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RMS Value of Alternating Current | What is the rms value of alternating current ?

RMS Value of Alternating Current | What is the rms value of alternating current ? :- The root mean square (r.m.s.) value of alternating current is that value of direct current which produces the same amount of heat in a given resistor over a specific period of time as is produced by the alternating current in the same resistor over the same period. It is denoted by I_rms. The rms value of alternating current is also called the effective value (I_eff) or virtual value (I_v).

The RMS value for alternating current is calculated as :

Square the instantaneous values of current I(t) over one cycle,
Find the mean (average) of these squared values over one cycle,
Take the square root of the obtained result.

Mathematically, the root mean square value of alternating current is defined as :

$\displaystyle {{I}_{rms}}=\sqrt{\frac{\int\limits_{0}^{T}{{{\left[ I\left( t \right) \right]}^{2}}dt}}{T}}$

To find the root mean square value of alternating current, suppose an alternating current is represented by the following function :-

$\displaystyle I={{I}_{0}}sin\text{ }\omega t$ …..(1)

Suppose this current flows through a resistance R. Then, the heat produced in the resistance during a small time interval dt is given by :

$\displaystyle dH={{I}^{2}}Rdt$

Total heat generated in resistance R in one complete cycle (t = 0 to t = T) :

$\displaystyle H=\int\limits_{0}^{T}{{{I}^{2}}Rdt}$

Substituting equation (1),

$\displaystyle H=\int\limits_{0}^{T}{{{\left( {{I}_{0}}sin\text{ }\omega t \right)}^{2}}Rdt}$

$\displaystyle \Rightarrow H=\int\limits_{0}^{T}{\left( I_{\text{0}}^{2}si{{n}^{2}}\omega t \right)Rdt}$

$\displaystyle \Rightarrow H=I_{\text{0}}^{2}R\int\limits_{0}^{T}{\left( si{{n}^{2}}\omega t \right)dt}$

$\displaystyle \Rightarrow H=I_{\text{0}}^{2}R\int\limits_{0}^{T}{\left( \frac{1-cos\text{ 2}\omega t}{2} \right)dt}$

$\displaystyle \Rightarrow H=\frac{I_{\text{0}}^{2}R}{2}\left[ \int\limits_{0}^{T}{1.dt-\int\limits_{0}^{T}{cos\text{ 2}\omega tdt}} \right]$

$\displaystyle \Rightarrow H=\frac{I_{\text{0}}^{2}R}{2}\left[ \left( t \right)_{0}^{T}-\left( \frac{sin\text{ 2}\omega t}{2\omega } \right)_{0}^{T} \right]$

$\displaystyle \Rightarrow H=\frac{I_{\text{0}}^{2}R}{2}\left[ \left( T-0 \right)-\left( \frac{sin\text{ 2}\omega T}{2\omega }-\frac{sin\text{ 2}\omega \times 0}{2\omega } \right) \right]$

$\displaystyle \Rightarrow H=\frac{I_{\text{0}}^{2}R}{2}\left[ T-\frac{sin\text{ 2}\times \frac{2\pi }{T}\times T}{2\omega } \right]$

$\displaystyle \Rightarrow H=\frac{I_{\text{0}}^{2}R}{2}\left[ T-\frac{sin\text{ 4}\pi }{2\omega } \right]$

$\displaystyle \Rightarrow H=\frac{I_{\text{0}}^{2}R}{2}\left[ T-\frac{0}{2\omega } \right]$

$\displaystyle \Rightarrow H=\frac{I_{\text{0}}^{2}RT}{2}$ …..(2)

If I_rms is the root mean square value of an alternating current, then the heat produced in the same resistor R during the same time interval T (from t = 0 to t = T) is,

$\displaystyle H=I_{rms}^{2}RT$ …..(3)

From equations (2) and (3),

$\displaystyle I_{rms}^{2}RT=\frac{I_{\text{0}}^{2}RT}{2}$

$\displaystyle \Rightarrow I_{rms}^{2}=\frac{I_{\text{0}}^{2}}{2}$

$\displaystyle \therefore {{I}_{rms}}=\frac{{{I}_{0}}}{\sqrt{2}}=0.707{{I}_{0}}$ …..(4)

Thus, the root mean square value (I_rms) or effective value (I_eff) of alternating current is 0.707 times its peak value (I₀), or 70.7% of its peak value.

Similarly, we can also find the root mean square value of alternating electromotive force :-

Root Mean Square Value of Alternating E.M.F.

The root mean square (RMS) value of alternating electromotive force (emf) is that value of direct voltage which produces the same amount of heat in a given resistor over a specified period of time as is produced by the alternating emf in the same resistor over the same period. It is denoted by E_rms. The RMS value of alternating emf is also known as the effective value (E_eff) or virtual value (E_v).

