Introduction

So far, we have discussed circuits with direct current (dc) sources where current flows in one direction and its magnitude may be constant or varying with time. This chapter focuses on **alternating current (ac)**, where current and voltage periodically change direction and magnitude with time. The main power supply in homes and offices is an ac voltage, typically varying sinusoidally with time. This is because ac voltages can be easily and efficiently changed to different voltage levels using transformers, making economic transmission over long distances possible.

AC circuits have unique characteristics, such as resonance, which are used in devices like radios and TVs.

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AC Voltage Applied To A Resistor

Consider a circuit with a pure resistor of resistance $R$ connected to an ac voltage source. Let the voltage source produce a sinusoidally varying potential difference given by:

$\mathbf{v = v_m \sin(\omega t)}$

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

Applying Kirchhoff's loop rule, the voltage across the resistor must equal the source voltage:

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

According to Ohm's law, the instantaneous current $i$ through the resistor is $i = v_R / R$.

$i = \frac{v_m \sin(\omega t)}{R}$

Since $R$ is constant, we can write this as:

$\mathbf{i = i_m \sin(\omega t)}$

where $i_m = v_m / R$ is the amplitude (peak current). This is the ac equivalent of Ohm's law for a resistor.

Comparing the voltage $v = v_m \sin(\omega t)$ and current $i = i_m \sin(\omega t)$, we see that both quantities vary sinusoidally with the same frequency $\omega$. They reach their zero, maximum, and minimum values at the same time instants. Thus, in a purely resistive ac circuit, the voltage and current are **in phase** with each other (phase difference is zero).

Although the average current over a complete cycle is zero, the instantaneous power dissipated in the resistor ($p = i^2 R$) is always non-negative ($i^2$ is always positive). So, there is Joule heating and energy dissipation.

The instantaneous power dissipated is $p = i^2 R = i_m^2 R \sin^2(\omega t)$.

The average power dissipated over one complete cycle is $\mathbf{P = \overline{p} = \overline{i^2 R} = i_m^2 R \overline{\sin^2(\omega t)}}$. Since the average value of $\sin^2(\omega t)$ over a full cycle is $1/2$, the average power is:

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

To express the average power in a form similar to dc power ($P=I^2R$), the concept of **root mean square (rms)** or **effective** value of ac current is introduced. The rms current $I$ is defined such that $P = I^2 R$.

$I^2 R = \frac{1}{2} i_m^2 R \implies I^2 = \frac{1}{2} i_m^2 \implies I = \sqrt{\frac{1}{2} i_m^2} = \frac{i_m}{\sqrt{2}}$.

$\mathbf{I_{rms} = I = \frac{i_m}{\sqrt{2}} \approx 0.707 i_m}$.

Similarly, the rms voltage is defined as $\mathbf{V_{rms} = V = \frac{v_m}{\sqrt{2}} \approx 0.707 v_m}$.

In terms of rms values, Ohm's law becomes $V = IR$, and average power becomes $P = IV = I^2R = V^2/R$, which are the same forms as in dc circuits. Unless otherwise specified, ac voltage and current values are given as rms values (e.g., household voltage 220 V is rms).

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Representation Of Ac Current And Voltage By Rotating Vectors — Phasors

Representing ac voltages and currents that vary sinusoidally with time using simple sine or cosine functions can become complex when dealing with circuits containing inductors, capacitors, and resistors due to phase differences. To visualise and analyse phase relationships, we use **phasors**.

A phasor is a vector that rotates counterclockwise about the origin in a diagram called a phasor diagram, with an angular speed equal to the angular frequency ($\omega$) of the ac quantity. The length of the phasor represents the amplitude (peak value) of the oscillating quantity, and its projection onto the vertical axis (or horizontal axis, by convention) represents the instantaneous value of the quantity.

For a resistor connected to an ac source $v = v_m \sin(\omega t)$, the current is $i = i_m \sin(\omega t)$.

