A Resistor Of Resistance 100ohm Is Connected To An Ac Source =(12v) Sin(250πs−1)T. Find The Energy Dissipated As Heat During T=0 To T=1.0ms

Given :

The peak voltage of AC source  E0 = 12 v.

The angular frequency,  [latex] \omega = 250 \pi s^{-1} [/latex].

The resistance of the resistor, R = 100[latex]  \Omega [/latex].

The energy dissipated as heat (H) is given by, H =[latex] \frac{E_{rms}}{R}T [/latex].

Where:

[latex] E_{rms} [/latex] = RMS value of voltage.

R = Resistance of the resistor.

T = Temperature.

Energy dissipated as heat during t = 0 to t = 1.0 ms.

[latex] H = \int_{0}^{10 – 3} d H [/latex].

[latex]  \Rightarrow \int \frac{E_{0}^{2}sin^{2} wt}{R} d t [/latex].

Since [latex] (E_{Rms} = E_{Rms}E_{0}sin wt) [/latex].

[latex] \Rightarrow \frac{144}{100}\int_{0}^{10-3} (sin^{2}wt) dt [/latex].

[latex] \Rightarrow 144 \int_{0}^{10-3} (\frac{1 – cos 2 wt}{2}) dt [/latex].

[latex] \Rightarrow \frac{144}{100}\int_{0}^{10-3}(sin^{2}wt)dt [/latex].

[latex] \Rightarrow 1.44\int_{0}^{10-3}(\frac{1-cos 2 wt}{2})dt [/latex].

[latex] \Rightarrow \frac{1.44}{2}[\int_{0}^{10-3} dt + \int_{0}^{10-3} cos 2 wt dt] [/latex].

[latex] \Rightarrow 0.72 [10^{-3} – (\frac{sin2wt}{2w})_{0}^{10^{-3}}] [/latex].

[latex] \Rightarrow 0.72 [\frac{1}{100} – \frac{1}{500\Pi}] [/latex].

[latex] \Rightarrow 0.72 [\frac{1}{1000} – \frac{1}{500\Pi}] [/latex].

[latex] \Rightarrow 0.72 [\frac{1}{1000} – \frac{2}{1000\Pi}] [/latex].

[latex] \Rightarrow (\frac{\Pi – 2}{1000\Pi}) * 0.72 [/latex].

[latex] \Rightarrow 0.0002614 [/latex].

[latex] \Rightarrow 2.61 * 10^{-4} J [/latex].

Therefore, the energy dissipated as heat during t=0 to t=1.0ms is [latex] 2.61 * 10^{-4} J [/latex].

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