What are the voltages generated from this current source?

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The discussion focuses on analyzing the voltages generated from a current source using Kirchhoff's Current Law (KCL). It establishes that for specific voltage ranges, the current equations yield different relationships, particularly noting that for positive current, voltage cannot be in the interval [0,1]. The analysis includes four cases based on the value of voltage, leading to a sinusoidal current source that varies between -1 and 1, resulting in voltage values between -1 and 2. The participants also discuss the notation differences between v-i and V-I characteristics, emphasizing the context in which each is used. The conversation concludes with a clarification on the appropriate use of symbols in circuit analysis.
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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
1702630454250.png


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

1702630726220.png

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

1702631251429.png


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##?
 
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@berkeman What difference does it make if I write v-i characteristics rather than V-I characteristics that you edited my title?

Agarwal, "Foundations of Analog and Digital Circuits":
1702691623949.png
 
zenterix said:
@berkeman What difference does it make if I write v-i characteristics rather than V-I characteristics that you edited my title?
Good question. For stand-alone symbols, it's more common in my experience to use V and I. However if you are writing the time functions explicitly, then ##v(t)## ##i(t)## would probably be more appropriate. Remember that lowercase "i" or "j" is often used for complex exponents.
 
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