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When VT is under on-state, this function is set up:

$$L\frac{di_d}{dt} + Ri_d = \sqrt{2}U_2\sin(\omega t) \quad\quad\quad\quad (1-1)$$

I can't solve this function :(

The instantaneous moment when VT was into on-state, $$\omega t = \alpha$$ $$i_d = 0$$ this is the Initial condition of the function above. Solving the function and substitude initial condition into the solution, we got:

$$i_d = -\frac{\sqrt{2}U_2}{Z}\sin(\alpha - \varphi)e^{-\frac{R}{\omega L}(\omega t-\alpha)}+\frac{\sqrt{2}U_2}{Z}\sin(\omega t-\varphi) \quad\quad (1-2)$$

where $$Z = \sqrt{R^2+(\omega L)^2}$$

$$\varphi = \arctan(\frac{\omega L}{R})$$

I tried to solve (1-1) and get (1-2) but got into a mess: $$L\frac{di_d}{dt} = \sqrt{2} U_2 \sin(\omega t) - Ri_d$$

$$\frac{di_d}{dt} = \frac{\sqrt{2} U_2 \sin(\omega t) - Ri_d}{L}$$

$$\frac{1}{\sqrt{2} U_2 \sin(\omega t) - Ri_d}di_d = \frac{1}{L}dt$$

$$\int\frac{1}{\sqrt{2} U_2 \sin(\omega t) - Ri_d}di_d = \int\frac{1}{L}dt$$

$$\frac{\sqrt{2}\ln(i_d)}{2R\cdot U_2 \sin(\omega t)}+C_1 = \frac{t}{L}+C_2$$ And I got stuck... ㅠㅠ

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$$L\frac{di_d}{dt} = \sqrt{2} U_2 \sin(\omega t) - Ri_d$$ $$\frac{di_d}{dt} + \frac{R}{L}i_d = \frac{\sqrt{2} U_2}{L} \sin(\omega t)$$ Multiplying both sides by e^(Rt/L), $$e^{\frac{Rt}{L}}\frac{di_d}{dt} + e^{\frac{Rt}{L}}\frac{R}{L}i_d = e^{\frac{Rt}{L}}\frac{\sqrt{2} U_2}{L} \sin(\omega t)$$ By product rule of differentiation, $$\frac{d(e^{\frac{Rt}{L}}i_d)}{dt} = e^{\frac{Rt}{L}}\frac{\sqrt{2} U_2}{L} \sin(\omega t)$$ $$e^{\frac{Rt}{L}}i_d = \int e^{\frac{Rt}{L}}\frac{\sqrt{2} U_2}{L} \sin(\omega t) dt$$ $$i_d = e^{\frac{-Rt}{L}}\int e^{\frac{Rt}{L}}\frac{\sqrt{2} U_2}{L} \sin(\omega t) dt$$

Integrating and substituting your initial conditions, you should get your solution for i.

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  • \$\begingroup\$ I solved the your equation by using SymPy, I got this: $$e^{\frac{Rt}{L}}i_d = \frac{\sqrt{2}U_2(-\frac{L^2\omega \cdot e^{\frac{Rt}{L}} \cos(\omega t)}{L^2\omega^2 + R^2} + \frac{LR\cdot e^{\frac{Rt}{L}}\sin(\omega t)}{L^2\omega^2 + R^2})}{L}$$ it's closer to the answer $$i_d = -\frac{\sqrt{2}U_2}{Z}\sin(\alpha - \varphi)e^{-\frac{R}{\omega L}(\omega t-\alpha)}+\frac{\sqrt{2}U_2}{Z}\sin(\omega t-\varphi)$$ than the mess I made, But I still stuck here, sorry my math is terrible ㅠ.ㅠ \$\endgroup\$ – Wulfric Lee Mar 18 '18 at 12:55
  • \$\begingroup\$ I rewrite it in this form, $$i_d\cdot e^{\frac{Rt}{L}} = -\frac{\sqrt{2}U_2}{Z^2}\cos(\alpha)\cdot L\omega \cdot e^{\frac{Rt}{L}} + \frac{\sqrt{2}U_2}{Z^2}\sin(\alpha)\cdot R \cdot e^{\frac{Rt}{L}}$$ but I don't know what's the next step? \$\endgroup\$ – Wulfric Lee Mar 18 '18 at 13:25
  • \$\begingroup\$ Sorry I got two questions: 1. I don't understand why did you multiplied $$e^{\frac{Rt}{L}}$$ in $$(2-1)$$ 2. In $$(2-2)$$ Why $$e^{\frac{Rt}{L}}\frac{R}{L}i_d$$ just been eliminated without reason? @Kevin Selva Prasanna \$\endgroup\$ – Wulfric Lee Mar 18 '18 at 14:38
  • \$\begingroup\$ This is the standard procedure for solving linear differential equations. Refer en.wikipedia.org/wiki/… for more details. \$\endgroup\$ – Kevin Selva Prasanna Mar 18 '18 at 15:20
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$$L\frac{di_d}{dt} = \sqrt{2} U_2 \sin(\omega t) - Ri_d $$

Its a transcendental equation whose closed form solution does not exist. You can also try to plug random values of id, but that's stupidity.

Google Newton Raphson method which can help to approximate the solution.

Edit: Its a first order linear differential equation as pointed by @Kevin Selva Prasanna

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    \$\begingroup\$ You can solve it using Laplace transform. \$\endgroup\$ – anhnha Mar 18 '18 at 11:45
  • \$\begingroup\$ It is a standard first order linear differential equation. A solution should exist. I don't think the equation is transcendental. \$\endgroup\$ – Kevin Selva Prasanna Mar 18 '18 at 11:50
  • \$\begingroup\$ Yes my bad its solution does exist. Lazy morning. \$\endgroup\$ – Transistor Mar 18 '18 at 11:57

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