor, current leads voltage by $\pi/2$ or 90°** — current reaches its peak one-quarter period before voltage.

Capacitive Reactance

**Definition**: **Capacitive reactance** ($X_C$) is the effective "resistance" offered by a capacitor to AC current:

$$X_C = \frac{1}{\omega C}$$

**Formula for Current Amplitude**:

$$i_m = \frac{v_m}{X_C}$$

**Properties**:

SI unit: Ohm (Ω)

Inversely proportional to frequency and capacitance

At DC (f = 0): $X_C = \infty$ (capacitor blocks DC, acts as open circuit)

At high frequency: $X_C$ becomes small (acts as short circuit)

Power in a Capacitor

**Instantaneous Power**:

$$p = iv = i_m \sin(\omega t + \frac{\pi}{2}) \times v_m \sin(\omega t)$$

$$p = i_m v_m \cos(\omega t) \sin(\omega t) = \frac{i_m v_m}{2} \sin(2\omega t)$$

**Average Power**:

$$\bar{P}_C = \left\langle \frac{i_m v_m}{2} \sin(2\omega t) \right\rangle = 0$$

**Important Conclusion**: **A capacitor consumes ZERO average power**, similar to an inductor.

**Example 7.4 Working**:

C = 15.0 µF, f = 50 Hz, V = 220 V (RMS)

$\omega = 2\pi f = 314.16 \text{ rad/s}$

$X_C = \frac{1}{\omega C} = \frac{1}{314.16 \times 15 \times 10^{-6}} = 212 \, \Omega$

$I = \frac{V}{X_C} = \frac{220}{212} = 1.04 \text{ A}$ (RMS)

$i_m = \sqrt{2} \times 1.04 = 1.47 \text{ A}$ (peak)

If frequency doubles: $X_C$ halves, current doubles

**Example 7.3 Insight**: A capacitor in series with a lamp and connected to DC source: lamp does **not glow** (capacitor blocks DC). With AC source: lamp **glows** (capacitor allows AC current). Reducing capacitance increases $X_C$, decreasing current and brightness.

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AC VOLTAGE APPLIED TO A SERIES LCR CIRCUIT

**Circuit Description**: Resistor (R), Inductor (L), and Capacitor (C) connected in series with an AC source.

Equation of Motion

Applying Kirchhoff's loop rule to a series LCR circuit:

$$v = L\frac{di}{dt} + iR + \frac{q}{C}$$

Since $i = \frac{dq}{dt}$:

$$v_m \sin(\omega t) = L\frac{d^2q}{dt^2} + R\frac{dq}{dt} + \frac{q}{C}$$

Or in terms of current:

$$v_m \sin(\omega t) = L\frac{di}{dt} + iR + \frac{q}{C}$$

**Solution Form**: The steady-state current is:

$$i = i_m \sin(\omega t + \phi)$$

where $\phi$ is the **phase angle** between voltage and current.

Phasor Representation of LCR Circuit

Since current is the same throughout the series circuit, we take **current as reference** and plot phasors for:

1. **Voltage across Resistor** ($V_R$):

$V_R = iR$ (in phase with current)

Phasor along positive x-axis (reference)

Amplitude: $V_R = I R$

2. **Voltage across Inductor** ($V_L$):

$V_L = i X_L$ (leads current by 90°)

Phasor along positive y-axis

Amplitude: $V_L = I X_L$

3. **Voltage across Capacitor** ($V_C$):

$V_C = i X_C$ (lags current by 90°)

