In this post we will create a circuit diagram using a using a Python package called SchemDraw. Then we'll solve a problem using this circuit diagram and Python.

circuit diagram 7 resistors 3 loops

In this post, we are going to solve a circuit diagram problem using Python and a package called SchemDraw. SchemDraw is a specialized Python package for drawing circuit diagrams. For SchemDraw documentation see:

https://cdelker.bitbucket.io/SchemDraw/SchemDraw.html

Given:

The circuit diagram below with a driving voltage $V_t = 5.20 V$ and resistor values in the table below.

7_resistors_3_loops.png

A table of resistance values is below:

Vt = 5.20 V
R1 = 13.2 mΩ
R2 = 21.0 mΩ
R3 = 3.60 mΩ
R4 = 15.2 mΩ
R5 = 11.9 mΩ
R6 = 2.20 mΩ
R7 = 7.40 mΩ

Find:

V6 and V7, the voltage drop across resistors R6 and R7

I3 and I6, the current running through resistors R3 and R6

P4 and P7, the power dissipated by resistors R4 and R7

Solution:

First we'll import the necessary packages. I'm using a jupyter notebook, so the %matplotlib inline command is included. If you want high-resolution circuit diagrams, include the line:

%config InlineBackend.figure_format = 'svg'

at the top of the notebook will ensure high-resolution images.

In [1]:
import matplotlib.pyplot as plt
# if using a jupyter notebook: include %matplotlib inline. If constructing a .py-file: comment out
%matplotlib inline
# if high-resolution images are desired: include %config InlineBackend.figure_format = 'svg'
%config InlineBackend.figure_format = 'svg'
import SchemDraw as schem
import SchemDraw.elements as e

Now we'll build the circuit diagram by creating a SchemDraw Drawing object and adding elements to it.

In [2]:
d = schem.Drawing(unit=2.5)
R7 = d.add(e.RES, d='right', botlabel='$R_7$')
R6 = d.add(e.RES, d='right', botlabel='$R_6$')
d.add(e.LINE, d='right', l=2)
d.add(e.LINE, d='right', l=2)
R5 = d.add(e.RES, d='up' , botlabel='$R_5$')
R4 = d.add(e.RES, d='up', botlabel='$R_4$')
d.add(e.LINE, d='left', l=2)
d.push()
R3 = d.add(e.RES, d='down', toy=R6.end, botlabel='$R_3$')
d.pop()
d.add(e.LINE, d='left', l=2)
d.push()
R2 = d.add(e.RES, d='down', toy=R6.end, botlabel='$R_2$')
d.pop()
R1 = d.add(e.RES, d='left', tox=R7.start, label='$R_1$')
Vt = d.add(e.BATTERY, d='up', xy=R7.start, toy=R1.end, label='$V_t$', lblofst=0.3)
d.labelI(Vt, arrowlen=1.5, arrowofst=0.5)
d.draw()
d.save('7_resistors_3_loops.png')
#d.save('7_resistors_3_loops.pdf')

Find Rt

Now we'll find the total resistance of the circuit Rt using the individual resistances. First, define the resistances and driving voltage as variables.

In [3]:
Vt = 5.2

R1 = 0.0132
R2 = 0.021
R3 = 0.00360
R4 = 0.0152
R5 = 0.0119
R6 = 0.0022
R7 = 0.00740

Find R45 and R67

To simplify the circuit diagram, we'll combine the resistors in series.

For resistors in a simple series circuit:

$$ R_t = R_1 + R_2 + R_3 ... + R_n $$

Since resistors $R_4$ and $R_5$ are in simple series:

$$ R_{45} = R_4 + R_5 $$

Since resistors $R_6$ and $R_7$ are in simple series:

$$ R_{67} = R_6 + R_7 $$

We can easily calculate this with Python. After the calculation, we can use an fstring to print the results. Note the round() function is used on the inside of the fstring curly braces { }, in case there are some floating point math errors that lead to the values printing out as long floats.

In [4]:
R45 = R4 + R5
R67 = R6 + R7

print(f'R45 = {round(R45,7)} Ohm, R67 = {round(R67,5)} Ohm')
R45 = 0.0271 Ohm, R67 = 0.0096 Ohm

Let's redraw our circuit diagram to show the combined resistors.

In [5]:
d = schem.Drawing(unit=2.5)
R67 = d.add(e.RES, d='right', botlabel='$R_{67}$')
d.add(e.LINE, d='right', l=2)
d.add(e.LINE, d='right', l=2)
R45 = d.add(e.RES, d='up', botlabel='$R_{45}$')
d.add(e.LINE, d='left', l=2)
d.push()
R3 = d.add(e.RES, d='down', toy=R67.end, botlabel='$R_3$')
d.pop()
d.add(e.LINE, d='left', l=2)
d.push()
R2 = d.add(e.RES, d='down', toy=R67.end, botlabel='$R_2$')
d.pop()
R1 = d.add(e.RES, d='left', tox=R67.start, label='$R_1$')
Vt = d.add(e.BATTERY, d='up', xy=R67.start, toy=R1.end, label='$V_t$', lblofst=0.3)
d.labelI(Vt, arrowlen=1.5, arrowofst=0.5)
d.draw()
d.save('5_resistors_3_loops.png')
#d.save('5_resistors_3_loops.pdf')

Find R2345

Next we can combine the resistors in parallel. The resistors in parallel are $R_2$, $R_3$ and $R_{45}$. For a resistors in a simple parallel circuit:

$$ \frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} ... + \frac{1}{R_n} $$

Since $R_2$, $R_3$ and $R_{45}$ are in parallel:

$$ \frac{1}{R_{2345}} = \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_{45}} $$

$$ R_{2345} = \frac{1}{\frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_{45}}} $$

We can code this calculation in Python. To find the reciprocal, raise the combined sum to the negative one power. Remember, exponentiation is performed with a double asterisk ** in Python.