To determine the RMS of an alternating emf, we consider that the alternating voltage is represented by the following function :

$\displaystyle E={{E}_{0}}sin\text{ }\omega t$ …..(5)

Let this alternating voltage be applied across a resistor R. Then, the heat produced in the resistor in a small time interval dt is :

$\displaystyle dH=\frac{{{E}^{2}}}{R}dt=\frac{{{\left( {{E}_{0}}sin\text{ }\omega t \right)}^{2}}}{R}dt$

$\displaystyle \Rightarrow dH=\frac{E_{0}^{2}}{R}\left( si{{n}^{2}}\omega t \right)dt$

The total heat produced in a resistor R during one complete cycle (from t = 0 to t = T) is given by :

$\displaystyle H=\int\limits_{0}^{T}{\frac{E_{0}^{2}}{R}\left( si{{n}^{2}}\omega t \right)dt}$

$\displaystyle \Rightarrow H=\frac{E_{0}^{2}}{R}\int\limits_{0}^{T}{\left( si{{n}^{2}}\omega t \right)dt}$

$\displaystyle \Rightarrow H=\frac{E_{0}^{2}}{R}\int\limits_{0}^{T}{\left( \frac{1-cos\text{ 2}\omega t}{2} \right)dt}$

$\displaystyle \Rightarrow H=\frac{E_{0}^{2}}{2R}\left[ \int\limits_{0}^{T}{1.dt-\int\limits_{0}^{T}{cos\text{ 2}\omega tdt}} \right]$

$\displaystyle \Rightarrow H=\frac{E_{0}^{2}}{2R}\left[ \left( t \right)_{0}^{T}-\left( \frac{sin\text{ 2}\omega t}{2\omega } \right)_{0}^{T} \right]$

$\displaystyle \Rightarrow H=\frac{E_{0}^{2}}{2R}\left[ \left( T-0 \right)-\left( \frac{sin\text{ 2}\omega T}{2\omega }-\frac{sin\text{ 2}\omega \times 0}{2\omega } \right) \right]$

$\displaystyle \Rightarrow H=\frac{E_{0}^{2}}{2R}\left[ T-\frac{sin\text{ 2}\times \frac{2\pi }{T}\times T}{2\omega } \right]$

$\displaystyle \Rightarrow H=\frac{E_{0}^{2}}{2R}\left[ T-\frac{sin\text{ 4}\pi }{2\omega } \right]$

$\displaystyle \Rightarrow H=\frac{E_{0}^{2}}{2R}\left[ T-\frac{0}{2\omega } \right]$

$\displaystyle H=\frac{E_{0}^{2}T}{2R}$ …..(6)

If E_rms is the root mean square value of the alternating voltage, then the heat produced in the same resistance R during the same time T (from t = 0 to t = T) will be :

$\displaystyle H=\frac{E_{rms}^{2}T}{R}$ …..(7)

From equations (6) and (7),

$\displaystyle \frac{E_{rms}^{2}T}{R}=\frac{E_{0}^{2}T}{2R}$

$\displaystyle \Rightarrow E_{rms}^{2}=\frac{E_{\text{0}}^{2}}{2}$

$\displaystyle \therefore {{E}_{rms}}=\frac{{{E}_{0}}}{\sqrt{2}}=0.707{{E}_{0}}$ …..(8)

Therefore, the RMS value of alternating electromotive force (E_rms) is equal to 0.707 times its peak value (E₀), or 70.7% of the peak value.

Some Important Waves and their Root Mean Square and Average Values

Note :-

(1). Measurement of alternating current and alternating emf is done using hot-wire ammeters and hot-wire voltmeters respectively. These instruments operate on the principle that when an electric current passes through a conducting wire, the heating effect of the current causes the wire to heat up and expand. This expansion results in the movement of the pointer of the ammeter or voltmeter, indicating the value of current or potential difference. Hot-wire ammeters measure the root mean square (RMS) value of both AC and DC currents, and since they work on the heating effect of current, they directly indicate the RMS or effective value (I_eff).

(2). In an AC circuit, the values of current and voltage that are generally specified are their root mean square (RMS) values. For example, when we say 220 Volt AC, it means E_rms = 220 Volt, and similarly, a 5 Ampere AC current means I_rms = 5 Ampere.

(3). If the electric current (or potential difference) is a combination of several components, the RMS value of the resultant current (or voltage) equals the square root of the sum of the squares of the RMS values of the individual components.