Phasor along negative y-axis

Amplitude: $V_C = I X_C$

Impedance and Current

**Net Reactance**:

$$X = X_L - X_C = \omega L - \frac{1}{\omega C}$$

**Impedance** (Z):

Total opposition to AC current, combining resistance and reactance:

$$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + X^2}$$

**RMS Current Amplitude**:

$$I = \frac{V}{Z}$$

$$I = \frac{V}{\sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}}$$

**Phase Angle** (between voltage and current):

$$\tan(\phi) = \frac{X_L - X_C}{R} = \frac{\omega L - \frac{1}{\omega C}}{R}$$

**Interpretation of Phase Angle**:

If $X_L > X_C$: $\phi > 0$ (current lags voltage) — **inductive circuit**

If $X_L < X_C$: $\phi < 0$ (current leads voltage) — **capacitive circuit**

If $X_L = X_C$: $\phi = 0$ (current in phase with voltage) — **resonance condition**

Peak Voltage Across Circuit

From phasor diagram, the total applied voltage is:

$$V_m = \sqrt{V_R^2 + (V_L - V_C)^2}$$

$$V_m = i_m \sqrt{R^2 + (X_L - X_C)^2}$$

$$V_m = i_m Z$$

$$v_m = \sqrt{(iR)^2 + (X_L - X_C)^2}$$

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RESONANCE IN SERIES LCR CIRCUIT

**Definition**: **Resonance** occurs when the impedance is minimum and current is maximum.

Condition for Resonance

At resonance, the net reactance is zero:

$$X_L - X_C = 0$$

$$\omega_r L = \frac{1}{\omega_r C}$$

$$\omega_r^2 = \frac{1}{LC}$$

**Resonant Angular Frequency**:

$$\omega_r = \frac{1}{\sqrt{LC}}$$

**Resonant Frequency** (in Hz):

$$f_r = \frac{1}{2\pi\sqrt{LC}}$$

Properties at Resonance

1. **Impedance is minimum**:

$$Z_{min} = R$$

2. **Current is maximum**:

$$I_{max} = \frac{V}{R}$$

3. **Voltage and current are in phase**: $\phi = 0°$

4. **Voltages across L and C are equal**: $V_L = V_C$, but they cancel out

5. **Power factor** = $\cos(0°) = 1$ (maximum)

6. **Power dissipation is maximum**: $P = I^2 R = \frac{V^2}{R}$

Quality Factor (Q-factor)

The **quality factor** measures the sharpness of resonance peak:

$$Q = \frac{\omega_r L}{R} = \frac{1}{\omega_r RC} = \frac{1}{R}\sqrt{\frac{L}{C}}$$

**High Q-factor**: Sharp resonance peak, narrow bandwidth

**Low Q-factor**: Broad resonance peak, wide bandwidth

**Bandwidth**:

$$\Delta f = \frac{f_r}{Q}$$

**Real-life Example**: In a **radio receiver**, tuning to different stations uses the resonance property of LC circuits. Varying capacitance C changes $f_r$, allowing different frequencies to resonate and be amplified.

Power in LCR Circuit

**Average Power Dissipated** (only in resistor):

$$P = VI \cos(\phi) = I^2 R$$

where $\cos(\phi)$ is the **power factor**.

**Power Factor**:

$$\cos(\phi) = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + (X_L - X_C)^2}}$$

**Reactive Power** (stored and returned, no net dissipation):

$$Q = VI \sin(\phi) = I^2 (X_L - X_C)$$

**Apparent Power**:

$$S = VI = I^2 Z$$

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WATTLESS OR REACTIVE CURRENT

When a circuit has reactance but no resistance (or negligible), the **power factor** $\cos(\phi)

eq 0$ but current exists with $\phi = ±90°$.

**Definition**: A **wattless current** (or **reactive current**) is one that produces **zero average power** while still flowing through the circuit.

**Examples**:

Pure inductor: Current lags voltage by 90°, no power dissipation

Pure capacitor: Current leads voltage by 90°, no power dissipation

**Real-life Impact**: In industrial AC power systems, inductive loads create wattless current, which:

Increases current without doing useful work

Increases transmission losses

Reduces power factor of the system

**Power Factor Correction**: Capacitors are added in parallel with inductive loads to increase power factor toward 1.

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AC GENERATOR (ELEMENTARY TREATMENT)

**Definition**: An **AC generator** converts mechanical energy into electrical energy by rotating a coil in a magnetic field.

