Physics: Class 12 (Book 1) — Chapter 7: Alternating Current

• Efficient Voltage Transformation: Alternating voltages can be easily and efficiently stepped up or stepped down using transformers .
• Economical Transmission: By stepping up the voltage, electrical energy can be transmitted over long distances at high voltage and low current, significantly reducing $I^2 R$ power losses in transmission lines.
• Widespread Device Compatibility: Most modern industrial and household appliances are designed to operate directly on ac voltage.

The instantaneous ac voltage is given by:

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

where $v_m$ is the amplitude of the oscillating potential difference and $\omega$ is its angular frequency.

Applying Kirchhoff's loop rule ($\sum \epsilon(t) = 0$) to the circuit:

$$v_m \sin \omega t = i R$$

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

Since $R$ is constant, we can write:

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

where the current amplitude $i_m$ is given by Ohm's law :

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

• Phase: Comparing the equations for $v$ and $i$, we see that both reach their zero, minimum, and maximum values at the same time. Therefore, the voltage and current in a pure resistor are in phase with each other.
• Average Current: The average value of current over one complete cycle is zero because the positive and negative half-cycles cancel out.
• Joule Heating: Although the average current is zero, the average power dissipated is not zero because heating depends on $i^2$, which is always positive.
• Instantaneous Power: $$p = i^2 R = i_m^2 R \sin^2 \omega t$$
• Average Power: Since the average value of $\sin^2 \omega t$ over a complete cycle is $\frac{1}{2}$, the average power is:
• $$P = \bar{p} = \frac{1}{2} i_m^2 R$$

To express ac power in the same mathematical form as dc power ($P = I^2 R$), we define the root mean square current ($I$ or $I_{rms}$):

$$I = \sqrt{\bar{i^2}} = \frac{i_m}{\sqrt{2}} \approx 0.707 i_m$$

Similarly, the rms voltage ($V$ or $V_{rms}$) is defined as:

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

Using these rms values, the average power can be written as:

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

Let the applied voltage be $v = v_m \sin \omega t$. Applying Kirchhoff's loop rule:

$$v - L \frac{di}{dt} = 0 \implies \frac{di}{dt} = \frac{v_m}{L} \sin \omega t$$

Integrating both sides with respect to time $t$:

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

$$i = -\frac{v_m}{\omega L} \cos(\omega t) + \text{constant}$$

Since the source voltage oscillates symmetrically about zero, the current must also oscillate symmetrically about zero, meaning the integration constant is zero.

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

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

where the current amplitude $i_m$ is:

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

• Inductive Reactance ($X_L$): The quantity $\omega L$ acts as the effective resistance of the inductor and is called inductive reactance :$$X_L = \omega L = 2\pi \nu L$$Its SI unit is the ohm ($\Omega$). It is directly proportional to both the inductance and the frequency of the ac source.
• Phase Relationship: Comparing $v = v_m \sin \omega t$ and $i = i_m \sin\left(\omega t - \frac{\pi}{2}\right)$, we see that the current lags the voltage by $\frac{\pi}{2}$ radians (or $90^\circ$, which is one-quarter of a period).

The instantaneous power supplied to the inductor is:

$$p_L = i v = \left[i_m \sin\left(\omega t - \frac{\pi}{2}\right)\right] \times [v_m \sin \omega t] = -i_m v_m \cos(\omega t) \sin(\omega t) = -\frac{i_m v_m}{2} \sin(2\omega t)$$

The average power over a complete cycle is:

$$P_L = -\frac{i_m v_m}{2} \langle\sin(2\omega t)\rangle = 0$$

Thus, the average power supplied to an inductor over a complete cycle is zero . No electrical energy is dissipated as heat in a pure inductor.

Let the applied voltage be $v = v_m \sin \omega t$. The instantaneous charge $q$ on the capacitor is:

$$q = C v = C v_m \sin \omega t$$

The current $i$ in the circuit is the rate of flow of charge:

$$i = \frac{dq}{dt} = \frac{d}{dt}(C v_m \sin \omega t) = \omega C v_m \cos(\omega t)$$

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

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

where the current amplitude $i_m$ is:

$$i_m = \omega C v_m = \frac{v_m}{1/(\omega C)}$$

• Capacitive Reactance ($X_C$): The quantity $\frac{1}{\omega C}$ acts as the effective resistance of the capacitor and is called capacitive reactance :$$X_C = \frac{1}{\omega C} = \frac{1}{2\pi \nu C}$$Its SI unit is the ohm ($\Omega$). It is inversely proportional to both the frequency and the capacitance.
• Phase Relationship: Comparing $v = v_m \sin \omega t$ and $i = i_m \sin\left(\omega t + \frac{\pi}{2}\right)$, we see that the current leads the voltage by $\frac{\pi}{2}$ radians (or $90^\circ$, which is one-quarter of a period).