Let

$\displaystyle I={{I}_{1}}+{{I}_{2}}+{{I}_{3}}+.....$

then the effective (RMS) value is,

$\displaystyle {{I}_{eff}}/{{I}_{rms}}=\sqrt{{{\left( {{I}_{1,rms}} \right)}^{2}}+{{\left( {{I}_{2,rms}} \right)}^{2}}+{{\left( {{I}_{3,rms}} \right)}^{2}}+.....}$

For (DC + AC) case

$\displaystyle I={{I}_{DC}}+{{I}_{AC}}$

Then

$\displaystyle {{I}_{eff}}/{{I}_{rms}}=\sqrt{{{\left( {{I}_{DC,rms}} \right)}^{2}}+{{\left( {{I}_{AC,rms}} \right)}^{2}}}$

Since

$\displaystyle {{I}_{DC,rms}}={{I}_{DC}}$

We can write

$\displaystyle {{I}_{eff}}/{{I}_{rms}}=\sqrt{{{\left( {{I}_{DC}} \right)}^{2}}+{{\left( {{I}_{AC,rms}} \right)}^{2}}}$

For example if I = ( a sin ωt + b cos ωt ), then the RMS value is equal to the square root of the sum of the squares of the RMS values of the component waves, i.e.

$\displaystyle {{I}_{rms}}=\sqrt{{{\left( \frac{a}{\sqrt{2}} \right)}^{2}}+{{\left( \frac{b}{\sqrt{2}} \right)}^{2}}}$

$\displaystyle \therefore {{I}_{rms}}=\frac{1}{\sqrt{2}}\left( \sqrt{{{a}^{2}}+{{b}^{2}}} \right)$

Proof :-

$\displaystyle dH={{I}^{\text{2}}}Rdt={{\left( a\text{ }sin\omega t\text{ + }b\text{ co}s\omega t \right)}^{2}}Rdt$

$\displaystyle \Rightarrow H=\int\limits_{0}^{T}{{{\left( a\text{ }sin\omega t\text{ + }b\text{ co}s\omega t \right)}^{2}}Rdt}$

$\displaystyle \Rightarrow H=R\int\limits_{0}^{T}{\left( {{a}^{2}}\text{ }si{{n}^{2}}\omega t\text{ + }{{b}^{2}}\text{ co}{{s}^{2}}\omega t\text{ + 2}ab\text{ }sin\omega t\text{ co}s\omega t \right)dt}$

$\displaystyle \Rightarrow H=R\left[ {{a}^{2}}\int\limits_{0}^{T}{\left( si{{n}^{2}}\omega t \right)dt+}\text{ }{{b}^{2}}\int\limits_{0}^{T}{\left( \text{co}{{s}^{2}}\omega t \right)dt+\text{ }}ab\int\limits_{0}^{T}{\left( \text{2 }sin\omega t\text{ co}s\omega t \right)dt\text{ }} \right]$

$\displaystyle \Rightarrow H=R\left[ {{a}^{2}}\times \left( \frac{T}{2} \right)\text{ + }{{b}^{2}}\times \left( \frac{T}{2} \right)\text{ + }ab\times \left( 0 \right) \right]$

$\displaystyle \Rightarrow H=\frac{RT}{2}\left( {{a}^{2}}+{{b}^{2}} \right)$ …..(9)

Now, if I_rms is the root mean square value of the alternating current, then

$\displaystyle H=I_{rms}^{2}RT$ …..(10)

From equations (9) and (10),

$\displaystyle I_{rms}^{2}RT=\frac{RT}{2}\left( {{a}^{2}}+{{b}^{2}} \right)$

$\displaystyle I_{rms}^{2}=\frac{1}{2}\left( {{a}^{2}}+{{b}^{2}} \right)$

$\displaystyle \therefore {{I}_{rms}}=\frac{1}{\sqrt{2}}\left( \sqrt{{{a}^{2}}+{{b}^{2}}} \right)$ …..(11)

Example 1.

If $\displaystyle I=2\sqrt{t}$ amperes, calculate the average value and the root mean square (RMS) value of the current from t = 2 seconds to t = 4 seconds.

Solution :

$\displaystyle {{I}_{avg}}=\frac{\int\limits_{2}^{4}{2\sqrt{t.}dt}}{\int\limits_{2}^{4}{dt}}=\frac{4}{3}\frac{({{t}^{\frac{3}{2}}})_{2}^{4}}{(t)_{2}^{4}}$

$\displaystyle \therefore {{I}_{avg}}=\frac{2}{3}\left[ 8-2\sqrt{2} \right]A$

$\displaystyle {{I}_{rms}}=\sqrt{\frac{\int_{2}^{4}{{{(2\sqrt{t})}^{2}}dt}}{\int_{2}^{4}{dt}}}=\sqrt{\frac{\int_{2}^{4}{4t\,dt}}{2}}=\sqrt{2\left[ \frac{{{t}^{2}}}{2} \right]_{2}^{4}}\,\,=\,\,2\sqrt{3}\,A$

Example 2.

For an alternating current of 50 Hz, determine the time taken for the current to change from zero to its root mean square (RMS) value.

