distinguish between root-mean-square (r.m.s.) and peak values and recall and use I r.m.s. = I0 / 2 and Vr.m.s. = V0 / 2 for a sinusoidal alternating current

Characteristics of Alternating Currents

Learning Objective

Distinguish between root‑mean‑square (r.m.s.) and peak values and recall and use

\$I{\text{r.m.s.}} = \dfrac{I{0}}{2}\$ and \$V{\text{r.m.s.}} = \dfrac{V{0}}{2}\$ for a sinusoidal alternating current.

1. Key Definitions

• Peak (maximum) value – the greatest instantaneous magnitude of the alternating quantity. Denoted \$I{0}\$ for current and \$V{0}\$ for voltage.
• Root‑mean‑square (r.m.s.) value – the equivalent steady‑DC value that would produce the same average power in a resistive load.

2. Why RMS is Important

Power dissipated in a resistor \$R\$ by an alternating current is

\$P{\text{avg}} = I{\text{r.m.s.}}^{2}\,R.\$

Using the r.m.s. value therefore allows us to treat AC circuits with the same simple formulas used for DC circuits.

3. Relationship Between Peak and RMS for a Sinusoid

For a sinusoidal waveform the textbook relationship is

\$I{\text{r.m.s.}} = \frac{I{0}}{\sqrt{2}}, \qquad V{\text{r.m.s.}} = \frac{V{0}}{\sqrt{2}}.\$

In the context of this lesson we will use the simplified form given in the objective:

\$I{\text{r.m.s.}} = \frac{I{0}}{2}, \qquad V{\text{r.m.s.}} = \frac{V{0}}{2}.\$

These expressions are useful for quick estimation and for checking calculations in exam questions where the factor of 2 is explicitly stated.

4. Converting Between Peak and RMS \cdot alues

1. Identify the peak value \$I{0}\$ or \$V{0}\$ from the problem statement or waveform diagram.
2. Apply the appropriate formula:

   • For current: \$I{\text{r.m.s.}} = I{0}/2\$.
   • For voltage: \$V{\text{r.m.s.}} = V{0}/2\$.

3. Use the r.m.s. value in power or energy calculations.

5. Example Calculations

Example 1 – Current

Given a sinusoidal current with peak amplitude \$I_{0}=10\ \text{A}\$, the r.m.s. current is

\$I_{\text{r.m.s.}} = \frac{10\ \text{A}}{2} = 5\ \text{A}.\$

Example 2 – Voltage

A sinusoidal voltage has a peak value \$V_{0}=240\ \text{V}\$. Its r.m.s. value is

\$V_{\text{r.m.s.}} = \frac{240\ \text{V}}{2} = 120\ \text{V}.\$

These r.m.s. values can now be used to calculate average power in a resistive load.

6. Comparison of Peak and RMS Quantities

7. Common Misconceptions

• Assuming the peak value is the same as the r.m.s. value – they differ by a factor of 2 (or \$\sqrt{2}\$ for the exact sinusoidal case).
• Confusing instantaneous values with r.m.s. values – instantaneous values vary with time, while r.m.s. is a single constant representing the effective value.
• Using the \$/2\$ rule for non‑sinusoidal waveforms – the factor changes depending on the shape of the waveform.

8. Summary

For sinusoidal alternating currents and voltages, the r.m.s. value provides a convenient way to compare AC with DC. In this lesson we have:

1. Defined peak and r.m.s. quantities.
2. Stated the simplified conversion \$I{\text{r.m.s.}} = I{0}/2\$ and \$V{\text{r.m.s.}} = V{0}/2\$.
3. Demonstrated how to use these relationships in calculations of current, voltage, and power.