To calculate the effective (RMS) current in an AC circuit when the maximum (peak) current is given, we use the formula:

$I_{\text{rms}} = \frac{I_{\text{max}}}{\sqrt{2}}$

where:

• $I_{\text{rms}}$ is the root mean square (RMS) current (effective current),
• $I_{\text{max}}$ is the maximum (peak) current.

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(1) Given $I_{\text{max}} = 8.4$ A

$I_{\text{rms}} = \frac{8.4}{\sqrt{2}}$ $I_{\text{rms}} = \frac{8.4}{1.414}$ $I_{\text{rms}} \approx 5.94 \, \text{A}$

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(2) Given $I_{\text{max}} = 8.8$ A

$I_{\text{rms}} = \frac{8.8}{\sqrt{2}}$ $I_{\text{rms}} = \frac{8.8}{1.414}$ $I_{\text{rms}} \approx 6.22 \, \text{A}$

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(3) Given $I_{\text{max}} = 65$ mA

Convert to Amperes: $I_{\text{max}} = 65 \times 10^{-3} \, \text{A}$

$I_{\text{rms}} = \frac{65 \times 10^{-3}}{\sqrt{2}}$

$I_{\text{rms}} = \frac{65 \times 10^{-3}}{1.414}$

$I_{\text{rms}} \approx 46.0 \, \text{mA}$

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Final Answers:

1. For $I_{\text{max}} = 8.4$ A → $I_{\text{rms}} \approx 5.94$ A
2. For $I_{\text{max}} = 8.8$ A → $I_{\text{rms}} \approx 6.22$ A
3. For $I_{\text{max}} = 65$ mA → $I_{\text{rms}} \approx 46.0$ mA

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Let me know if you need any clarifications!

Related Questions:

1. Why do we use RMS values in AC circuits instead of peak values?
2. How does the RMS current relate to power calculations in AC circuits?
3. What is the difference between RMS, peak, and peak-to-peak values?
4. How would the RMS current change if the waveform was not sinusoidal?
5. How can we measure RMS current using an oscilloscope?

Tip:

RMS values are used in AC circuits because they provide an equivalent DC power effect, making it easier to compare with DC systems.