AC Circuits with Individual Components (R, L, C)

Ac Voltage Applied To A Resistor

In DC circuits, we use Ohm's Law ($V=IR$) to relate voltage, current, and resistance. In AC circuits, the voltage and current are not constant; they vary sinusoidally with time. When an alternating voltage is applied across a circuit component, an alternating current flows through it. We need to understand the relationship between the instantaneous voltage and instantaneous current in AC circuits, including their magnitudes and phase difference.

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Circuit Setup

Consider a circuit containing a pure resistor of resistance $R$ connected across an AC voltage source. The alternating voltage supplied by the source is represented by:

$ v(t) = V_0 \sin(\omega t) $

Where $v(t)$ is the instantaneous voltage at time $t$, $V_0$ is the peak voltage (amplitude), and $\omega$ is the angular frequency of the AC source.

A pure resistor connected to an AC voltage source.

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Instantaneous Current through the Resistor

According to Ohm's Law, the instantaneous current $i(t)$ through the resistor at any instant is given by the instantaneous voltage across it divided by its resistance:

$ i(t) = \frac{v(t)}{R} $

Substitute the expression for $v(t)$:

$ i(t) = \frac{V_0 \sin(\omega t)}{R} = \left(\frac{V_0}{R}\right) \sin(\omega t) $

The term $V_0/R$ represents the maximum value (amplitude) of the instantaneous current. Let $I_0 = V_0/R$.

$ i(t) = I_0 \sin(\omega t) $

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Relationship between Voltage and Current

Comparing the expressions for instantaneous voltage and current:

$ v(t) = V_0 \sin(\omega t) $

$ i(t) = I_0 \sin(\omega t) $

Both $v(t)$ and $i(t)$ are sinusoidal functions of time with the same angular frequency $\omega$. Importantly, the arguments of the sine functions are the same ($\omega t$). This means that the current reaches its maximum value, zero value, and minimum value exactly at the same time instants as the voltage. There is no phase difference between the voltage and the current in a pure resistive AC circuit.

In a pure resistor, the AC voltage and the AC current are in phase.

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Peak and RMS Values

The relationship between the peak voltage ($V_0$) and peak current ($I_0$) is still given by Ohm's Law:

$ I_0 = \frac{V_0}{R} \quad \text{or} \quad V_0 = I_0 R $

For AC quantities, the Root Mean Square (RMS) values are often used because they represent the equivalent DC values that would produce the same heating effect. For a sinusoidal waveform, the RMS value is the peak value divided by $\sqrt{2}$.

$ V_{rms} = \frac{V_0}{\sqrt{2}} $

$ I_{rms} = \frac{I_0}{\sqrt{2}} $

Dividing the peak value relationship by $\sqrt{2}$:

$ \frac{I_0}{\sqrt{2}} = \frac{V_0/R}{\sqrt{2}} \implies I_{rms} = \frac{V_{rms}}{R} $

$ V_{rms} = I_{rms} R $

Ohm's Law is valid for both peak values and RMS values in a resistive AC circuit.

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Power Dissipation in a Resistor

The instantaneous power dissipated in the resistor is given by $p(t) = v(t) i(t)$.

$ p(t) = (V_0 \sin(\omega t)) (I_0 \sin(\omega t)) = V_0 I_0 \sin^2(\omega t) $

Using $V_0 = I_0 R$, $p(t) = I_0^2 R \sin^2(\omega t)$.

The power pulsates between 0 and $V_0 I_0$ (or $I_0^2 R$) and is always positive (since $\sin^2(\omega t)$ is always $\ge 0$), meaning energy is always being delivered to the resistor and dissipated as heat.

The average power dissipated over one complete cycle is:

$ P_{avg} = \langle V_0 I_0 \sin^2(\omega t) \rangle $

The average value of $\sin^2(\omega t)$ over a cycle is $1/2$.

$ P_{avg} = V_0 I_0 \left\langle \sin^2(\omega t) \right\rangle = V_0 I_0 \left(\frac{1}{2}\right) $

$ P_{avg} = \left(\frac{V_0}{\sqrt{2}}\right) \left(\frac{I_0}{\sqrt{2}}\right) = V_{rms} I_{rms} $

Using $V_{rms} = I_{rms} R$:

$ P_{avg} = I_{rms}^2 R = \frac{V_{rms}^2}{R} $

The average power dissipated in a resistor in an AC circuit is given by the same formulas as in DC circuits, but using the RMS values of voltage and current. This is why RMS values are widely used in AC power calculations. For example, the 220 V mains supply in India refers to the RMS voltage.

