Timeline for Resistor in RC circuit
Current License: CC BY-SA 4.0
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| Oct 3, 2023 at 17:26 | comment | added | periblepsis | @HV16 You could work out that the current pulse would be roughly triangular, requiring \$5\tau\$ of time on the x-axis and starting at a peak of \$18\:\text{A}\$. This would occur twice per cycle. The triangle is actually an RC decay, so you'd need to adjust down somewhat the value you'd get assuming a triangle, though. But the power could be nearing 3 to 4 watts depending on just how square the wave is. Certainly it could go above 2 watts. The shape of the wave matters. | |
| Oct 3, 2023 at 17:00 | comment | added | periblepsis | @HV16 Power in the resistor could be very much higher as a result. And here I mean higher than what you calculated. | |
| Oct 3, 2023 at 16:57 | comment | added | periblepsis | @HV16 No, it's no longer true. Worse, you need to then define the exact shape of the `square wave'. There is no such thing as a perfect square wave. If perfect it would have frequencies that rise towards infinity without limit. The analysis process would change depending upon the Fourier transform of your source wave shape. But the process is similar. Instead of \$\frac{V_p}{R\,C}\cdot\sin\left(\omega_{_0}\,t\right)\$ on the right side of the starting equation, you'd need to replace that with the new equivalent. An infinite number of annihilators may also be required for a perfect square wave. | |
| Oct 3, 2023 at 16:31 | comment | added | HV16 | the signal is a square wave with Vmin = 0Vdc and Vmax = 90Vdc. Is this still true? | |
| Oct 3, 2023 at 16:02 | history | edited | periblepsis | CC BY-SA 4.0 |
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| Oct 3, 2023 at 4:45 | history | answered | periblepsis | CC BY-SA 4.0 |