The voltage phasor $\vec{V}$ has magnitude $v_m$ and makes an angle $\omega t$ with the horizontal axis (at time $t$). Its vertical projection is $v_m \sin(\omega t) = v$.

The current phasor $\vec{I}$ has magnitude $i_m$ and also makes an angle $\omega t$ with the horizontal axis. Its vertical projection is $i_m \sin(\omega t) = i$.

In the case of a resistor, since voltage and current are in phase, their phasors $\vec{V}$ and $\vec{I}$ are always in the same direction at any instant $t$. As they rotate together, their vertical projections generate the sinusoidal waveforms for $v$ and $i$.

Using phasors simplifies the addition of voltages and currents with phase differences, as vector addition rules can be applied to the rotating phasors.

Ac Voltage Applied To An Inductor

Consider a pure inductor with self-inductance $L$ and negligible resistance connected to an ac voltage source $v = v_m \sin(\omega t)$.

Applying Kirchhoff's loop rule, the sum of voltages around the loop is zero. The voltage across the inductor is $v_L = L (di/dt)$.

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

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

To find the current $i$, we integrate both sides with respect to time:

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

$i = \int \frac{v_m}{L} \sin(\omega t) dt = \frac{v_m}{L} \int \sin(\omega t) dt = \frac{v_m}{L} \left(-\frac{\cos(\omega t)}{\omega}\right) + C$.

Since the current oscillates symmetrically about zero, the integration constant $C$ is zero.

$i = -\frac{v_m}{\omega L} \cos(\omega t)$.

Using $-\cos(\omega t) = \sin(\omega t - \pi/2)$, we have:

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

where $i_m = \frac{v_m}{\omega L}$ is the amplitude of the current.

The term $\omega L$ plays a role similar to resistance in limiting the current amplitude and is called the **inductive reactance ($X_L$)**:

$\mathbf{X_L = \omega L}$.

The amplitude of the current is $i_m = v_m / X_L$. The dimension of inductive reactance is ohm ($\Omega$). It is directly proportional to inductance $L$ and frequency $\omega$.

Comparing $v = v_m \sin(\omega t)$ and $i = i_m \sin(\omega t - \pi/2)$, the current lags the voltage by $\pi/2$ (or 90° or one-quarter cycle). This is represented in the phasor diagram by the current phasor $\vec{I}$ being $\pi/2$ behind the voltage phasor $\vec{V}$ in their counterclockwise rotation (Fig. 7.6).

Instantaneous power supplied to the inductor: $p = vi = (v_m \sin(\omega t)) (i_m \sin(\omega t - \pi/2)) = v_m i_m \sin(\omega t) (-\cos(\omega t)) = -\frac{1}{2} v_m i_m \sin(2\omega t)$.

The average power over a complete cycle is $\mathbf{P = \overline{p} = -\frac{1}{2} v_m i_m \overline{\sin(2\omega t)} = 0}$, since the average of $\sin(2\omega t)$ over a cycle is zero.

Thus, a pure inductor consumes **zero average power** over a complete cycle. Energy is stored in the inductor's magnetic field during one quarter cycle and returned to the source during the next quarter cycle.

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AC Voltage Applied To A Capacitor

Consider a pure capacitor with capacitance $C$ connected to an ac voltage source $v = v_m \sin(\omega t)$.

Applying Kirchhoff's loop rule, the voltage across the capacitor $v_C = q/C$ must equal the source voltage $v$.

$v = v_C = q/C \implies q = Cv = Cv_m \sin(\omega t)$.

To find the instantaneous current $i$, we use $i = dq/dt$.

$i = \frac{d}{dt} (Cv_m \sin(\omega t)) = Cv_m \frac{d}{dt}(\sin(\omega t)) = Cv_m (\omega \cos(\omega t))$.

$i = \omega C v_m \cos(\omega t)$.

Using $\cos(\omega t) = \sin(\omega t + \pi/2)$, we have:

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

where $i_m = \omega C v_m$ is the amplitude of the current.

