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Jan Eerland
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The peak can be found by using:

$$\text{V}_\text{R}=\text{V}_\text{in}=\text{I}_\text{R}\cdot\text{R}=\text{I}_\text{in}\cdot\text{R}\tag1$$

And the RMS can be found by finding:

$$\overline{\text{I}}_\text{R}=\overline{\text{I}}_\text{in}=\sqrt{\frac{1}{\frac{1}{4}-0}\int_0^\frac{1}{4}\text{I}_\text{R}^2\space\text{d}x}=\sqrt{\frac{1}{\frac{1}{4}-0}\int_0^\frac{1}{4}\text{I}_\text{in}^2\space\text{d}x}=$$$$\frac{1}{100}=\overline{\text{I}}_\text{R}=\overline{\text{I}}_\text{in}=\sqrt{\frac{1}{\frac{1}{4}-0}\int_0^\frac{1}{4}\text{I}_\text{R}^2\space\text{d}x}=\sqrt{\frac{1}{\frac{1}{4}-0}\int_0^\frac{1}{4}\text{I}_\text{in}^2\space\text{d}x}=$$ $$\sqrt{\frac{1}{\frac{1}{4}-0}\int_0^\frac{1}{4}\left(\frac{\text{V}_\text{R}}{\text{R}}\right)^2\space\text{d}x}=\sqrt{\frac{1}{\frac{1}{4}-0}\int_0^\frac{1}{4}\left(\frac{\text{V}_\text{in}}{\text{R}}\right)^2\space\text{d}x}\tag2$$

The peak can be found by using:

$$\text{V}_\text{R}=\text{V}_\text{in}=\text{I}_\text{R}\cdot\text{R}=\text{I}_\text{in}\cdot\text{R}\tag1$$

And the RMS can be found by finding:

$$\overline{\text{I}}_\text{R}=\overline{\text{I}}_\text{in}=\sqrt{\frac{1}{\frac{1}{4}-0}\int_0^\frac{1}{4}\text{I}_\text{R}^2\space\text{d}x}=\sqrt{\frac{1}{\frac{1}{4}-0}\int_0^\frac{1}{4}\text{I}_\text{in}^2\space\text{d}x}=$$ $$\sqrt{\frac{1}{\frac{1}{4}-0}\int_0^\frac{1}{4}\left(\frac{\text{V}_\text{R}}{\text{R}}\right)^2\space\text{d}x}=\sqrt{\frac{1}{\frac{1}{4}-0}\int_0^\frac{1}{4}\left(\frac{\text{V}_\text{in}}{\text{R}}\right)^2\space\text{d}x}\tag2$$

The peak can be found by using:

$$\text{V}_\text{R}=\text{V}_\text{in}=\text{I}_\text{R}\cdot\text{R}=\text{I}_\text{in}\cdot\text{R}\tag1$$

And the RMS can be found by finding:

$$\frac{1}{100}=\overline{\text{I}}_\text{R}=\overline{\text{I}}_\text{in}=\sqrt{\frac{1}{\frac{1}{4}-0}\int_0^\frac{1}{4}\text{I}_\text{R}^2\space\text{d}x}=\sqrt{\frac{1}{\frac{1}{4}-0}\int_0^\frac{1}{4}\text{I}_\text{in}^2\space\text{d}x}=$$ $$\sqrt{\frac{1}{\frac{1}{4}-0}\int_0^\frac{1}{4}\left(\frac{\text{V}_\text{R}}{\text{R}}\right)^2\space\text{d}x}=\sqrt{\frac{1}{\frac{1}{4}-0}\int_0^\frac{1}{4}\left(\frac{\text{V}_\text{in}}{\text{R}}\right)^2\space\text{d}x}\tag2$$

Source Link
Jan Eerland
  • 8.2k
  • 13
  • 23

The peak can be found by using:

$$\text{V}_\text{R}=\text{V}_\text{in}=\text{I}_\text{R}\cdot\text{R}=\text{I}_\text{in}\cdot\text{R}\tag1$$

And the RMS can be found by finding:

$$\overline{\text{I}}_\text{R}=\overline{\text{I}}_\text{in}=\sqrt{\frac{1}{\frac{1}{4}-0}\int_0^\frac{1}{4}\text{I}_\text{R}^2\space\text{d}x}=\sqrt{\frac{1}{\frac{1}{4}-0}\int_0^\frac{1}{4}\text{I}_\text{in}^2\space\text{d}x}=$$ $$\sqrt{\frac{1}{\frac{1}{4}-0}\int_0^\frac{1}{4}\left(\frac{\text{V}_\text{R}}{\text{R}}\right)^2\space\text{d}x}=\sqrt{\frac{1}{\frac{1}{4}-0}\int_0^\frac{1}{4}\left(\frac{\text{V}_\text{in}}{\text{R}}\right)^2\space\text{d}x}\tag2$$