Working Principle

**Faraday's Law of Electromagnetic Induction**:

When a rectangular coil of area A rotates with constant angular velocity $\omega$ in a uniform magnetic field B, the magnetic flux through it varies as:

$$\Phi = NBA \cos(\omega t)$$

**Induced EMF** (by Faraday's law):

$$\mathcal{E} = -\frac{d\Phi}{dt} = NBA\omega \sin(\omega t) = \mathcal{E}_m \sin(\omega t)$$

where $\mathcal{E}_m = NBA\omega$ is the **peak EMF**.

RMS EMF

The RMS value of the generated EMF:

$$\mathcal{E}_{rms} = \frac{\mathcal{E}_m}{\sqrt{2}}$$

**Important Components**:

**Coil/Armature**: The rotating coil that cuts magnetic field lines

**Magnetic field**: Usually from permanent magnets or electromagnets

**Slip rings and brushes**: Mechanical contacts to extract current from the rotating coil

**Commutator**: In DC generator; absent in AC generator

**Frequency of Generated AC**:

$$f = \frac{\omega}{2\pi} = \frac{N \times \omega}{2\pi}$$

where N is the number of poles.

**Real-life Example**: Household electricity is generated by rotating coils in the stator of power generators at power plants, producing 50 Hz in India.

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TRANSFORMER

**Definition**: A **transformer** is a device that changes the voltage and current of an AC signal while keeping power nearly constant, using electromagnetic induction.

Construction

A transformer consists of:

1. **Primary coil** (input): N₁ turns, carries input voltage V₁ and current I₁

2. **Secondary coil** (output): N₂ turns, carries output voltage V₂ and current I₂

3. **Soft iron core**: Provides magnetic coupling between coils

Principle of Operation

When AC voltage V₁ is applied to the primary coil:

$$\Phi = \Phi_m \sin(\omega t)$$

**Induced EMF in Primary**:

$$\mathcal{E}_1 = -N_1 \frac{d\Phi}{dt} = -N_1 \Phi_m \omega \cos(\omega t)$$

**Induced EMF in Secondary**:

$$\mathcal{E}_2 = -N_2 \frac{d\Phi}{dt} = -N_2 \Phi_m \omega \cos(\omega t)$$

Voltage Transformation Equation

For an **ideal transformer** (no losses):

$$\frac{V_2}{V_1} = \frac{N_2}{N_1}$$

**Transformation Ratio**:

$$k = \frac{N_2}{N_1}$$

If **N₂ > N₁** (k > 1): **Step-up transformer** — increases voltage, decreases current

If **N₂ < N₁** (k < 1): **Step-down transformer** — decreases voltage, increases current

Current Relationship

For an ideal transformer, **power is conserved**:

$$V_1 I_1 = V_2 I_2$$

$$\frac{I_2}{I_1} = \frac{V_1}{V_2} = \frac{N_1}{N_2} = \frac{1}{k}$$

Impedance Transformation

A transformer changes the **apparent resistance/impedance** seen by the primary:

$$Z_{eq} = \left(\frac{N_1}{N_2}\right)^2 Z_{load}$$

This is the **impedance matching** property, useful in audio and RF systems.

Energy Losses in Real Transformer

Real transformers have losses:

1. **Copper Loss** (I²R loss): Due to resistance of coil windings

$$P_{Cu} = I_1^2 R_1 + I_2^2 R_2$$

2. **Core Loss** (Iron Loss): Due to hysteresis and eddy currents in iron core

$$P_{core} = P_{hysteresis} + P_{eddy}$$

3. **Dielectric Loss**: Negligible

Efficiency of Transformer

**Definition**:

$$\eta = \frac{\text{Output Power}}{\text{Input Power}} = \frac{V_2 I_2 \cos(\phi_2)}{V_1 I_1 \cos(\phi_1)}$$

For an ideal transformer: **η = 100%**

For a real transformer: **η = 95-98%** (typically)