The term $\frac{1}{\omega C}$ plays a role similar to resistance in limiting the current amplitude and is called the **capacitive reactance ($X_C$)**:

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

The amplitude of the current is $i_m = v_m / X_C$. The dimension of capacitive reactance is ohm ($\Omega$). It is inversely proportional to capacitance $C$ and frequency $\omega$.

Comparing $v = v_m \sin(\omega t)$ and $i = i_m \sin(\omega t + \pi/2)$, the current leads the voltage by $\pi/2$ (or 90° or one-quarter cycle). This is represented in the phasor diagram by the current phasor $\vec{I}$ being $\pi/2$ ahead of the voltage phasor $\vec{V}$ (Fig. 7.9).

Instantaneous power supplied to the capacitor: $p_C = vi = (v_m \sin(\omega t)) (i_m \sin(\omega t + \pi/2)) = v_m i_m \sin(\omega t) (\cos(\omega t)) = \frac{1}{2} v_m i_m \sin(2\omega t)$.

The average power over a complete cycle is $\mathbf{P_C = \overline{p_C} = \frac{1}{2} v_m i_m \overline{\sin(2\omega t)} = 0}$.

Thus, a pure capacitor also consumes **zero average power** over a complete cycle. Energy is stored in the capacitor's electric field during one quarter cycle and returned to the source during the next quarter cycle.

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Ac Voltage Applied To A Series Lcr Circuit

Consider a series circuit containing a resistor ($R$), an inductor ($L$), and a capacitor ($C$) connected to an ac voltage source $v = v_m \sin(\omega t)$.

Applying Kirchhoff's loop rule, the sum of voltage drops across $R, L, C$ must equal the source voltage $v$. Let $q$ be the instantaneous charge on the capacitor and $i$ be the instantaneous current ($i = dq/dt$).

$v_R = iR$

$v_L = L \frac{di}{dt}$

$v_C = \frac{q}{C} = \frac{\int i dt}{C}$

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

This is a second-order linear differential equation. We can solve it using phasor diagrams or analytical methods.

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Phasor-Diagram Solution

In a series circuit, the instantaneous current $i$ is the same through all elements. Let's assume $i = i_m \sin(\omega t + \phi)$, where $\phi$ is the phase difference between the current and the source voltage. The rms current $I = i_m/\sqrt{2}$.

The voltage drops across each element are:

• Resistor $R$: $v_R = iR$. $v_R$ is in phase with $i$. Amplitude $v_{Rm} = i_m R$. Phasor $\vec{V}_R$ is parallel to $\vec{I}$.
• Inductor $L$: $v_L = L(di/dt)$. $v_L$ leads $i$ by $\pi/2$. Amplitude $v_{Lm} = i_m X_L = i_m (\omega L)$. Phasor $\vec{V}_L$ is $\pi/2$ ahead of $\vec{I}$.
• Capacitor $C$: $v_C = q/C = (1/C) \int i dt$. $v_C$ lags $i$ by $\pi/2$. Amplitude $v_{Cm} = i_m X_C = i_m (1/(\omega C))$. Phasor $\vec{V}_C$ is $\pi/2$ behind $\vec{I}$.

The instantaneous source voltage $v = v_R + v_L + v_C$. The corresponding phasor equation is $\vec{V} = \vec{V}_R + \vec{V}_L + \vec{V}_C$.

Since $\vec{V}_L$ and $\vec{V}_C$ are along the same line and in opposite directions, their vector sum is $\vec{V}_L + \vec{V}_C$ with magnitude $|v_{Lm} - v_{Cm}|$. The phasors $\vec{V}_R$ and $(\vec{V}_L + \vec{V}_C)$ are perpendicular.

Using the Pythagorean theorem on the phasor diagram triangle (Fig. 7.13(b)):

$\mathbf{v_m^2 = v_{Rm}^2 + (v_{Lm} - v_{Cm})^2}$.

Substituting amplitudes in terms of $i_m$:

$v_m^2 = (i_m R)^2 + (i_m X_L - i_m X_C)^2 = i_m^2 [R^2 + (X_L - X_C)^2]$.

