Question

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.

Answer 1

Answer:

Proof is as follows

Proof:

Given that , $V = V_{ac} + V_{dc}$

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

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

In our case, function is $V = V_{ac} + V_{dc}$

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

Now open the square term as follows

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

Rearranging  terms

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

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  $V_{ac}V_{dc} = 0$

so

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

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

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