Root Mean Square Value of AC Signal - Practice Problems, FAQs

The values of sinusoidal AC signals are constantly changing. Therefore, the issue of how to describe a voltage or signal magnitude arises when working with alternating voltages (or currents). Using the waveform's peak values is one straightforward technique. The root mean square value is a popular way to express AC current or voltage. Voltage and current ratings for every appliance are by default expressed as RMS values. We examine how to compute the RMS value for specific signals in this post.

Table of Contents

RMS Value
RMS value of sinusoidal function
RMS value of square waveform
Practice Problems
FAQs

RMS Value

While finding the average value of sinusoidal AC over a complete cycle, we found that it was zero. So we can not differentiate two sinusoidal signals of different amplitudes using average values because the average value will be zero in both cases. For that purpose, we need to find the RMS value of AC.

The RMS value is defined according to heat created by the AC in the resistor. It is the equivalent value of AC voltage at which heat dissipation in an AC circuit is equal to heat dissipation in a DC circuit.

The procedure of finding the RMS value of any function is just doing the mathematical operation in the reverse order of the name i.e., square of function ⇒ find its mean ⇒ find square root.

The RMS value of any function f(x) from x1 to x2 is given by,

$R M S V a l u e = \sqrt{\frac{\int_{x_{1}}^{x_{2}} f (x)^{2} d x}{\int_{x_{1}}^{x_{2}} d x}}$

RMS value of sinusoidal function for half and complete period is the same.So let’s calculate it.

RMS value of sinusoidal function

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After squaring it will become like this:

${i_{r m s}}^{2} = \frac{1}{T} \int_{0}^{T} {i_{o}}^{2} s i n^{2} (ω t) d t$

${i_{r m s}}^{2} = \frac{{i_{o}}^{2}}{T} \int_{0}^{T} \frac{1 - c o s (2 ω t)}{2} d t$

${i_{r m s}}^{2} = \frac{{i_{o}}^{2}}{2 T} [\int_{0}^{T} d t - \int_{0}^{T} c o s (2 ω t) d t]$

${i_{r m s}}^{2} = \frac{{i_{o}}^{2}}{2} [\frac{1}{T} \int_{0}^{T} d t - \frac{1}{T} \int_{0}^{T} c o s (2 ω t) d t]$

The second term will be zero, as integration of cosine function over the complete period is zero.

${i_{r m s}}^{2} = \frac{{i_{o}}^{2}}{2}$

$i_{r m s} = \frac{i_{o}}{\sqrt{2}}$ …….for half as well as full cycle.

RMS value of square waveform

Consider the following square waveform and let’s calculate its root mean square value using the formula discussed.

$R M S V a l u e = \sqrt{\frac{\int_{x_{1}}^{x_{2}} f (x)^{2} d x}{\int_{x_{1}}^{x_{2}} d x}}$

$V_{r m s} = \sqrt{\frac{1}{T} (\int_{0}^{\frac{T}{2}} {V_{o}}^{2} d t + \int_{\frac{T}{2}}^{T} (- V_{o})^{2} d t)}$

$V_{r m s} = \sqrt{\frac{1}{T} (\int_{0}^{\frac{T}{2}} {V_{o}}^{2} d t + \int_{\frac{T}{2}}^{T} {V_{0}}^{2} d t)}$

$V_{r m s} = \sqrt{\frac{{V_{o}}^{2}}{T} (\int_{0}^{\frac{T}{2}} d t + \int_{\frac{T}{2}}^{T} d t)}$

$V_{r m s} = \sqrt{\frac{{V_{o}}^{2}}{T} \int_{0}^{T} d t}$

$V_{r m s} = \sqrt{\frac{{V_{o}}^{2}}{T} T}$

$V_{r m s} = V_{0}$

Practice Problems

Q. The electric current in a circuit is given by i(t)=2io(tT) for some time. Calculate the RMS current for the periods t=0 to t=T.