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

By analogy to resistance, the term in the denominator is called the **impedance ($Z$)** of the series LCR circuit:

$\mathbf{Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}}$.

The amplitude of the current is $\mathbf{i_m = \frac{v_m}{Z}}$. This is the ac equivalent of Ohm's law relating voltage and current amplitudes via impedance.

The phase angle $\phi$ between the source voltage $\vec{V}$ and the current $\vec{I}$ (which is in phase with $\vec{V}_R$) is given by the impedance diagram (Fig. 7.14):

$\mathbf{\tan\phi = \frac{v_{Lm} - v_{Cm}}{v_{Rm}} = \frac{i_m X_L - i_m X_C}{i_m R} = \frac{X_L - X_C}{R}}$.

$\mathbf{\phi = \tan^{-1}\left(\frac{\omega L - 1/(\omega C)}{R}\right)}$.

If $X_L > X_C$, $\phi > 0$, the circuit is inductive, current lags voltage. If $X_C > X_L$, $\phi < 0$, circuit is capacitive, current leads voltage. If $X_L = X_C$, $\phi = 0$, circuit is resistive, current is in phase with voltage.

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Analytical Solution

The voltage equation $iR + L\frac{di}{dt} + \frac{q}{C} = v_m \sin(\omega t)$ is a differential equation for $i(t)$ or $q(t)$. Assuming a steady-state solution of the form $q(t) = q_m \sin(\omega t + \theta)$, we can find $i(t) = dq/dt$ and $di/dt = d^2q/dt^2$, substitute into the equation, and solve for $q_m$ and $\theta$. This process yields the same expressions for current amplitude and phase angle as the phasor method.

The steady-state current is $i(t) = i_m \sin(\omega t + \phi)$, where $i_m = v_m/Z$ and $\tan\phi = (X_L - X_C)/R$. The phase angle $\phi$ determines the phase difference between the voltage $v(t) = v_m \sin(\omega t)$ and the current $i(t) = i_m \sin(\omega t + \phi)$.

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Resonance

A key phenomenon in a series LCR circuit is **resonance**, which occurs when the driving frequency $\omega$ is such that the inductive reactance $X_L$ equals the capacitive reactance $X_C$.

$X_L = X_C \implies \omega L = \frac{1}{\omega C} \implies \omega^2 = \frac{1}{LC}$.

The frequency at which this occurs is called the **resonant frequency ($\omega_0$)**: $\mathbf{\omega_0 = \frac{1}{\sqrt{LC}}}$.

At resonance, the impedance $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + 0^2} = R$. The impedance is at its minimum value, equal to the resistance $R$.

The amplitude of the current at resonance is $i_m = v_m / Z = v_m / R$. This is the maximum possible current amplitude for the given source voltage amplitude $v_m$ and resistance $R$.

The phase angle $\tan\phi = (X_L - X_C)/R = 0/R = 0$. So, $\phi = 0$. At resonance, the current is **in phase** with the applied voltage. The voltages across the inductor and capacitor are equal in magnitude ($v_{Lm} = i_m X_L$, $v_{Cm} = i_m X_C$, and $X_L=X_C$) and exactly out of phase ($\pi$ apart), so their sum is zero. The entire source voltage appears across the resistor.

Resonance occurs only if both $L$ and $C$ are present in the circuit. Resonant circuits are used in tuning circuits (e.g., radios) to select a specific frequency signal by adjusting $L$ or $C$ to match the signal frequency with the circuit's resonant frequency.

Sharpness of resonance: The sharpness of the resonance peak (how quickly the current amplitude drops off as the frequency moves away from resonance) is measured by the **quality factor ($Q$)**. A higher $Q$ indicates a sharper resonance and greater selectivity of the circuit to frequencies near resonance.

The width of the resonance peak is often defined by the bandwidth, $2\Delta\omega$, where $\Delta\omega$ is the difference between the resonant frequency $\omega_0$ and the frequencies $\omega_1, \omega_2$ at which the current amplitude is $1/\sqrt{2}$ times the maximum current amplitude (where power is half the maximum). At $\omega_1$ and $\omega_2$, $|X_L - X_C| = R$.