Representation Of Ac Current And Voltage By Rotating Vectors — Phasors

Analysing AC circuits with components like inductors and capacitors becomes more complicated because of the phase differences between voltage and current. Simple scalar algebra used in DC circuits is insufficient. Adding and subtracting sinusoidal quantities that are out of phase requires using trigonometry, which can be cumbersome. A very convenient and powerful technique to represent and analyse sinusoidal AC voltages and currents is by using phasors.

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What are Phasors?

A phasor is a rotating vector that represents a sinusoidally varying quantity like AC voltage or current. The length of the phasor represents the amplitude (peak value) of the sinusoidal quantity, and its angle with respect to a reference axis (usually the horizontal axis) represents its phase.

Consider a sinusoidal voltage $v(t) = V_0 \sin(\omega t + \phi)$. This voltage can be represented by a phasor $\vec{V}$ of length $V_0$. This vector is assumed to rotate in a counter-clockwise direction in an imaginary plane (called the phasor diagram) with an angular velocity $\omega$ equal to the angular frequency of the AC source.

A voltage phasor $\vec{V}$ rotating in a phasor diagram.

At any instant of time $t$, the projection of this rotating phasor onto the vertical axis (or y-axis) gives the instantaneous value of the voltage:

$ v(t) = V_0 \sin(\omega t + \phi) $

If the initial phase is $\phi=0$, then $v(t) = V_0 \sin(\omega t)$, and the phasor is along the horizontal axis at $t=0$. Its projection on the vertical axis is $V_0 \sin(\omega t)$. Some representations use the projection on the horizontal axis ($V_0 \cos(\omega t)$), but the vertical projection convention aligns with the sine function often used for AC voltage sources.

Similarly, an AC current $i(t) = I_0 \sin(\omega t + \phi_i)$ can be represented by a phasor $\vec{I}$ of length $I_0$, rotating with the same angular velocity $\omega$. The angle of the current phasor relative to the voltage phasor represents the phase difference between the current and voltage.

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Phasor Diagram

A phasor diagram is a snapshot of the rotating phasors at a particular instant in time, usually at $t=0$ or when one of the quantities is at its peak. The relative angles between the phasors represent the phase differences between the quantities they represent.

For example, if voltage is $v(t) = V_0 \sin(\omega t)$ and current is $i(t) = I_0 \sin(\omega t - \pi/2)$ (meaning current lags voltage by $90^\circ$), the phasor diagram at $t=0$ would show the voltage phasor along the horizontal axis (since $\sin(0)=0$, but phase is 0) and the current phasor $90^\circ$ clockwise from the voltage phasor (since $\sin(-\pi/2)=-1$, phase is $-\pi/2$). Both phasors rotate counter-clockwise at angular speed $\omega$.

Phasor diagram showing voltage phasor $\vec{V}$ and current phasor $\vec{I}$ with current lagging voltage by phase angle $\phi$.

The angle between the phasors remains constant as they rotate because they rotate at the same angular velocity $\omega$.

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Using Phasors for AC Circuit Analysis

Phasors allow us to represent sinusoidal quantities as complex numbers or simply as vectors in a plane. Addition and subtraction of sinusoidal quantities with phase differences can be done by vector addition/subtraction of their corresponding phasors.

For example, if $v_1(t) = V_{01} \sin(\omega t + \phi_1)$ and $v_2(t) = V_{02} \sin(\omega t + \phi_2)$, the sum $v(t) = v_1(t) + v_2(t)$ can be found by adding the corresponding voltage phasors $\vec{V}_1$ and $\vec{V}_2$ vectorially. The resultant phasor $\vec{V} = \vec{V}_1 + \vec{V}_2$ will have a magnitude $V_0$ and angle $\phi$ such that $v(t) = V_0 \sin(\omega t + \phi)$.

This greatly simplifies the analysis of AC circuits, especially series and parallel combinations of R, L, and C components, where phase relationships are crucial. We can use vector diagrams (phasor diagrams) instead of complex trigonometric identities.

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Answer:

Ac Voltage Applied To An Inductor

Now let's analyse the behaviour of a pure inductor (a coil with negligible resistance) when connected to an AC voltage source. This is important for understanding circuits containing inductive components.

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Circuit Setup

Consider a circuit containing a pure inductor of inductance $L$ connected across an AC voltage source. The alternating voltage supplied is $v(t) = V_0 \sin(\omega t)$.

A pure inductor connected to an AC voltage source.

