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In [1]:

    
from lcapy import Circuit, V
cct = Circuit()
cct.add("""
V1 1 0_1 8; down
R1 1 x 3; right, size=2, i=I
R2 x 2 1; right, size=2
E1 2 0_2 x 0 4; down, l = 4 V_x
W 0_1 0; right
W 0 0_2; right
Ox x 0; down, v=V_x
""")
cct.draw(label_ids=True)

Let's determine the current through R1. There are many ways to solve this; the easiest is to combine the sources, combine the resistances, and then use Ohm's law. The result is a function of $V_x$:



In [2]:

    
Vx = V('V_x').Voc
I = (cct.V1.V - 4 * Vx) / (cct.R1.Z + cct.R2.Z)

Now given the current, we can use Ohm's law to determine the voltage drop across R1.



In [3]:

    
I * cct.R1.Z









    Out[3]:





$$\left\{ t : 6 - 3 V_{x}\right\}$$



In [4]:

    
cct.V1.V - I * cct.R1.Z









    Out[4]:





$$\left\{ t : 3 V_{x} + 2\right\}$$

Thus we know that $V_x = 3 V_x + 2$ or $V_x = -1$. Of course, Lcapy can determine this directly. Here Ox is the name of the open circuit over which we wish to determine the voltage difference:



In [5]:

    
cct.Ox.V









    Out[5]:





$$\left\{ t : -1\right\}$$

Alternatively, we can query Lcapy for the voltage at node 'x' with respect to ground. This gives the same result.



In [6]:

    
cct['x'].V









    Out[6]:





$$\left\{ t : -1\right\}$$

Let's check the current with Lcapy:



In [7]:

    
cct.R1.I









    Out[7]:





$$\left\{ t : 3\right\}$$