The relationship between bandwidth and resistance is $2\Delta\omega = R/L$. The quality factor $Q$ is defined as:

$\mathbf{Q = \frac{\omega_0}{2\Delta\omega} = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 C R} = \frac{1}{R}\sqrt{\frac{L}{C}}}$.

A higher $Q$ (lower $R$ or higher $L$) means smaller bandwidth ($2\Delta\omega$), i.e., sharper resonance and better selectivity.

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Power In Ac Circuit: The Power Factor

The instantaneous power $p$ supplied by an ac source is the product of the instantaneous voltage $v$ and current $i$. For a series LCR circuit, if $v = v_m \sin(\omega t)$ and $i = i_m \sin(\omega t + \phi)$, the instantaneous power is $p = v_m i_m \sin(\omega t) \sin(\omega t + \phi)$.

The **average power** $\mathbf{P}$ over a complete cycle is the average of this instantaneous power. Calculating the average of $v_m i_m \sin(\omega t) \sin(\omega t + \phi)$ over a cycle leads to:

$\mathbf{P = \frac{1}{2} v_m i_m \cos\phi}$.

In terms of rms values ($V = v_m/\sqrt{2}$, $I = i_m/\sqrt{2}$), this becomes:

$\mathbf{P = V I \cos\phi}$.

Here, $\phi$ is the phase angle between the voltage and current, $\tan\phi = (X_L - X_C)/R$. The term $\mathbf{\cos\phi}$ is called the **power factor**. It is the cosine of the phase angle between the voltage across the source and the current in the circuit. From the impedance diagram (Fig. 7.14), $\cos\phi = R/Z$.

So, average power can also be written as $\mathbf{P = V I (R/Z)}$. Since $V=IZ$, $P = (IZ)I(R/Z) = I^2 R$. This confirms that the power is dissipated only in the resistance $R$ of the circuit.

Cases based on power factor:

• Purely resistive circuit ($\phi=0$): $\cos\phi=1$. $P=VI$. Maximum power dissipation.
• Purely inductive or capacitive circuit ($\phi=\pm\pi/2$): $\cos\phi=0$. $P=0$. No power dissipation. The current is called **wattless current**.
• LCR circuit: $\cos\phi = R/Z$. Power dissipation $P=I^2R$.
• Resonance in LCR circuit ($\phi=0$): $\cos\phi=1$. $P=I^2R$. Maximum power dissipation for a given $R$.

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Sharpness Of Resonance

The sharpness of the resonance peak in a series LCR circuit is related to the **quality factor ($Q$)**. A higher $Q$ means a sharper peak, and the circuit is more selective in responding to frequencies close to the resonant frequency. The amplitude of the current is maximum ($i_m = v_m/R$) at resonance $\omega_0$. At frequencies $\omega_1$ and $\omega_2$ where the current amplitude is $1/\sqrt{2}$ times the maximum value (half-power points), the difference $\Delta\omega = |\omega - \omega_0|$ is related to the circuit parameters. The bandwidth is $2\Delta\omega = R/L$.

The quality factor $Q$ is defined as:

$\mathbf{Q = \frac{\omega_0}{2\Delta\omega} = \frac{\omega_0 L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}}}$.

A high $Q$ (low $R$ or high $L$) leads to a narrow bandwidth and a sharper resonance, making the circuit more selective to frequencies near $\omega_0$. This is important in tuning circuits where a specific frequency signal needs to be selected from many signals.

Lc Oscillations

An LC circuit, consisting of a capacitor and an inductor, can exhibit electrical oscillations. If a charged capacitor is connected to an inductor, energy stored in the capacitor's electric field is transferred to the inductor's magnetic field, and vice versa, resulting in oscillations of charge and current in the circuit. This is analogous to the mechanical oscillation of a mass-spring system.

