Timeline for Equivalent "capacitative reactance" for calculating rms current under mixed-frequency AC voltage waveform
Current License: CC BY-SA 3.0
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| Jun 11, 2020 at 15:10 | history | edited | CommunityBot |
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| Jun 30, 2016 at 12:46 | history | edited | user115412 | CC BY-SA 3.0 |
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| Jun 30, 2016 at 12:37 | vote | accept | CommunityBot | ||
| Jun 30, 2016 at 12:34 | history | edited | user115412 | CC BY-SA 3.0 |
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| Jun 30, 2016 at 12:30 | comment | added | user115412 | @Captainj2001 Thanks...I'm not an electrical engineer (applied/engineering mathematics) , so I was writing this as a monitoring point...I'll correct. | |
| Jun 30, 2016 at 12:29 | comment | added | Captainj2001 | @Bey Your ammeter will not observe anything if it is not in series with the current you would like to measure. | |
| Jun 30, 2016 at 12:28 | history | edited | user115412 | CC BY-SA 3.0 |
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| Jun 30, 2016 at 12:27 | answer | added | Captainj2001 | timeline score: 0 | |
| Jun 30, 2016 at 12:25 | comment | added | user115412 | @Chu where did you comment go? It was helpful. | |
| Jun 30, 2016 at 12:24 | comment | added | user115412 | @Chu ah, thanks so much. Yes, I was wondering where additivity came into play. I was hoping to be able to reduce a complex situation like I describe into an "rms-equivalent" circuit driven by a single-frequency AC voltage and some capacitance. From both your and Andy's comments, it looks like this integration has to happen at the actual RMS value itself, and not at the underlying reactance (i.e., maybe no analog of Thevenin's theorem for dc circuits for this situation). | |
| Jun 30, 2016 at 12:09 | history | edited | user115412 | CC BY-SA 3.0 |
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| Jun 30, 2016 at 11:42 | comment | added | user115412 | @Chu in that case, \$f(\omega)\$ would have the form \$f(\omega)=\alpha 1(\omega)_{\omega_1} + (1-\alpha) 1(\omega)_{\omega_2}\$ where \$0\leq \alpha \leq 1\$ and the resulting capacitance would be \$X_c(f) = \alpha X_c(\omega_1)+(1-\alpha) X_c(\omega_2)\$ | |
| Jun 30, 2016 at 11:02 | answer | added | Andy aka | timeline score: 0 | |
| Jun 30, 2016 at 7:29 | comment | added | Chu | To clarify your question, if you take the simple case where \$f(\omega)\$ is composed of two discrete sinusoids at, say, \$\omega_1\$ and \$\omega_2\$, that will give two values of \$X_c\$. In what sense would these two reactance values be additive? In other words, what would \$X_c(f)\$ look like? | |
| Jun 30, 2016 at 3:30 | history | edited | user115412 | CC BY-SA 3.0 |
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| Jun 30, 2016 at 2:55 | review | First posts | |||
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| Jun 30, 2016 at 2:55 | history | asked | user115412 | CC BY-SA 3.0 |