Peak And R.M.S Value Of An Alternating Current / Voltage

What is an Alternating current?

Alternating current can be defined as the current whose magnitude may change with time and also reverses the direction periodically.

The general equation is given by,

\(I\) = \(I_o ~sin~ ωt\)

\(I\) = \(I_o ~cos~ ωt\)

Where, \(I_o\) is termed as peak value of an alternating current.

Alternating Current

Considering the above equation the current changes at any instantaneous time, if the current is passed through the circuit and it can be assumed to remain constant for any small time dt. As current is passed for short time a small amount of charge is flown through the circuit in time dt and it is represented as:

If the current ‘\(I\)’ is indicated as a sine function then,

\(I\) = \(I_o ~sin~ ωt\)

\(dq\) = \(I ~dt\)

\(dq\) = \(I_o~ sin~ ωt~dt\)

At half the period of an alternating current, the amount of charge passed through the circuit at time

\(\frac{T}{2}\) is given by:

\(q\) = \(\int\limits_{0}^{\frac{T}{2}} I_0 sin~ωt~dt\)

\(q\) = \(I_0 \int\limits_{0}^{\frac{T}{2}} sin~ωt~dt\)

\(q\) = \(I_0 [\frac{-cos ~ωt}{ω}]_{0}^{\frac{T}{2}}\)

\(q\) = \(\frac{-I_0}{ω} [cos~ωt]_{0}^{\frac{T}{2}}\)

\(q\) =  \( \frac{-I_0 ~T}{2\pi} [cos~\pi ~-~ cos~0]\)

\(q\) =  \(\frac{-I_0~T}{2\pi}[-1 -1]\)

\(q\) = \(\frac{I_0~T}{\pi}\)      ……….equation (1)

The mean value of the alternating current is given by,

\(q\) = \( I_m. ~\frac{T}{2}\)      ……..equation (2)

On equating equation (1) and equation (2), we get:

\(I_m.~\frac{T}{2} \) = \( \frac{I_0~T}{\pi}\)

\(I_m\) = \(\frac{2I_0}{\pi}\) = \(0.636~ I_0\)

Do you know what will happen to the mean square value of an alternating current when it completes its full cycle?

The mean value of an alternating current for completing the full cycle will be zero.

 Root Mean Square value of an Alternating Current:

When the alternating current is passed through the resistance, for a small amount of time, heat is produced in the resistance at that particular time.

It is denoted by \(I_{r.m.s}\) or \(I_v\).

\(I\) = \(I_o~ sin~ ωt\)

\(dH\) = \(I^2~ R~dt\)

\(dH\) = \((I_0~sin~ωt)^2~R~dt\)

\(dH\) = \(I_0^2~R~sin^2~ωt~dt\)

The heat produced in a time \(\frac{T}{2}\) is given by:

\(H\) = \(\int\limits_{0}^{\frac{T}{2}}~{I_0}^2~R~sin^2~ωt~dt\)

\(H\) = \({I_0}^2~R~\int\limits_{0}^{\frac{T}{2}} sin^2~ωt~dt\)

\(H\) = \(\frac{{I_0}^2~R}{2}[\frac{T}{2}~-~0]\) = \(\frac{{I_0}^2~R}{2}.\frac{T}{2}\)         ……equation (3)

The rms value of AC is represented as:

\(H\) = \( {I_{r.m.s}}^2~R.~\frac{T}{2}\)        ………equation (4)

By equating equation (3) and equation (4), we get:

\({I_{r.m.s}^2}~R.~\frac{T}{2}\) = \(\frac{{I_0}^2 ~R}{2}. \frac{T}{2}\)

\({I_{r.m.s}^2}\) = \(\frac{{I_0}^2}{2}\)

\(I_{r.m.s}\) = \(\frac{I_0}{√2} \)= \(0.707 ~{I_0}\)

These values are measured by Ammeter and Voltmeter that are used in the circuit.

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Practise This Question

One mole of an ideal gas with CpCv=γ at absolute temperature T1 is adiabatically compressed from an initial pressure
P1 to a final pressure P2. The resulting temperature T2 of the gas is given by