I have a circuit with AC square wave at 10V and 4 Hz. There is only a single resistor of 1000 ohm in the circuit. I would like to calculate the rms current and also the peak to peak current of this circuit.
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1seems like a school assignment ... please show your work that you have done so far – jsotola Mar 05 '18 at 07:33
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It is not a school assignment. Am on DIY circuit. Am new to electronics – iamnamrud Mar 05 '18 at 07:34
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1Draw waveforms, search on how to find rms and average value of any signal – Deep Mar 05 '18 at 08:31
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@iamnamrud: Have a look at my answer to https://electronics.stackexchange.com/questions/356562/ac-effect-value/356611#356611 and see if that helps. – Transistor Mar 05 '18 at 08:54
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I have a circuit with AC square wave at 10V and 4 Hz. There is only a single resistor of 1000 ohm in the circuit. I would like to calculate the rms current and also the peak to peak current of this circuit.
- So calculate RMS current based on the peak voltage of 10 volts and 1000 ohms.
- The peak current is also the same as the RMS current
- The peak to peak current is twice the peak current.
Andy aka
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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:
$$\frac{1}{100}=\overline{\text{I}}_\text{R}=\overline{\text{I}}_\text{in}=\sqrt{\frac{1}{\frac{1}{4}-0}\int_0^\frac{1}{4}\text{I}_\text{R}^2\space\text{d}x}=\sqrt{\frac{1}{\frac{1}{4}-0}\int_0^\frac{1}{4}\text{I}_\text{in}^2\space\text{d}x}=$$ $$\sqrt{\frac{1}{\frac{1}{4}-0}\int_0^\frac{1}{4}\left(\frac{\text{V}_\text{R}}{\text{R}}\right)^2\space\text{d}x}=\sqrt{\frac{1}{\frac{1}{4}-0}\int_0^\frac{1}{4}\left(\frac{\text{V}_\text{in}}{\text{R}}\right)^2\space\text{d}x}\tag2$$
Jan Eerland
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@iamnamrud It does, where do you think that the $\frac{1}{4}$ comes from?! – Jan Eerland Mar 05 '18 at 09:10
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