Consider a capacitor initially charged to $q_m$, connected to an inductor. Applying Kirchhoff's loop rule: $\frac{q}{C} + L\frac{di}{dt} = 0$. Since $i = -dq/dt$ (charge decreases as current increases), $di/dt = -d^2q/dt^2$.

$\frac{q}{C} + L\left(-\frac{d^2q}{dt^2}\right) = 0 \implies \frac{d^2q}{dt^2} + \frac{1}{LC} q = 0$.

This is the equation of simple harmonic motion for the charge $q$. The angular frequency of these oscillations is $\mathbf{\omega_0 = \frac{1}{\sqrt{LC}}}$. The charge and current vary sinusoidally with this frequency. If $q=q_m\cos(\omega_0 t)$ (assuming $q=q_m$ at $t=0$), then $i = dq/dt = -q_m \omega_0 \sin(\omega_0 t)$.

The total energy in the LC circuit is the sum of the energy stored in the capacitor's electric field ($U_E = q^2/(2C)$) and the energy stored in the inductor's magnetic field ($U_B = \frac{1}{2}LI^2$).

During oscillations, energy is continuously transferred between the capacitor and the inductor. At times when the capacitor is fully charged, $i=0$ and $U_B=0$, so $U_{total} = U_E = q_m^2/(2C)$. At times when the current is maximum, $q=0$ and $U_E=0$, so $U_{total} = U_B = \frac{1}{2}Li_m^2$. The total energy remains constant if there is no resistance or radiation loss.

Analogy between mechanical (mass-spring) and electrical (LC) oscillations:

Mechanical system	Electrical system
Mass $m$	Inductance $L$
Spring constant $k$	Reciprocal capacitance $1/C$
Displacement $x$	Charge $q$
Velocity $v=dx/dt$	Current $i=dq/dt$
Mechanical energy $E = \frac{1}{2}kx^2 + \frac{1}{2}mv^2$	Electromagnetic energy $U = \frac{1}{2}\frac{q^2}{C} + \frac{1}{2}Li^2$
Natural frequency $\omega_0 = \sqrt{k/m}$	Natural frequency $\omega_0 = \sqrt{1/LC}$

In reality, resistance (even in wiring) causes damping, dissipating energy as heat, and oscillations eventually die out. Energy is also radiated away as electromagnetic waves.

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Transformers

A **transformer** is a device used to change (transform) an alternating voltage from one value to another, either stepped up (increased) or stepped down (decreased). It operates on the principle of **mutual induction**.

A transformer consists of a soft-iron core on which two coils are wound: a **primary coil** with $N_p$ turns and a **secondary coil** with $N_s$ turns. The coils are insulated from each other and from the core. The primary coil is connected to the ac voltage source, and the secondary coil is connected to the load or output circuit.

When an ac voltage $v_p$ is applied to the primary coil, it drives a current that produces a continually changing magnetic flux $\phi$ in the core. This flux passes through the core and links both the primary and secondary windings. According to Faraday's law of induction, this changing flux induces an emf in both coils.

The induced emf in the secondary coil ($e_s$) with $N_s$ turns is $e_s = -N_s \frac{d\phi}{dt}$. If the secondary is connected to a load, the voltage across it is approximately $v_s \approx e_s$.

The induced emf in the primary coil ($e_p$) with $N_p$ turns is $e_p = -N_p \frac{d\phi}{dt}$. In an ideal transformer (negligible primary resistance, all flux links both coils), the applied primary voltage $v_p$ is almost equal and opposite to the induced back emf in the primary, $v_p \approx -e_p$.

So, $v_s \approx -N_s \frac{d\phi}{dt}$ and $v_p \approx -(-N_p \frac{d\phi}{dt}) = N_p \frac{d\phi}{dt}$.

Dividing the two equations (assuming $d\phi/dt$ is the same for both):

$\mathbf{\frac{v_s}{v_p} = \frac{-N_s (d\phi/dt)}{N_p (d\phi/dt)} = \frac{N_s}{N_p}}$.