**Efficiency formula**:

$$\eta = \frac{P_{out}}{P_{in}} = \frac{P_{in} - P_{losses}}{P_{in}} = 1 - \frac{P_{losses}}{P_{in}}$$

**Practical Example**: A 220 V/110 V step-down transformer supplying a 10 Ω load.

Primary voltage: 220 V

Secondary voltage: 110 V

Transformation ratio: 220/110 = 2

If secondary current is 5 A, primary current = 5/2 = 2.5 A

Power: 220 × 2.5 = 550 W (approximately)

Transformer Equation (Faraday's Form)

$$V_1 = N_1 \frac{d\Phi}{dt} \quad \text{and} \quad V_2 = N_2 \frac{d\Phi}{dt}$$

This directly gives the turns ratio relationship.

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ENERGY LOSS IN TRANSMISSION AND DISTRIBUTION

**Why AC is Preferred for Transmission**:

1. **Voltage step-up at generating station** using step-up transformer: High voltage, low current

2. **Transmission at high voltage**: Reduces I²R losses in transmission lines

3. **Voltage step-down at distribution** using step-down transformer: Safe voltage for consumers

**Transmission Loss Calculation**:

Power loss in transmission lines:

$$P_{loss} = I^2 R = \left(\frac{P}{V}\right)^2 R = \frac{P^2 R}{V^2}$$

where P is power transmitted, V is transmission voltage, R is line resistance.

**Key Insight**: To minimize losses, voltage should be as **high as possible** (reduces current for same power).

**Example**:

Transmit 1000 kW at 100 V (current = 10,000 A): Loss = (10,000)² × R = huge

Transmit 1000 kW at 10,000 V (current = 100 A): Loss = (100)² × R = 1% of power (if R is appropriate)

This is why power grids use stepping up/down transformers.

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AC GENERATOR AND TRANSFORMER (QUALITATIVE)

Both devices are based on **electromagnetic induction** and **Faraday's law**.

**AC Generator**: Converts mechanical rotation → AC voltage

**Transformer**: Converts one AC voltage level → another AC voltage level

Both require:

Changing magnetic flux

Multiple coil turns to accumulate induced EMF

AC signal (DC doesn't work in transformers; a DC motor is different)

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BOARD-IMPORTANT FORMULAS AND RESULTS

| Quantity | Formula | Remarks |

|----------|---------|----------|

| Inductive Reactance | $X_L = \omega L$ | Increases with frequency |

| Capacitive Reactance | $X_C = \frac{1}{\omega C}$ | Decreases with frequency |

| Impedance (LCR) | $Z = \sqrt{R^2 + (X_L - X_C)^2}$ | Vector combination |

| RMS Current | $I = \frac{V}{Z}$ | For LCR circuit |

| Phase Angle | $\tan(\phi) = \frac{X_L - X_C}{R}$ | Relative to voltage |

| Power Factor | $\cos(\phi) = \frac{R}{Z}$ | Always ≤ 1 |

| Average Power | $P = VI\cos(\phi) = I^2 R$ | Only in resistor |

| Resonant Frequency | $f_r = \frac{1}{2\pi\sqrt{LC}}$ | Minimum impedance |

| Maximum Current | $I_{max} = \frac{V}{R}$ | At resonance |

| Transformer Ratio | $\frac{V_2}{V_1} = \frac{N_2}{N_1}$ | Voltage transformation |

| Transformer Current | $\frac{I_2}{I_1} = \frac{N_1}{N_2}$ | Inverse ratio |

| Transformer Efficiency | $\eta = \frac{P_{out}}{P_{in}}$ | Real: 95-98% |

| RMS Value | $V = \frac{V_m}{\sqrt{2}} = 0.707 V_m$ | From peak value |

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KEY POINTS FOR EXAMINATION

1. **Voltage-Current Phase Relationships**:

Resistor: In phase (φ = 0°)