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Instantaneous Current through the Inductor

The voltage across an inductor is related to the rate of change of current through it by the self-induction formula:

$ v(t) = L \frac{di(t)}{dt} $ (Magnitude; ignoring the negative sign related to back EMF direction for simplicity in calculation here)

Substitute the expression for $v(t)$:

$ V_0 \sin(\omega t) = L \frac{di(t)}{dt} $

To find the current $i(t)$, we need to integrate this equation with respect to time:

$ di(t) = \frac{V_0}{L} \sin(\omega t) \, dt $

$ \int di(t) = \int \frac{V_0}{L} \sin(\omega t) \, dt $

$ i(t) = \frac{V_0}{L} \int \sin(\omega t) \, dt $

The integral of $\sin(\omega t)$ is $-\frac{1}{\omega}\cos(\omega t)$.

$ i(t) = \frac{V_0}{L} \left(-\frac{1}{\omega}\cos(\omega t)\right) + \text{constant} $

In AC circuits, the DC component (constant) is zero, so we can ignore the constant of integration.

$ i(t) = -\frac{V_0}{\omega L} \cos(\omega t) $

We can express $-\cos(\omega t)$ in terms of sine using the identity $-\cos x = \sin(x - \pi/2)$ or $\sin(x + \pi/2) = \cos x$. Using $\sin(x - \pi/2) = -\cos x$:

$ i(t) = \frac{V_0}{\omega L} \sin(\omega t - \pi/2) $

The term $\frac{V_0}{\omega L}$ represents the maximum value (amplitude) of the instantaneous current. Let $I_0 = \frac{V_0}{\omega L}$.

$ i(t) = I_0 \sin(\omega t - \pi/2) $

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Relationship between Voltage and Current (Phase)

Comparing the expressions for instantaneous voltage and current:

$ v(t) = V_0 \sin(\omega t) $

$ i(t) = I_0 \sin(\omega t - \pi/2) $

Both $v(t)$ and $i(t)$ are sinusoidal functions of time with the same angular frequency $\omega$. However, their phases are different. The phase of the current is $\omega t - \pi/2$, which is $\pi/2$ radians ($90^\circ$) less than the phase of the voltage $\omega t$.

In a pure inductor, the AC current lags the AC voltage by a phase angle of $\pi/2$ radians ($90^\circ$). Equivalently, the voltage leads the current by $\pi/2$ radians ($90^\circ$).

Voltage and current waveforms for a pure inductor. Current lags voltage by 90°.

Phasor diagram for a pure inductor. $\vec{I}$ lags $\vec{V}$ by 90°.

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Inductive Reactance ($X_L$)

From the peak value relationship $I_0 = \frac{V_0}{\omega L}$, we can write $V_0 = I_0 (\omega L)$. This looks similar to Ohm's Law ($V_0 = I_0 R$), but the term $\omega L$ replaces resistance $R$.

The quantity $\omega L$ is called the inductive reactance ($X_L$). It represents the opposition offered by an inductor to the flow of alternating current.

$ X_L = \omega L = 2\pi f L $

The unit of inductive reactance is the same as resistance, the ohm ($\Omega$).

The relationship between peak voltage and current (or RMS voltage and current) in a pure inductive circuit is:

$ V_0 = I_0 X_L \quad \text{or} \quad V_{rms} = I_{rms} X_L $

Inductive reactance is directly proportional to both the frequency of the AC source ($\omega$ or $f$) and the inductance ($L$). At higher frequencies or larger inductance, the inductor offers more opposition to the AC current. For DC current ($\omega=0$), $X_L = 0$, meaning a pure inductor offers zero opposition to steady DC current (it behaves like a short circuit after any transient effects).

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Power in an Inductor

The instantaneous power delivered to the inductor is $p(t) = v(t) i(t)$.

$ p(t) = (V_0 \sin(\omega t)) (I_0 \sin(\omega t - \pi/2)) $

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

$ p(t) = V_0 I_0 \sin(\omega t) (-\cos(\omega t)) = -V_0 I_0 \sin(\omega t) \cos(\omega t) $

Using the identity $\sin(2x) = 2 \sin x \cos x$, so $\sin(\omega t)\cos(\omega t) = \frac{1}{2}\sin(2\omega t)$:

$ p(t) = -\frac{1}{2} V_0 I_0 \sin(2\omega t) $

The instantaneous power varies sinusoidally at twice the frequency of the voltage or current. The average power delivered to a pure inductor over one complete cycle is the average value of $\sin(2\omega t)$, which is zero.

$ P_{avg} = \langle -\frac{1}{2} V_0 I_0 \sin(2\omega t) \rangle = 0 $

In a pure inductor, energy is alternately stored in the magnetic field (when power is positive) and returned to the source (when power is negative) over each cycle. There is no net consumption of energy. This is because a pure inductor has no resistance to dissipate heat. Energy is conserved; it is just exchanged between the source and the inductor's magnetic field.

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Answer:

Ac Voltage Applied To A Capacitor

Finally, let's analyse the behaviour of a pure capacitor (an ideal capacitor with no resistance) when connected to an AC voltage source. Capacitors behave differently in AC circuits compared to DC circuits.

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Circuit Setup

Consider a circuit containing a pure capacitor of capacitance $C$ connected across an AC voltage source. The alternating voltage supplied is $v(t) = V_0 \sin(\omega t)$.

A pure capacitor connected to an AC voltage source.

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Instantaneous Current through the Capacitor

The voltage across a capacitor is related to the charge stored on it by $q(t) = C v(t)$.

Substitute the expression for $v(t)$:

$ q(t) = C V_0 \sin(\omega t) $

The instantaneous current $i(t)$ through the capacitor is the rate of change of charge on its plates:

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

Differentiate $q(t)$ with respect to time:

$ i(t) = \frac{d}{dt}(C V_0 \sin(\omega t)) $

Since $C$ and $V_0$ are constants:

$ i(t) = C V_0 \frac{d}{dt}(\sin(\omega t)) $

The derivative of $\sin(\omega t)$ with respect to $t$ is $\cos(\omega t) \cdot \omega$.

$ i(t) = C V_0 \omega \cos(\omega t) $

$ i(t) = \omega C V_0 \cos(\omega t) $

We can express $\cos(\omega t)$ in terms of sine using the identity $\cos x = \sin(x + \pi/2)$.

$ i(t) = \omega C V_0 \sin(\omega t + \pi/2) $

The term $\omega C V_0$ represents the maximum value (amplitude) of the instantaneous current. Let $I_0 = \omega C V_0$.

$ i(t) = I_0 \sin(\omega t + \pi/2) $

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Relationship between Voltage and Current (Phase)

Comparing the expressions for instantaneous voltage and current:

$ v(t) = V_0 \sin(\omega t) $

$ i(t) = I_0 \sin(\omega t + \pi/2) $

Both are sinusoidal with the same frequency. The phase of the current is $\omega t + \pi/2$, which is $\pi/2$ radians ($90^\circ$) greater than the phase of the voltage $\omega t$.

In a pure capacitor, the AC current leads the AC voltage by a phase angle of $\pi/2$ radians ($90^\circ$). Equivalently, the voltage lags the current by $\pi/2$ radians ($90^\circ$).

Voltage and current waveforms for a pure capacitor. Current leads voltage by 90°.

Phasor diagram for a pure capacitor. $\vec{I}$ leads $\vec{V}$ by 90°.

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Capacitive Reactance ($X_C$)

From the peak value relationship $I_0 = \omega C V_0$, we can write $V_0 = I_0 \left(\frac{1}{\omega C}\right)$. This term $\frac{1}{\omega C}$ replaces resistance $R$.

The quantity $\frac{1}{\omega C}$ is called the capacitive reactance ($X_C$). It represents the opposition offered by a capacitor to the flow of alternating current.

$ X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} $

The unit of capacitive reactance is also the ohm ($\Omega$).

The relationship between peak voltage and current (or RMS voltage and current) in a pure capacitive circuit is:

$ V_0 = I_0 X_C \quad \text{or} \quad V_{rms} = I_{rms} X_C $

Capacitive reactance is inversely proportional to both the frequency of the AC source ($\omega$ or $f$) and the capacitance ($C$). At higher frequencies or larger capacitance, the capacitor offers less opposition to the AC current. For DC current ($\omega=0$), $X_C = \infty$, meaning a pure capacitor offers infinite opposition to steady DC current (it blocks DC).

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Power in a Capacitor

The instantaneous power delivered to the capacitor is $p(t) = v(t) i(t)$.

$ p(t) = (V_0 \sin(\omega t)) (I_0 \sin(\omega t + \pi/2)) $

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

$ p(t) = V_0 I_0 \sin(\omega t) \cos(\omega t) $

Using the identity $\sin(2x) = 2 \sin x \cos x$, so $\sin(\omega t)\cos(\omega t) = \frac{1}{2}\sin(2\omega t)$:

$ p(t) = \frac{1}{2} V_0 I_0 \sin(2\omega t) $

The instantaneous power varies sinusoidally at twice the frequency, just like in an inductor, but with a phase shift. The average power delivered to a pure capacitor over one complete cycle is the average value of $\sin(2\omega t)$, which is zero.

$ P_{avg} = \left\langle \frac{1}{2} V_0 I_0 \sin(2\omega t) \right\rangle = 0 $

In a pure capacitor, energy is alternately stored in the electric field (when power is positive) and returned to the source (when power is negative) over each cycle. There is no net consumption of energy. Energy is exchanged between the source and the capacitor's electric field.

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