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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.

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usmanqadeer

Answer:

Proof is as follows

Proof:

Given that , [tex]V = V_{ac} + V_{dc}[/tex]

for any function f with period T, RMS is given by

[tex]RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {[f(t)]^{2} } \, dt }[/tex]

In our case, function is [tex]V = V_{ac} + V_{dc}[/tex]

[tex]RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {[V_{ac} + V_{dc}]^{2} } \, dt }[/tex]

Now open the square term as follows

[tex]RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {[V_{ac}^{2} + V_{dc}^{2} + 2V_{dc}V_{ac}] } \, dt }[/tex]

Rearranging  terms

[tex]RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {V_{dc}^{2} } \, dt + \frac{1}{T}\int\limits^T_0 {V_{ac}^{2} } \, dt + \frac{1}{T}\int\limits^T_0 {2V_{dc}V_{ac} } \, dt }[/tex]

You can see that

  • second term is square of RMS value of Vac
  • Third terms is average of VdcVac and given is that                      average of  [tex]V_{ac}V_{dc} = 0[/tex]

so

[tex]RMS = \sqrt{\frac{1}{T}TV_{dc}^{2} + [RMS~~ of~~ V_{ac}]^2 }[/tex]

[tex]RMS = \sqrt{V_{dc}^{2} + [RMS~~ of~~ V_{ac}]^2 }[/tex]

So it has been proved that given expression for root mean square (RMS) is valid

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