Engineering, 14.02.2020 05:26 lLavenderl
Consider a voltage v = Vdc + vac where Vdc = a constant and the average value of vac = 0. Apply the integral definition of RMS to prove the formula RMS =√ [Vdc 2 +(RMS of vac) 2 ]. Hint: v2 = [(Vdc) 2 + 2 *vac* Vdc +(vac) 2 ] and the average value of vac* Vdc = 0.
Answers: 2
![RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {[f(t)]^{2} } \, dt }](/tpl/images/0511/2883/560ee.png)
![RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {[V_{ac} + V_{dc}]^{2} } \, dt }](/tpl/images/0511/2883/af6bf.png)
![RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {[V_{ac}^{2} + V_{dc}^{2} + 2V_{dc}V_{ac}] } \, dt }](/tpl/images/0511/2883/e4796.png)
![RMS = \sqrt{\frac{1}{T}TV_{dc}^{2} + [RMS~~ of~~ V_{ac}]^2 }](/tpl/images/0511/2883/dcb97.png)
![RMS = \sqrt{V_{dc}^{2} + [RMS~~ of~~ V_{ac}]^2 }](/tpl/images/0511/2883/0789a.png)
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Consider a voltage v = Vdc + vac where Vdc = a constant and the average value of vac = 0. Apply the...
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