00:01
Hello, so here we have an lr series circuit that has a variable inductor with the inductance is defined by l of t is equal to either 1 minus 110th t if 0 is less than equal to t, which is less than equal to 10, or is just equal to zero if t is greater than 0.
00:19
And the resistance r is 0 .2 oms, and the voltage is, well, e of t is equal to 4, i of 0 is equal to 0.
00:27
We're going to find the current i of t so the linear differential equation then is going to be l of t times d i d t plus r i is equal to e of t so here we are going to get that um di d t plus 0 .2 divided by 1 minus 1 tenth t i is going to be equal to 4 over 1 .1 .1 .2.
00:59
Minus one tenth t so we solve um the equation here to get our integrating factor is going to be the integral of 0 .2 over one minus one tenth t d t um which gives us that our integrating factor is one minus one tenth t to the negative two multiplying through by that integrating factor and then integrating both sides is going to give us that i is going to be equal to 20 plus constant c times 1 minus 1 tenth t quantity squared.
01:36
We have that i of 0 is equal to 0, giving us that c then is equal to negative 20.
01:42
So we get that i is going to be equal to 20 minus 20 times 1 minus 1 tenth t squared...