This is the **transformer equation**, relating the primary and secondary voltages to the number of turns in each coil. Since the voltage and current are sinusoidal, this relationship also holds for their rms values ($V_s/V_p = N_s/N_p$).

For an ideal transformer (100% efficient, no energy losses), the power input in the primary equals the power output in the secondary ($P_{in} = P_{out}$). $P = vi$, so $v_p i_p = v_s i_s$.

$\mathbf{v_p i_p = v_s i_s}$ or $\mathbf{V_p I_p = V_s I_s}$ (using rms values).

Combining the voltage and power relations:

$\mathbf{\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}}$.

• If $N_s > N_p$, then $V_s > V_p$ (voltage is stepped up), and $I_s < I_p$ (current is stepped down). This is a **step-up transformer**.
• If $N_s < N_p$, then $V_s < V_p$ (voltage is stepped down), and $I_s > I_p$ (current is stepped up). This is a **step-down transformer**.

Actual transformers have energy losses due to: (i) Flux leakage (not all flux links both coils), (ii) Resistance of windings (Joule heating), (iii) Eddy currents in the core (induced currents dissipate heat), (iv) Hysteresis (energy loss due to repeated magnetisation/demagnetisation of the core). These losses are minimised by design (e.g., laminated core, using materials with low hysteresis loss).

Transformers are essential for long-distance power transmission. Voltage is stepped up at the generator (reducing current, thus reducing $I^2R$ loss in lines), transmitted at high voltage, and stepped down at sub-stations for distribution and use.

Summary

This chapter covers the principles and applications of alternating current (AC).

• AC voltage/current varies sinusoidally with time. $v = v_m \sin(\omega t)$, $i = i_m \sin(\omega t + \phi)$. rms values $V=v_m/\sqrt{2}, I=i_m/\sqrt{2}$ are typically used.
• Resistor ($R$): $v_R = iR$, $V_R = IR$. $v_R$ is in phase with $i$. $X_R = R$. Average power $P = I^2 R$.
• Pure Inductor ($L$): $v_L = L(di/dt)$, $V_L = IX_L$. $i$ lags $v_L$ by $\pi/2$. $X_L = \omega L$. Average power $P = 0$.
• Pure Capacitor ($C$): $v_C = q/C$, $V_C = IX_C$. $i$ leads $v_C$ by $\pi/2$. $X_C = 1/(\omega C)$. Average power $P = 0$.
• Series LCR circuit: $v = v_R + v_L + v_C$. Impedance $Z = \sqrt{R^2 + (X_L - X_C)^2}$. $I = V/Z$. Phase angle $\phi = \tan^{-1}((X_L - X_C)/R)$ between $v$ and $i$.
• Phasors are rotating vectors representing ac quantities for visualising phase relationships.
• Resonance in series LCR: $X_L = X_C \implies \omega_0 = 1/\sqrt{LC}$. Minimum impedance $Z=R$, maximum current $I_{max} = V/R$. $\phi=0$. Sharpness measured by Quality factor $Q = \omega_0 L/R$.
• Power in AC circuit: Average power $P = VI\cos\phi = I^2R$. $\cos\phi$ is the power factor ($R/Z$). Wattless current ($\phi = \pm \pi/2$) dissipates no power.
• LC oscillations: Energy oscillates between $L$ and $C$ at natural frequency $\omega_0 = 1/\sqrt{LC}$. Total energy is conserved (ideally).
• Transformer: Changes ac voltage using mutual induction. $\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}$ (ideal). Step-up ($N_s > N_p$) or step-down ($N_s < N_p$). Losses occur in real transformers. Essential for power transmission.

Exercises

Questions covering calculations involving resistance, reactance, impedance, current, voltage, power, and phase angles in ac circuits with resistors, inductors, capacitors, and series LCR combinations. Includes topics like resonance, LC oscillations, and transformer principles. Problems often involve applying Ohm's law equivalent, impedance calculations, power factor, and energy storage/dissipation in ac circuits.

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