Solved Problem on RLC Circuits
Português
English
advertisement
The current flowing in a circuit is given by
\[
\begin{gather}
i=2\sin 4t
\end{gather}
\]
Determine:a) The average current;
b) The rms current;
Solution
a) The average value of a function is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\langle f(t)\rangle =\frac{1}{T}\int_{0}^{T}f(t)\;dt}
\end{gather}
\]
where f(t) = i(t)
\[
\begin{gather}
\langle i\rangle =\frac{1}{T}\int_{0}^{T}2\sin 4t\;dt
\end{gather}
\]
The period of oscillation is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{T=\frac{2\pi}{\omega}}
\end{gather}
\]
the function given in the problem is of type
\( i=i_{o}\sin \omega t \),
we have ω = 4, and the period will be
\[
\begin{gather}
T=\frac{2\pi}{4}\\[5pt]
T=\frac{\pi}{2}
\end{gather}
\]
\[
\begin{gather}
\langle i\rangle =\frac{1}{\frac{\pi}{2}}\int_{0}^{{\frac{\pi}{2}}}2\sin 4t\;dt\\[5pt]
\langle i\rangle =\frac{2}{\pi}\times 2\int_{0}^{{\frac{\pi}{2}}}\sin 4t\;dt\\[5pt]
\langle i\rangle =\frac{4}{\pi}\int_{0}^{{\frac{\pi}{2}}}\sin 4t\;dt
\end{gather}
\]
Integral of
\( \displaystyle \int_{0}^{{\frac{\pi}{2}}}\sin 4t\;dt \)
changing the variable
for t = 0
we have u = 0
for \( t=\dfrac{\pi}{2} \)
we have \( u=4\times\dfrac{\pi}{2}=2\pi \)
changing the variable
\[
\begin{array}{l}
u=4t\\[5pt]
\dfrac{du}{dt}=4\Rightarrow dt=\dfrac{du}{4}
\end{array}
\]
changing the limits of integration
for t = 0
we have u = 0
for \( t=\dfrac{\pi}{2} \)
we have \( u=4\times\dfrac{\pi}{2}=2\pi \)
\[
\begin{split}
\int_{0}^{{\frac{\pi}{2}}}\sin 4t\;dt &=\int_{0}^{{2\pi}}\sin u\frac{du}{4}=\left.-{\frac{1}{4}}\;\cos u\;\right|_{\;0}^{\;2\pi}=\\[5pt]
&=-\frac{1}{4}\;\left(\cos2\pi -\cos 0\right)=-{\frac{1}{4}}\;(1-1)=0
\end{split}
\]
\[
\begin{gather}
\langle i\rangle =\frac{4}{\pi}\times 0
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\langle i\rangle =0}
\end{gather}
\]
The rms value of a quantity that oscillates with to sine or cosine is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{f_{rms}=\left[\frac{1}{T}\int_{0}^{T}f(t)^{2}\;dt\right]^{1/2}}
\end{gather}
\]
substituting the square of the expression given in the problem
\[
\begin{gather}
i_{rms}=\left[\frac{1}{T}\int_{0}^{T}2^{2}\sin ^{2}4t\;dt\right]^{1/2}
\end{gather}
\]
substituting the period of oscillation
\[
\begin{gather}
i_{rms}=\left[\frac{1}{\frac{\pi}{2}}\int_{0}^{{\frac{\pi}{2}}}4\sin ^{2}4t\;dt\right]^{1/2}\\[5pt]
i_{rms}=\left[\frac{2}{\pi}\times 4\int_{0}^{{\frac{\pi}{2}}}\sin ^{2}4t\;dt\right]^{1/2}\\[5pt]
i_{rms}=\left[\frac{8}{\pi}\int_{0}^{{\frac{\pi}{2}}}\sin ^{2}4t\;dt\right]^{1/2}
\end{gather}
\]
Integral of
\( \displaystyle \int_{0}^{{\frac{\pi}{2}}}\sin ^{2}4t\;dt \)
substituting \( \sin ^{2}x=\dfrac{1}{2}-\dfrac{1}{2}\cos 2x \), with x = 4t
for t = 0
we have u = 0
for \( t=\frac{\pi}{2} \)
we have \( u=8\times\frac{\pi}{2}=4\pi \)
substituting \( \sin ^{2}x=\dfrac{1}{2}-\dfrac{1}{2}\cos 2x \), with x = 4t
\[
\begin{gather}
\int_{0}^{{\frac{\pi}{2}}}\sin ^{2}4t\;dt=\int_{0}^{{\frac{\pi}{2}}}\left(\frac{1}{2}-\frac{1}{2}\cos2\times 4t\right)\;dt=\frac{1}{2}\left(\int_{0}^{{\frac{\pi}{2}}}dt-\int_{0}^{{\frac{\pi}{2}}}\cos 8t\;dt\right)
\end{gather}
\]
in the 2nd integral in parentheses, changing the variable
\[
\begin{array}{l}
u=8t\\[5pt]
\dfrac{du}{dt}=8\Rightarrow dt=\dfrac{du}{8}
\end{array}
\]
changing the limits of integration
for t = 0
we have u = 0
for \( t=\frac{\pi}{2} \)
we have \( u=8\times\frac{\pi}{2}=4\pi \)
\[
\begin{split}
\int_{0}^{{\frac{\pi}{2}}}\sin ^{2}4t\;dt &=\frac{1}{2}\left(\left.u\;\right|_{\;0}^{\;\frac{\pi}{2}}-\int_{0}^{{4\pi}}\cos u\frac{du}{8}\right)=\\[5pt]
&=\frac{1}{2}\left[\left.u\;\right|_{\;0}^{\;\frac{\pi}{2}}-\frac{1}{8}\left(\left.\sin u\;\right|_{\;0}^{\;4\pi}\right)\right]=\\[5pt]
&=\frac{1}{2}\left[\frac{\pi}{2}-0-\frac{1}{8}\left(\sin 4\pi-\sin 0\right)\right]=\\[5pt]
&=\frac{1}{2}\left[\frac{\pi}{2}-\frac{1}{8}\times 0\right]=\frac{\pi}{4}
\end{split}
\]
\[
\begin{gather}
i_{rms}=\left[\frac{8}{\pi}\times\frac{\pi}{4}\right]^{1/2}\\[5pt]
i_{rms}=\left[2\right]^{1/2}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{i_{rms}=\sqrt{2}=1,41\;\mathrm{A}}
\end{gather}
\]
advertisement
Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .