Question

A resistor of resistance 100ohm is connected to an AC source \[E=12\sin 250\pi {{s}^{-1}}\]. Find the energy dissipated as heat during t=0 to t=1.0ms

Answer 1

Hint: In this problem, we are given AC, so to find out the value of heat we will have to integrate the heat formula over the given value of time. The time is given in millisecond and you have to convert it into seconds

Complete step by step answer:
Given values, E= 12 V
Resistance = 100 \[\Omega \]
We need to find the energy dissipated for the given time.
Time is 0 to 1 ms that is \[{{10}^{-3}}s\]
Using the formula, \[H=\int\limits_{0}^{{{10}^{-3}}}{\frac{{{E}^{2}}}{R}}dt\]
 We are given that \[E=12\sin 250\pi {{s}^{-1}}\] comparing this with the standard equation we get,
\[\omega =250\pi {{s}^{-1}}\]
So, \[H=\int\limits_{0}^{{{10}^{-3}}}{\frac{{{E}^{2}}}{R}}dt\]
\[\begin{align}
 & H=\int\limits_{0}^{{{10}^{-3}}}{\frac{{{E}^{2}}{{\sin }^{2}}\omega t}{R}}dt \\
 & =\frac{144}{100}\int\limits_{0}^{{{10}^{-3}}}{{{\sin }^{2}}wt}dt \\
 & =\frac{144}{100}\int\limits_{0}^{{{10}^{-3}}}{(\frac{1-\cos 2\omega t}{2}})dt \\
 & =\frac{1.44}{2}\{\int\limits_{0}^{{{10}^{-3}}}{{}}dt-\int\limits_{0}^{{{10}^{-3}}}{\cos 2\omega t}dt\} \\
 & =0.72\{\int\limits_{0}^{{{10}^{-3}}}{{}}dt-\int\limits_{0}^{{{10}^{-3}}}{\frac{\sin 2\omega t}{2\omega }}dt\} \\
 & =0.72[0.001-\frac{1}{500\pi }] \\
 & =0.0002614J \\
\end{align}\].

So, the value of heat dissipated comes out to be 0.0002614 J.

Note:
 Electrical power consumed by resistance in an AC circuit is different from the power consumed by a reactance as reactances do not dissipate energy. In a DC circuit, the power consumed is the product of the DC voltage times the DC, given in watts. However, for AC circuits this is not the case because the AC changes its direction after every half cycle. It is easy and simple to calculate power in a given DC circuit. There is another term which comes into play in AC circuits called power factor, which involves trigonometric cosine. It is defined as the ratio of the real power to the apparent power and gives the true value of heat dissipated in the AC circuits.