Inductor: Current lags by 90°

Capacitor: Current leads by 90°

2. **Average Power Dissipation**:

Only resistor dissipates power: P = I²R

Inductor and capacitor: Zero average power

3. **Impedance Concept**:

Combines resistance and reactance

Must use vector (Pythagorean) combination, not simple addition

4. **Resonance**:

Occurs when $X_L = X_C$

Impedance minimum, current maximum, power factor = 1

Used in radio tuning and frequency selection

5. **Transformer Principles**:

Voltage inversely proportional to turns ratio

Current proportional to turns ratio

Power conserved (ideal case)

Only works with AC

6. **RMS vs Peak Values**:

RMS = Peak/√2

RMS used in power calculations and AC ratings

Household mains: 220 V is RMS value

7. **Power Factor Importance**:

$\cos(\phi)$ determines how much of apparent power is actually used

Industries correct low power factor using capacitors

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NUMERICAL PROBLEMS WITH FULL WORKING

**Problem 1**: An inductor of 100 mH, resistor of 100 Ω, and capacitor of 10 µF are connected in series with a 220 V, 50 Hz source. Find: (a) impedance; (b) current; (c) power dissipated; (d) power factor; (e) whether circuit is inductive or capacitive.

**Solution**:

Given: L = 100 mH = 0.1 H, R = 100 Ω, C = 10 µF = 10 × 10⁻⁶ F, V = 220 V, f = 50 Hz

$\omega = 2\pi f = 2\pi(50) = 314.16$ rad/s

(a) **Inductive Reactance**:

$$X_L = \omega L = 314.16 × 0.1 = 31.42 \, \Omega$$

**Capacitive Reactance**:

$$X_C = \frac{1}{\omega C} = \frac{1}{314.16 × 10 × 10^{-6}} = \frac{1}{0.003142} = 318.3 \, \Omega$$

**Impedance**:

$$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{(100)^2 + (31.42 - 318.3)^2}$$

$$Z = \sqrt{10,000 + (-286.88)^2} = \sqrt{10,000 + 82,300} = \sqrt{92,300} = 303.8 \, \Omega$$

(b) **Current**:

$$I = \frac{V}{Z} = \frac{220}{303.8} = 0.724 \text{ A}$$

(c) **Power Dissipated**:

$$P = I^2 R = (0.724)^2 × 100 = 0.525 × 100 = 52.5 \text{ W}$$

(d) **Power Factor**:

$$\cos(\phi) = \frac{R}{Z} = \frac{100}{303.8} = 0.329$$

(e) Since $X_C > X_L$: Circuit is **capacitive** (current leads voltage)

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**Problem 2**: At resonance in a series LCR circuit with L = 1 H and C = 100 µF, the impedance is 10 Ω. Find: (a) resonant frequency; (b) resistance; (c) quality factor.

**Solution**:

At resonance: Z = R (impedance = resistance only)

So, **R = 10 Ω**

(a) **Resonant Frequency**:

$$f_r = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{1 × 100 × 10^{-6}}}$$

$$f_r = \frac{1}{2\pi\sqrt{10^{-4}}} = \frac{1}{2\pi × 0.01} = \frac{1}{0.0628} = 15.92 \text{ Hz}$$

(b) **Resistance = 10 Ω** (given as impedance at resonance)

(c) **Quality Factor**:

$$\omega_r = 2\pi f_r = 2\pi(15.92) = 100 \text{ rad/s}$$

$$Q = \frac{\omega_r L}{R} = \frac{100 × 1}{10} = 10$$

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**Problem 3**: A transformer has 500 turns in the primary and 10 turns in the secondary. If primary is connected to 220 V supply, find: (a) secondary voltage; (b) if secondary supplies 10 A current to a load, what is primary current (ideal transformer)?

**Solution**:

$N_1 = 500$, $N_2 = 10$, $V_1 = 220$ V, $I_2 = 10$ A

(a) **Secondary Voltage**:

$$\frac{V_2}{V_1} = \