A. Given, $i (t) = 2 i_{o} (\frac{t}{T})$

$i_{r m s} = \sqrt{\frac{1}{T} \int_{0}^{T} i (t)^{2} d t}$

$i_{r m s} = \sqrt{\frac{1}{T} \int_{0}^{T} 4 {i_{o}}^{2} (\frac{t}{T})^{2} d t}$

$i_{r m s} = \sqrt{\frac{4 {i_{o}}^{2}}{T^{3}} \int_{0}^{T} t^{2} d t}$

$i_{r m s} = \frac{4 {i_{o}}^{2}}{T^{3}} [t 33] 0 T$

$i_{r m s} = 2 \sqrt{\frac{{i_{o}}^{2}}{T^{3}} \frac{T^{3}}{3}}$

$i_{r m s} = 2 \sqrt{\frac{{i_{o}}^{2}}{3}}$

$i_{r m s} = \frac{2 i_{o}}{\sqrt{3}}$

Q. In the previous question, find the average value of the current.

A. $i_{a v g} = \frac{1}{Δt} \int_{t_{1}}^{t_{2}} i d t$

$i_{a v g} = \frac{1}{T} \int_{0}^{T} i (t) d t$

$i_{a v g} = \frac{1}{T} \int_{0}^{T} 2 i_{o} (\frac{t}{T}) d t$

$i_{a v g} = \frac{2 i_{o}}{T^{2}} \int_{0}^{T} t d t$

$i_{a v g} = \frac{2 i_{o}}{T^{2}} [t 22] 0 T$

$i_{a v g} = \frac{2 i_{o}}{T^{2}} \frac{T^{2}}{2}$

$i_{a v g} = i_{0}$

Q. The electric current in a circuit is given by i(t)=io(tT). Find the form factor of the waveform.

A. We have already calculated the RMS and Average value of the given waveform in previous questions for similar expression.

$i_{r m s} = \frac{i_{o}}{\sqrt{3}}$

$i_{a v g} = \frac{i_{o}}{2}$

$F o r m F a c t o r = \frac{R M S V a l u e}{A v e r a g e V a l u e}$

$F o r m F a c t o r = \frac{\frac{i_{o}}{\sqrt{3}}}{\frac{i_{o}}{2}}$

$F o r m F a c t o r = \frac{2}{\sqrt{3}}$

$F o r m F a c t o r = 1.15$

Q. A 30 Ω resistance is connected to a source of 220 V, 50 Hz. Find RMS current and maximum current?

A. $V_{r m s} = 220 V$

$R = 30 o h m$

$i_{r m s} = \frac{V_{r m s}}{R}$

$i_{r m s} = \frac{220}{30}$

$i_{r m s} = 7.33 A$

$i_{r m s} = \frac{i_{o}}{\sqrt{2}}$

$i_{o} = i_{r m s} \sqrt{2}$

$i_{o} = 7.33 \sqrt{2}$

$i_{o} = 10.36 A$

FAQs

Q. What is the significance of average value?
A. Average value of AC voltage is defined based on the charge transfer. It is the equivalent AC voltage at which the charge transfer in the AC circuit is equal to charge transfer in the DC circuit.

Q. What is the significance of RMS value?
A. RMS value is defined based on the heating effect of the waveform. It is the equivalent value of AC voltage at which heat dissipation in an AC circuit is equal to heat dissipation in a DC circuit.

Q. Can the form factor be less than unity?
A. No. As $F o r m F a c t o r = \frac{R M S V a l u e}{A v e r a g e V a l u e}$ . RMS value is always greater than or equal to one. So the form factor is always greater than or equal to one. Form factor is unity for the square waveform.

Q. Why is the average value of sinusoidal signal calculated in half cycle?
A. The average value of a whole sinusoidal waveform over one complete cycle is zero as the two halves cancel each other out, so the average value is taken over half a cycle.