Draw a mathematical expression for RMS value of a sinusoidal voltage \(v=V_{m}\sin{{\omega t}}\).
Let us determine the RMS value of a sinusoidally varying current, i(t) flowing through a resistance R. At any instant of time the heat generated of power dissipated in the resistance is \(\left(i(t)\right)^2R\). The average heat produced or power absorbed during a full cycle of period T is given by
\(P_{av}=\frac{1}{T}\int_{0}^{T}{\left(i(t)\right)^2R}dt\) …(1)
Thus according to the definition of RMS current \(I_{RMS}\), the heat produced or power absorbed is given by
\(P_{av}={I_{RMS}}^2R\) …(2)
From equation (1)and (2), we have
\({I_{RMS}}^2=\frac{1}{T}\int_{0}^{T}\left(i(t)\right)^2dt\) …(3)
Now equation (3) is replaced by a voltage source, we have
\({I_{RMS}}^2=\frac{1}{T}\int_{0}^{T}\left(v(t)\right)^2dt\) …(4)
Now consider a sinusoidal voltage source is
\(v(t)=V_m\sin{\omega t}\) …(5)
\({I_{RMS}}^2=\frac{1}{T}\int_{0}^{2\pi}\left(V_m\sin{\omega t}\right)^2d(\omega t)\)
\({I_{RMS}}^2=\frac{\left(V_m\right)^2}{2\pi}\int_{0}^{2\pi}\left(\sin{\omega t}\right)^2d(\omega t)\)
\({I_{RMS}}^2=\frac{\left(V_m\right)^2}{4\pi}\int_{0}^{2\pi}{(1-\cos{2\omega t)}}d(\omega t)\)
\({I_{RMS}}^2=\frac{\left(V_m\right)^2}{4\pi}\left[\left[(\omega t-\frac{sin2\omega t{)}}{2}\right]\right]_0^{2\pi}=\frac{\left(V_m\right)^2}{2}\)
\({I_{RMS}}^2=\frac{\left(V_m\right)^2}{2}\)
\(I_{RMS}=\frac{V_m}{\sqrt2}\)
\(I_{RMS}=0.707V_m\) …(6)
Thus the RMS value of alternating voltage =\( 0.707\times\) peak value of votage
Find VAB from the circuit if all the resistances are of same value of 1 Ω.
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Fig. 20(a)
Write short notes on:
State and prove the Parseval’s theorem for power.
