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In [1]:

    
from lcapy import Circuit
cct = Circuit("""
V1 1 0_1 dc; down
R1 1 x; right, size=1.5, i=I
R2 x 2; right, size=1.5
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()

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.



In [2]:

    
(cct.V1.V - cct.E1.V) / (cct.R1.Z + cct.R2.Z)









    Out[2]:





$$\left\{ t : \frac{\frac{4 R_{2} V_{1}}{3 R_{1} - R_{2}} + V_{1}}{R_{1} + R_{2}}\right\}$$

This can be simplified:



In [3]:

    
((cct.V1.V - cct.E1.V) / (cct.R1.Z + cct.R2.Z)).simplify()









    Out[3]:





$$\left\{ t : \frac{3 V_{1}}{3 R_{1} - R_{2}}\right\}$$

Not surprisingly due to KCL, this is the same as the current through R2. Lcapy can solve this directly:



In [4]:

    
cct.R2.I









    Out[4]:





$$\left\{ t : \frac{3 V_{1}}{3 R_{1} - R_{2}}\right\}$$

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



In [5]:

    
cct.R1.Z * cct.R1.I









    Out[5]:





$$\left\{ t : \frac{3 R_{1} V_{1}}{3 R_{1} - R_{2}}\right\}$$

Lcapy can determine this directly.



In [6]:

    
cct.R1.V









    Out[6]:





$$\left\{ t : \frac{3 R_{1} V_{1}}{3 R_{1} - R_{2}}\right\}$$

This can be simplified...



In [7]:

    
cct.R1.V.simplify()









    Out[7]:





$$\left\{ t : \frac{3 R_{1} V_{1}}{3 R_{1} - R_{2}}\right\}$$

Using KVL, the unknown voltage is thus the voltage generated by the source V1 minus the voltage drop across R1.



In [8]:

    
cct.V1.V - cct.R1.V









    Out[8]:





$$\left\{ t : - \frac{3 R_{1} V_{1}}{3 R_{1} - R_{2}} + V_{1}\right\}$$

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 [9]:

    
cct.Ox.V









    Out[9]:





$$\left\{ t : - \frac{R_{2} V_{1}}{3 R_{1} - R_{2}}\right\}$$

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



In [10]:

    
cct['x'].V









    Out[10]:





$$\left\{ t : - \frac{R_{2} V_{1}}{3 R_{1} - R_{2}}\right\}$$

This can be displayed in the time domain using:



In [11]:

    
cct['x'].v









    Out[11]:





$$- \frac{R_{2} V_{1}}{3 R_{1} - R_{2}}$$

Or alternatively using:



In [12]:

    
from lcapy import t
cct['x'].V(t)









    Out[12]:





$$- \frac{R_{2} V_{1}}{3 R_{1} - R_{2}}$$