Solution :

$\displaystyle \because I={{I}_{0}}sin\text{ }\omega t$

$\displaystyle \therefore \frac{{{I}_{0}}}{\sqrt{2}}={{I}_{0}}\sin \omega t$

$\displaystyle \Rightarrow \sin \omega t=\frac{1}{\sqrt{2}}$

$\displaystyle \Rightarrow \omega t=\frac{\pi }{4}$

$\displaystyle \ \Rightarrow \left( \frac{2\pi }{T} \right)\ t\ \ =\ \ \frac{\pi }{4}$

$\displaystyle \Rightarrow t=\frac{T}{8}$

$\displaystyle \therefore t=\frac{1}{8\times 50}=2.5\text{ }ms$

Example 3.

A periodic voltage waveform is shown in the figure.

RMS Value of Alternating Current | What is the rms value of alternating current ? - Curio Physics

Find :
(a) The frequency of the waveform, and
(b) The average value.

Solution :

(a) After 100 ms, the wave repeats. Therefore, the time period is T = 100 ms. Hence, the frequency of the waveform is :

$\displaystyle f=\frac{1}{T}=\frac{1}{100\times {{10}^{-3}}}=10Hz$

(b) Since the average value = Area / Time period, therefore,

$\displaystyle {{V}_{avg}}=\frac{(1/2)\,\,\times \,\,100\,\,\times \,\,10}{(100)}=5\text{ }Volt$

Example 4.

If a direct current of magnitude a amperes flows through a wire and is superimposed on an alternating current I = b sin ωt, what will be the effective (RMS value) of the resultant current in the circuit ?

Solution :

Current in the circuit at any instant t,

$\displaystyle I={{I}_{DC}}+{{I}_{AC}}=a+b\sin \omega t$

$\displaystyle {{I}_{eff}}=\sqrt{\frac{1}{T}\int\limits_{0}^{T}{{{I}^{2}}dt}}=\sqrt{\frac{1}{T}\int\limits_{0}^{T}{{{(a+b\sin \omega t)}^{2}}dt}}$

$\displaystyle \Rightarrow {{I}_{eff}}=\sqrt{\frac{1}{T}\int\limits_{0}^{T}{({{a}^{2}}+2ab\sin \omega t+{{b}^{2}}{{\sin }^{2}}\omega t)dt}}$

$\displaystyle \Rightarrow {{I}_{eff}}=\sqrt{\frac{1}{T}\left[ {{a}^{2}}\int\limits_{0}^{T}{dt}+2ab\int\limits_{0}^{T}{\sin \omega tdt+}{{b}^{2}}\int\limits_{0}^{T}{{{\sin }^{2}}\omega tdt} \right]}$

$\displaystyle \Rightarrow {{I}_{eff}}=\sqrt{\frac{1}{T}\left[ {{a}^{2}}\left( t \right)_{0}^{T}+2ab\times \left( 0 \right)+{{b}^{2}}\left( \frac{T}{2} \right) \right]}$

$\displaystyle \Rightarrow {{I}_{eff}}=\sqrt{\frac{1}{T}\left[ {{a}^{2}}\left( T-0 \right)+{{b}^{2}}\left( \frac{T}{2} \right) \right]}$

$\displaystyle \therefore {{I}_{eff}}=\sqrt{{{a}^{2}}+\frac{{{b}^{2}}}{2}}$

Alternative Method :-

As per Note 3,

$\displaystyle {{I}_{eff}}/{{I}_{rms}}=\sqrt{{{\left( {{I}_{DC}} \right)}^{2}}+{{\left( {{I}_{AC,rms}} \right)}^{2}}}$

Therefore

$\displaystyle {{I}_{eff}}=\sqrt{{{a}^{2}}+{{\left( \frac{b}{\sqrt{2}} \right)}^{2}}}$

$\displaystyle \therefore {{I}_{eff}}=\sqrt{{{a}^{2}}+\frac{{{b}^{2}}}{2}}$

Example 5.

In an AC circuit, the peak value of alternating current is 8.0 A. What will be the readings of (i) an AC ammeter and (ii) a DC ammeter connected in the circuit ?

Solution :

(i) An AC ammeter measures the root mean square (RMS) value of current; therefore, the reading of an AC ammeter is the RMS value of the alternating current.

$\displaystyle {{I}_{rms}}=\frac{{{I}_{0}}}{\sqrt{2}}=\frac{8}{\sqrt{2}}=5.66Amp.$

(ii) A DC ammeter measures the average value of current, and we know that the average value of alternating current is zero. Therefore, the reading of a DC ammeter connected in an AC circuit will be zero.

Next Topic :- Why is AC More Dangerous than DC of same voltage

Previous Topic :- Mean or Average Value of Alternating emf

Complete List of Topics :-

RMS Value of Alternating Current | What is the rms value of alternating current ?

Root Mean Square Value of Alternating E.M.F.

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