## 20.1 ALTERNATING CURRENT AND VOLTAGE This section introduces alternating current (AC) and voltage, defining their sinusoidal nature and key terminologies such as peak value, RMS value, frequency, and period. It also covers the calculation of power in AC circuits with resistive loads. An AC generator produces an alternating current or voltage that varies sinusoidally with time. ### Sinusoidal Waveform The general form for an alternating quantity (current or voltage) is: $$ x = x_0 \sin(\omega t) $$ where: - $x$ is the instantaneous value at time $t$ - $x_0$ is the maximum or peak value - $\omega$ is the angular frequency of the generator Specifically for current and voltage: - **Alternating current:** $I = I_0 \sin(\omega t)$ - **Alternating voltage:** $V = V_0 \sin(\omega t)$ A sinusoidal waveform has the following characteristics: - It changes direction (polarity) at regular intervals - Its magnitude changes continuously - The change is smooth, being most rapid at the zero-crossing points and slowest at the peaks ### AC Terminologies - **Cycle:** One complete set of positive and negative values of an alternating quantity. - **Time Period (T):** The time taken to complete one cycle. - **Frequency (f):** The number of cycles completed in one second, measured in Hertz (Hz). In Pakistan, the standard AC frequency is 50 Hz. - **Angular Frequency (ω):** The angular frequency is related to frequency by $\omega = 2\pi f$. The time period is $T = 1/f = 2\pi/\omega$. - **Peak Value ($x_0$):** The maximum value (positive or negative) of the alternating quantity. - **Peak-to-Peak Value ($p-p$):** The sum of the positive and negative peak values ($V_{p-p} = 2V_0$). - **Average Value:** The average of all values over a period. For a sinusoidal waveform over one complete cycle, the average value is zero because the positive and negative halves cancel each other out. $$ \text{Average value} = \frac{\text{Total (net) area under the curve for time T}}{\text{Time T}} $$ - **Root-Mean-Square (r.m.s.) Value:** The r.m.s. value of an AC current is the equivalent steady DC current that would produce the same heating effect in a resistor. It is also known as the effective value. The relationship between r.m.s. and peak values is: | Quantity | Formula | | :--- | :--- | | r.m.s. current | $I_{rms} = \frac{I_0}{\sqrt{2}} \approx 0.707 I_0$ | | r.m.s. voltage | $V_{rms} = \frac{V_0}{\sqrt{2}} \approx 0.707 V_0$ | ### Mean Power and Maximum Power When an alternating current $I = I_0 \sin(\omega t)$ flows through a resistor $R$, the instantaneous power dissipated is: $$ P = I^2 R = (I_0 \sin(\omega t))^2 R = I_0^2 R \sin^2(\omega t) $$ Since the current is squared, the power is always positive. The value of $\sin^2(\omega t)$ varies between 0 and 1, with an average value of $1/2$. Therefore, the average (or mean) power delivered to the resistor is: $$ \langle P \rangle = \frac{1}{2} I_0^2 R = I_{rms}^2 R $$ This shows that the mean power in a resistive load is half the maximum power ($P_{max} = I_0^2 R$). ### Worked Examples **Example 1:** An AC circuit consists of a pure resistance of $20\ \Omega$ and is connected across an AC supply of $220\ \text{V}$, $50\ \text{Hz}$. Calculate (a) the peak value of voltage, (b) the peak value of current, and (c) the equations for voltage and current. **Given:** $R = 20\ \Omega$, $V_{rms} = 220\ \text{V}$, $f = 50\ \text{Hz}$ **(a) Peak Voltage ($V_0$):** $$ V_0 = \sqrt{2}\ V_{rms} = \sqrt{2} \times 220\ \text{V} \approx 311.1\ \text{V} $$ **(b) Peak Current ($I_0$):** $$ I_0 = \frac{V_0}{R} = \frac{311.1\ \text{V}}{20\ \Omega} \approx 15.55\ \text{A} $$ **(c) Equations for Voltage and Current:** First, find the angular frequency: $\omega = 2\pi f = 2\pi(50\ \text{Hz}) = 314\ \text{rad/s}$ The equation for voltage is $V = V_0 \sin(\omega t) \implies V = 311.1 \sin(314t)$ The equation for current is $I = I_0 \sin(\omega t) \implies I = 15.55 \sin(314t)$ --- **Example 2:** The peak voltage of an AC supply is 320 V. What is the r.m.s. value of this voltage? **Given:** Peak Voltage $V_0 = 320\ \text{V}$ **Formula:** $V_{rms} = V_0 / \sqrt{2}$ **Calculation:** $V_{rms} = 320\ \text{V} / \sqrt{2} \approx 226.3\ \text{V}$

20.1 Alternating Current And Voltage | Alternating Current | Physics 12