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When t<0 Components in circuit are Jt = 4 A, R1 = R2 = 2 Ω, R3 = 4 Ω ja L = 10 H

At t=0 switch K will be closed and i need to figure out what iL(t) is when t=4

So i tried to figure out iL(t) with differential equation.
First i combined R1 and R2 to get R12=1Ω Then i transformed power supply from current to voltage Et=Jt*R12=J*1Ω =J
Then i combined R12 and R3 to get Rz=5Ω
Now i can create equation which is $$E_t=L*\frac{di(t)}{dt}+R_ti(t) $$ After adding constant i get
$$J_t=10*\frac{di(t)}{dt}+5i(t) $$ First i found the roots $$10r^t+5r=0 -> r=-\frac{1}{2}$$ So for homogeneous part i get $$y^h=C_1*(-\frac{1}{2})^t$$ And for nonhomogeneous part $$J_t=A$$ $$ \frac{dJ_t}{dt}=0 , \frac{d^2J_t}{dt^2}=0$$ So i get $$ 10*0+5=A->A=5 $$ For complete equation i get $$ y(t)=C_1*(-\frac{1}{2})^t+5 $$ To figure out what is C1 i use t=0 $$ y(0)=C_1*(-\frac{1}{2})^0+5=4 ->C_1=-1 $$ And now for t=4 i get $$ y(4)=-1*(-\frac{1}{2})^4+5=4.5 $$ I know this answer isn't correct but i don't know what went wrong
Also sorry if its little bit hard to read this, i'm new to this

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    \$\begingroup\$ You've solved the differential equation incorrectly. The solution should be an exponential function. The simple method is to analyze the circuit at steady-state before and after the switch closes. In steady-state before t=0, what's the current \$i_L\$? \$\endgroup\$ – Shamtam Sep 18 '17 at 14:40
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The proper way to form the homogeneous solution is (not what you have given in the fourth equation): $${i_L}_h=C_1*e^{(-\frac{1}{2})t}$$ (Your differential equation is written for \$i_L(t)\$, it is better not to change it to \$y\$ midstream.)

For t>0, without knowing \$J\$, you cannot solve \$i_L(t)\$. The only specification of \$J\$ given is that it is 4. So a forced assumption is that \$J = 4\$ for all time if you want a solution from what are given.

With \$J\$ being constant, by inspection, $$i_L(t) = \frac{J_t}{5} $$ is a particular solution to the differential equation. You can plug that into your second equation to confirm.

Now the complete solution becomes: $$i_L(t)=C_1*e^{(-\frac{1}{2})t} + \frac{J_t}{5}$$

Now you need the initial condition of \$i_L(0)\$ to solve for the constant \$C_1\$. Given \$J\$ is constant and assuming steady state at t=0, \$L\$ behaves as a short circuit at t=0. Therefore, the only elements left in the schematic for figuring out the initial condition are \$J,\ R_2,\ R_3\$, a simple current divider.

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