Question

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

Answer 1

Hint: When a current flows through a resistor, the electrical energy is converted into heat energy, and this heat is dissipated by all those components which possess at least some resistance.
In the question, voltage and resistance are given so we can calculate the power that the resistor will dissipate over an interval of time by using the formula \[H = \dfrac{{{{\left( {{V_{rms}}} \right)}^2}}}{R}t\], where t is the time.

Complete step by step answer:
Resistance \[R = 100\Omega \]
Voltage in the AC source \[V = \left( {12V} \right)\sin \left( {250\pi {s^{ - 1}}} \right)t\]
So the peak voltage of the AC source will be \[{V_0} = 12V\]
Also, the RMS voltage will be \[{V_{rms}} = {E_0}\sin \omega t\]
We know energy dissipated as heat is given as
\[H = \dfrac{{{{\left( {{V_{rms}}} \right)}^2}}}{R}t - - (i)\] where, \[t\] is the time
Since energy is dissipated from\[t = 0\] to \[t = 1.0ms\]
Hence the heat dissipated from time \[t = 0\] to \[t = 1.0ms\] will be:
\[H = \int\limits_0^{1 \times {{10}^{ - 3}}} {dH} - - (ii)\]
Now let’s differentiate equation (i) w.r.t to time t:
\[dH = \dfrac{{{{\left( {{V_{rms}}} \right)}^2}}}{R}dt - - - - (iii)\]
Now substituting equation (iii) in the equation (ii) we get,
\[
  H = \int\limits_0^{1 \times {{10}^{ - 3}}} {dH} \\
   = \int\limits_0^{1 \times {{10}^{ - 3}}} {\dfrac{{{{\left( {{V_{rms}}} \right)}^2}}}{R}dt} \\
 \]
Where \[{V_{rms}} = {E_0}\sin \omega t\]
\[
  H = \int\limits_0^{1 \times {{10}^{ - 3}}} {\dfrac{{{{\left( {{V_{rms}}} \right)}^2}}}{R}dt} \\
   = \int\limits_0^{1 \times {{10}^{ - 3}}} {\dfrac{{E_0^2{{\sin }^2}\omega t}}{R}dt} \\
   = \dfrac{{E_0^2}}{R}\int\limits_0^{1 \times {{10}^{ - 3}}} {{{\sin }^2}\omega tdt} \\
 \]
Since\[{\sin ^2}\omega t = \dfrac{{1 - \cos 2\omega t}}{2}\], hence by further solving, we can write
\[
  H = \dfrac{{E_0^2}}{R}\int\limits_0^{1 \times {{10}^{ - 3}}} {\left( {\dfrac{{1 - \cos 2\omega t}}{2}} \right)dt} \\
   = \dfrac{{E_0^2}}{R}\left[ {\int\limits_0^{\left( {1 \times {{10}^{ - 3}}} \right)} {\dfrac{1}{2}dt - \int\limits_0^{\left( {1 \times {{10}^{ - 3}}} \right)} {\dfrac{{\cos 2\omega t}}{2}} dt} } \right] \\
   = \dfrac{{E_0^2}}{R}\left[ {\dfrac{1}{2}\left[ t \right]_0^{\left( {1 \times {{10}^{ - 3}}} \right)} - \dfrac{1}{2}\left[ {\dfrac{{\sin 2\omega t}}{{2\omega }}} \right]_0^{\left( {1 \times {{10}^{ - 3}}} \right)}} \right] \\
   = \dfrac{{E_0^2}}{{2R}}\left[ {\left( {{{10}^{ - 3}} - 0} \right) - \dfrac{1}{{2\omega }}\left( {\sin 2\omega \times {{10}^{ - 3}} - 0} \right)} \right] \\
 \]
Now by substituting ${E_0} = 12$ and $\omega = 250\pi $ in the above equation, we get
\[H = \dfrac{{{{\left( {12} \right)}^2}}}{{2 \times 100}}\left[ {\left( {{{10}^{ - 3}}} \right) - \dfrac{1}{{2 \times 250\pi }}\left( {\sin 2 \times 250\pi \times {{10}^{ - 3}}} \right)} \right]\]
By further solving
\[
  H = \dfrac{{144}}{{200}}\left[ {\left( {{{10}^{ - 3}}} \right) - \dfrac{1}{{2 \times 250\pi }}\left( {\sin \left( {2 \times 250\pi \times {{10}^{ - 3}}} \right)} \right)} \right] \\
   = 0.72\left[ {\left( {{{10}^{ - 3}}} \right) - \dfrac{1}{{500\pi }}\left( {\sin \left( {0.5\pi } \right)} \right)} \right] \\
   = 0.72\left[ {\left( {{{10}^{ - 3}}} \right) - \dfrac{1}{{500\pi }}\left( {\sin \dfrac{\pi }{2}} \right)} \right] \\
   = 0.72\left[ {\left( {{{10}^{ - 3}}} \right) - \dfrac{1}{{500\pi }}} \right] \\
   = 0.72\left[ {\dfrac{1}{{1000}} - \dfrac{1}{{500\pi }}} \right] \\
   = 0.72\left[ {\dfrac{{\pi - 2}}{{1000\pi }}} \right] \\
 \]
As we know \[\pi = 3.14\], hence we get
\[
  H = 0.72\left[ {\dfrac{{\pi - 2}}{{1000\pi }}} \right] \\
   = 0.72\left[ {\dfrac{{3.14 - 2}}{{1000 \times 3.14}}} \right] \\
   = 0.0002614 \\
   = 2.61 \times {10^{ - 4}}J \\
 \]
Therefore, the energy dissipated as heat during \[t = 0\] to \[t = 1.0{\text{ }}ms\] is \[2.61 \times {10^{ - 4}}J\]_.

Note: It is interesting to note here that the energy is dissipated only in the passive elements (resistance, inductance and capacitance) and not in the active elements. Passive elements are those which consume energy and don’t have their own source of power.