Determine and draw the graph of the current \( i _{s}(t) \)
The work I have so far is basically using the relationship of voltage and current in an inductor
1. \( V=L \frac {di}{dt}
\frac {di}{dt} = \frac {V} {L}
\)
Now I found the current going through the inductor to be
\( i = 2V_{s}(t)+i_{s}(t) \)
Take the derivative of both sides
\( \frac {di}{dt} = 2V_{s}'(t)+i_{s}'(t) \)
Now substitute in what we found in equation 1 here
\( \frac {V} {L}=2V_{s}'(t)+i_{s}'(t) \)
Here V is \( V_{s}(t) \) and L = \( \frac{1}{2} \)
\( \frac {V_{s}(t)}{\frac{1}{2}} =2V_{s}'(t)+i_{s}'(t)
2V_{s}=2V_{s}'+i_{s}'(t) \)
And I'm pretty much stuck here not sure how I would get \( i_{s}(t) \) much less graph it...
The work I have so far is basically using the relationship of voltage and current in an inductor
1. \( V=L \frac {di}{dt}
\frac {di}{dt} = \frac {V} {L}
\)
Now I found the current going through the inductor to be
\( i = 2V_{s}(t)+i_{s}(t) \)
Take the derivative of both sides
\( \frac {di}{dt} = 2V_{s}'(t)+i_{s}'(t) \)
Now substitute in what we found in equation 1 here
\( \frac {V} {L}=2V_{s}'(t)+i_{s}'(t) \)
Here V is \( V_{s}(t) \) and L = \( \frac{1}{2} \)
\( \frac {V_{s}(t)}{\frac{1}{2}} =2V_{s}'(t)+i_{s}'(t)
2V_{s}=2V_{s}'+i_{s}'(t) \)
And I'm pretty much stuck here not sure how I would get \( i_{s}(t) \) much less graph it...