- #1
zenterix
- 708
- 84
- Homework Statement
- The circuit shown below contains two nonlinear devices and a current source. The characteristics of the two devices are given. Determine the voltage ##v## for
- Relevant Equations
- (a) ##i_S=1##A, (b) ##i_S=10##A, (c) ##i_S=1\cos{t}##.
Here is the circuit and the v-i characteristics
KCL gives us ##i_S=i_1+i_2##.
Thus, (a) and (b) are solved quickly by noting that for ##v\in [0,1]## we have ##i_1+i_2=0## so ##v## can't be in this interval for a positive current.
If we try ##v\in [1,\infty)## then we get ##i_S=1+(-2+v)=v-1##.
Thus, ##i_S=1##A gives ##v=2##V and ##i_S=10##A gives ##v=11##V.
My question is about item (c) where ##i_S=\cos{(t)}##.
What I did was consider four different cases related to the possible value of ##v##. All I did in each case was consider the KCL equation in the context of a restriction on values of ##v##. Note that only cases 2 and 3 are relevant to the solution of (c).
Case 1: ##v\in [1,\infty)##
From the KCL equation, ##i_S=-1+v##, which graphically is
and we see that ##v## satisfies our constraint when ##i_S\geq 1##.
Case 2: ##v\in [0,1)##
Here we have simply ##i_S=0##.
Case 3: ##v\in [-1,0)##
Then, ##i_S=v##
Case 4: ##v \in (-\infty, -1)##
Then, ##i_S=-1##
Putting all of this together we have
We know that ##i_S## is a sinusoid that varies between -1 and 1. Thus, ##v## takes on values between -1 and 2.
It seems that
$$v(i_S)=\begin{cases} i_S+1,\ \ \ \ \ i_S\in (0,1) \\ i_S,\ \ \ \ \ i_S\in [-1,0] \end{cases}$$
$$v(t)=\begin{cases} \cos{(t)}+1,\ \ \ \ \ t\in (-\pi/2,\pi/2) \\ \cos{(t)},\ \ \ \ \ t\in (\pi/2,3\pi/2) \end{cases}$$
My question is what happens when ##i_S=0##?
KCL gives us ##i_S=i_1+i_2##.
Thus, (a) and (b) are solved quickly by noting that for ##v\in [0,1]## we have ##i_1+i_2=0## so ##v## can't be in this interval for a positive current.
If we try ##v\in [1,\infty)## then we get ##i_S=1+(-2+v)=v-1##.
Thus, ##i_S=1##A gives ##v=2##V and ##i_S=10##A gives ##v=11##V.
My question is about item (c) where ##i_S=\cos{(t)}##.
What I did was consider four different cases related to the possible value of ##v##. All I did in each case was consider the KCL equation in the context of a restriction on values of ##v##. Note that only cases 2 and 3 are relevant to the solution of (c).
Case 1: ##v\in [1,\infty)##
From the KCL equation, ##i_S=-1+v##, which graphically is
and we see that ##v## satisfies our constraint when ##i_S\geq 1##.
Case 2: ##v\in [0,1)##
Here we have simply ##i_S=0##.
Case 3: ##v\in [-1,0)##
Then, ##i_S=v##
Case 4: ##v \in (-\infty, -1)##
Then, ##i_S=-1##
Putting all of this together we have
We know that ##i_S## is a sinusoid that varies between -1 and 1. Thus, ##v## takes on values between -1 and 2.
It seems that
$$v(i_S)=\begin{cases} i_S+1,\ \ \ \ \ i_S\in (0,1) \\ i_S,\ \ \ \ \ i_S\in [-1,0] \end{cases}$$
$$v(t)=\begin{cases} \cos{(t)}+1,\ \ \ \ \ t\in (-\pi/2,\pi/2) \\ \cos{(t)},\ \ \ \ \ t\in (\pi/2,3\pi/2) \end{cases}$$
My question is what happens when ##i_S=0##?