Pallad
New member
- Joined
- Jul 23, 2018
- Messages
- 1
Hello guys,
I solved this equation but I'm not sure if I did it right? Someone can help me check it out?
\(\displaystyle L\frac{di_d}{dt} + Ri_d = \sqrt{2}U_2\sin(\omega t) \quad\quad\quad\quad\)
\(\displaystyle \quad\rightarrow\quad \frac{di_d}{dt} + \frac{R}{L} i_d = \frac{\sqrt{2}U_2}{L}\sin (\omega t)\)
\(\displaystyle \quad\rightarrow\quad e^{\frac{R}{L}} \frac{d(i_d)}{dt} + e^{\frac{R}{L}}\frac{R}{L}i_d = e^{\frac{R}{L}}\frac{\sqrt{2}U_2}{L}\sin (\omega t)\)
\(\displaystyle \quad\rightarrow\quad 0 + \frac{d}{dt}(\frac{R}{L}\cdot t\cdot e^{\frac{R}{L}}\cdot i_d) = e^{\frac{R}{L}}\frac{\sqrt{2}U_2}{L}\sin (\omega t)\)
\(\displaystyle \quad\rightarrow\quad \frac{R}{L}\cdot t\cdot e^{\frac{R}{L}}\cdot i_d = e^{\frac{R}{L}}\cdot \int\frac{\sqrt{2}U_2}{L}\sin (\omega t) dt =e^{\frac{R}{L}}\cdot\frac{\sqrt{2}U_2}{L} \int\sin (\omega t) dt\)
\(\displaystyle \quad\rightarrow\quad \frac{R}{L}\cdot t\cdot i_d =\frac{\sqrt{2}U_2}{L} \int\sin (\omega t) dt\)
set \(\displaystyle \omega t = k\), so \(\displaystyle \frac{dk}{dt} = \omega \quad\rightarrow\quad dk = \omega\ dt\)
\(\displaystyle \quad\rightarrow\quad \frac{R}{L}\cdot t\cdot i_d = \frac{\sqrt{2}U_2}{L}\int\frac{1}{\omega} \sin (k) \cdot \omega \ dt = \frac{\sqrt{2}U_2}{\omega L} \int \sin (k)\ dk\)
\(\displaystyle \quad\rightarrow\quad \frac{R}{L}\cdot t \cdot i_d = \frac{\sqrt{2}U_2}{\omega L}\cdot (-\cos (k) + C)\)
\(\displaystyle \quad\rightarrow\quad i_d = \frac{\frac{\sqrt{2}U_2}{\omega L}\cdot (-\cos (\omega t)) + \frac{\sqrt{2}U_2}{\omega L}C}{\frac{R}{L}\cdot t }\)
I solved this equation but I'm not sure if I did it right? Someone can help me check it out?
\(\displaystyle L\frac{di_d}{dt} + Ri_d = \sqrt{2}U_2\sin(\omega t) \quad\quad\quad\quad\)
\(\displaystyle \quad\rightarrow\quad \frac{di_d}{dt} + \frac{R}{L} i_d = \frac{\sqrt{2}U_2}{L}\sin (\omega t)\)
\(\displaystyle \quad\rightarrow\quad e^{\frac{R}{L}} \frac{d(i_d)}{dt} + e^{\frac{R}{L}}\frac{R}{L}i_d = e^{\frac{R}{L}}\frac{\sqrt{2}U_2}{L}\sin (\omega t)\)
\(\displaystyle \quad\rightarrow\quad 0 + \frac{d}{dt}(\frac{R}{L}\cdot t\cdot e^{\frac{R}{L}}\cdot i_d) = e^{\frac{R}{L}}\frac{\sqrt{2}U_2}{L}\sin (\omega t)\)
\(\displaystyle \quad\rightarrow\quad \frac{R}{L}\cdot t\cdot e^{\frac{R}{L}}\cdot i_d = e^{\frac{R}{L}}\cdot \int\frac{\sqrt{2}U_2}{L}\sin (\omega t) dt =e^{\frac{R}{L}}\cdot\frac{\sqrt{2}U_2}{L} \int\sin (\omega t) dt\)
\(\displaystyle \quad\rightarrow\quad \frac{R}{L}\cdot t\cdot i_d =\frac{\sqrt{2}U_2}{L} \int\sin (\omega t) dt\)
set \(\displaystyle \omega t = k\), so \(\displaystyle \frac{dk}{dt} = \omega \quad\rightarrow\quad dk = \omega\ dt\)
\(\displaystyle \quad\rightarrow\quad \frac{R}{L}\cdot t\cdot i_d = \frac{\sqrt{2}U_2}{L}\int\frac{1}{\omega} \sin (k) \cdot \omega \ dt = \frac{\sqrt{2}U_2}{\omega L} \int \sin (k)\ dk\)
\(\displaystyle \quad\rightarrow\quad \frac{R}{L}\cdot t \cdot i_d = \frac{\sqrt{2}U_2}{\omega L}\cdot (-\cos (k) + C)\)
\(\displaystyle \quad\rightarrow\quad i_d = \frac{\frac{\sqrt{2}U_2}{\omega L}\cdot (-\cos (\omega t)) + \frac{\sqrt{2}U_2}{\omega L}C}{\frac{R}{L}\cdot t }\)