The instantaneous power supplied to the capacitor is:

$$p_C = i v = \left[i_m \cos(\omega t)\right] \times [v_m \sin \omega t] = \frac{i_m v_m}{2} \sin(2\omega t)$$

The average power over a complete cycle is:

$$P_C = \frac{i_m v_m}{2} \langle\sin(2\omega t)\rangle = 0$$

Thus, the average power supplied to a capacitor over a complete cycle is zero .

Since the elements are in series, the instantaneous current $i$ is the same at any point in the circuit. Let it be:

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

where $\phi$ is the phase difference between the voltage of the source and the current in the circuit.

Let $\mathbf{V}_R$, $\mathbf{V}_L$, $\mathbf{V}_C$, and $\mathbf{V}$ be the phasors representing the voltages across the resistor, inductor, capacitor, and source respectively. Their amplitudes are:

• $v_{Rm} = i_m R$ (in phase with current phasor $\mathbf{I}$)
• $v_{Lm} = i_m X_L$ (leads $\mathbf{I}$ by $\frac{\pi}{2}$)
• $v_{Cm} = i_m X_C$ (lags $\mathbf{I}$ by $\frac{\pi}{2}$)

The phasor relation is:

$$\mathbf{V}_L + \mathbf{V}_R + \mathbf{V}_C = \mathbf{V}$$

Since $\mathbf{V}_C$ and $\mathbf{V}_L$ are along the same line but in opposite directions, they combine to form a single phasor of magnitude $|v_{Cm} - v_{Lm}|$. Using the Pythagorean theorem on the right-triangle formed by $\mathbf{V}_R$, $(\mathbf{V}_C + \mathbf{V}_L)$, and $\mathbf{V}$:

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

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

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

where $Z$ is the impedance of the circuit:

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

The phase angle $\phi$ is given by:

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

An interesting characteristic of a series $RLC$ circuit is the phenomenon of resonance , where the current amplitude is maximized at a specific frequency.

• Resonance Condition: The current amplitude $i_m = \frac{v_m}{Z}$ is maximum when the impedance $Z$ is minimum. This occurs when:$$X_C = X_L \implies \frac{1}{\omega_0 C} = \omega_0 L$$
• Resonant Frequency ($\omega_0$): $$\omega_0 = \frac{1}{\sqrt{LC}}$$
• Impedance at Resonance: At $\omega_0$, $Z = R$, which is purely resistive. The current amplitude is at its maximum value:$$i_{m,\text{max}} = \frac{v_m}{R}$$
• Requirement for Resonance: Resonance can only occur if both $L$ and $C$ are present in the circuit. In $RL$ or $RC$ circuits, the voltages across the reactive elements cannot cancel each other out, so resonance is impossible.

• Coils: It consists of two coils wound on a laminated soft-iron core:
  • Primary Coil: Connected to the input ac voltage source, having $N_p$ turns.
  • Secondary Coil: Connected to the output load, having $N_s$ turns.
• Working: An alternating current in the primary coil creates a continuously changing magnetic flux in the iron core, which passes through the secondary coil and induces an alternating electromotive force (emf) in it.

Assuming an ideal transformer with zero flux leakage and negligible coil resistance, the induced emfs are:

$$\epsilon_s = -N_s \frac{d\phi}{dt}$$

$$\epsilon_p = -N_p \frac{d\phi}{dt}$$

Since $\epsilon_s = v_s$ and $\epsilon_p = v_p$ (where $v_s$ and $v_p$ are terminal voltages):

$$\frac{v_s}{v_p} = \frac{N_s}{N_p}$$

For an ideal 100% efficient transformer, input power equals output power ($i_p v_p = i_s v_s$):

$$\frac{i_p}{i_s} = \frac{v_s}{v_p} = \frac{N_s}{N_p}$$

• Step-up Transformer: Increases voltage ($V_s > V_p$) by having more turns in the secondary coil ($N_s > N_p$). Consequently, the secondary current is lower than the primary current ($I_s < I_p$).
• Step-down Transformer: Decreases voltage ($V_s < V_p$) by having fewer turns in the secondary coil ($N_s < N_p$). Consequently, the secondary current is higher than the primary current ($I_s > I_p$).

1. Flux Leakage: Not all magnetic flux from the primary coil passes through the secondary coil. Minimized by winding the primary and secondary coils over one another.
2. Copper Loss (Resistance of Windings): Heat produced ($I^2 R$) due to the resistance of the copper wire. Minimized by using thick wires for high-current windings.
3. Iron Loss (Eddy Currents): Alternating magnetic flux induces circulating currents (eddy currents) in the iron core, causing heating. Minimized by using a laminated iron core.
4. Hysteresis Loss: Energy lost as heat during the continuous magnetization and demagnetization of the core. Minimized by using a core material with low hysteresis loss, such as soft iron.