In [6]:
Vt = 5.2

R1 = 0.0132
R2 = 0.021
R3 = 0.00360
R4 = 0.0152
R5 = 0.0119
R6 = 0.0022
R7 = 0.00740

R45 = R4 + R5
R67 = R6 + R7

R2345 = ((1/R2)+(1/R3)+(1/R45))**(-1)
print(f'R2345 = {round(R2345,7)} Ohm')
R2345 = 0.0027602 Ohm

OK, now let's construct a new SchemDraw diagram of the simplified the circuit. In this diagram, we'll combine $R_2$, $R_3$ and $R_{45}$ into one big resistor, $R_{2345}$.

In [7]:
d = schem.Drawing(unit=2.5)
R67 = d.add(e.RES, d='right', botlabel='$R_{67}$')
R345 = d.add(e.RES, d='up' , botlabel='$R_{2345}$')
R1 = d.add(e.RES, d='left', tox=R67.start, label='$R_1$')
Vt = d.add(e.BATTERY, d='up', xy=R67.start, toy=R1.end, label='$V_t$', lblofst=0.3)
d.labelI(Vt, arrowlen=1.5, arrowofst=0.5)
d.draw()
d.save('3_resistors_1_loop.png')
#d.save('3_resistors_1_loop.pdf')

Find Rt

To find $R_t$, we again combine the resistors in series. The remaining resistors $R_1$, $R_{2345}$ and $R_{67}$ are in series:

$$ R_{1234567} = R_1 + R_{2345} + R_{67} $$

We'll call the total resistance of the circuit $R_t$ which is equal to $R_{1234567}$

$$ R_t = R_{1234567} $$

Another calculation in Python.

In [8]:
Vt = 5.2

R1 = 0.0132
R2 = 0.021
R3 = 0.00360
R4 = 0.0152
R5 = 0.0119
R6 = 0.0022
R7 = 0.00740

R45 = R4 + R5
R67 = R6 + R7

R2345 = ((1/R2)+(1/R3)+(1/R45))**(-1)

Rt = R1 + R2345 + R67
print(f'Rt = {round(Rt,7)} Ohm')
Rt = 0.0255602 Ohm

Last circuit diagram. The simplest one. This SchemDraw diagram just includes $V_t$ and $R_t$.

In [9]:
d = schem.Drawing(unit=2.5)
L2 = d.add(e.LINE, d='right')
Rt = d.add(e.RES, d='up' , botlabel='$R_{t}$')
L1 = d.add(e.LINE, d='left', tox=L2.start)
Vt = d.add(e.BATTERY, d='up', xy=L2.start, toy=L1.end, label='$V_t$', lblofst=0.3)
d.labelI(Vt, arrowlen=1.5, arrowofst=0.5)
d.draw()
d.save('1_resistor_no_loops.png')
#d.save('1_resistor_no_loops.pdf')

Find V6 and V7

Now that we've solved for the total resistance of the circuit $R_t$, we can find the total current running through the circuit using Ohm's Law $V = IR $.

$$ V = IR $$

$$ I = \frac{V}{R} $$

$$ I_t = \frac{V_t}{R_t} $$

In [10]:
Vt = 5.2

R1 = 0.0132
R2 = 0.021
R3 = 0.00360
R4 = 0.0152
R5 = 0.0119
R6 = 0.0022
R7 = 0.00740

R45 = R4 + R5
R67 = R6 + R7

R2345 = ((1/R2)+(1/R3)+(1/R45))**(-1)
Rt = R1 + R2345 + R67

It = Vt/Rt
print(f'It = {round(It,2)} A')
It = 203.44 A

The total current of the circuit, $I_t$ is the same as the current running through resistor $R_6$ and resistor $R_7$.

$$ I_t = I_6 = I_7 $$

We can apply Ohm's law to find $V_6$ now that we have $I_6$ and $I_7$.

$$ V_6 = I_6 R_6 $$ $$ V_7 = I_7 R_7 $$

In [11]:
I6 = It
I7 = It
V6 = I6 * R6
V7 = I7 * R7
print(f'V6 = {round(V6,5)} V, V7 = {round(V7,5)} V')
V6 = 0.44757 V, V7 = 1.50547 V

Find I3 and I6

The total current of the circuit, $I_t$ is the same as the current running through resistor $R_{2345}$.

$$ I_t = I_{2345} $$

We can apply Ohm's law to find $V_{2345}$ now that we have $I_{2345}$.

$$ V_{2345} = I_{2345} R_{2345} $$

In [12]:
I2345 = It
V2345 = I2345 * R2345
print(f'V2345 = {round(V2345,5)} V')
V2345 = 0.56153 V

The voltage drop across resistor $R_3$ is the same as the voltage drop across resistor $R_{2345}$.

$$ V_3 = V_{2345} $$

Since $V_3$ and $R_3$ are known, we can solve for $I_3$ using Ohm's law:

$$ V = IR $$

$$ I = \frac{V}{R} $$

$$ I_3 = \frac{V_3}{R_3} $$

The current $I_6$ running through resistor $R_6$ is the same as the total current $I_t$.

$$ I_6 = I_t $$

In [13]:
V3 = V2345
I3 = V3 / R3

I6 = It

print(f'I3 = {round(I3,2)} A, I6 = {round(I6,2)} A')
I3 = 155.98 A, I6 = 203.44 A

Find P7 and P4

Power is equal to voltage times current:

$$ P = VI $$

According to Ohm's law:

$$V = IR$$

If we substitute $V$ as $IR$ in the power equation we get:

$$ P = (IR)(I) $$

$$ P = I^2 R $$

With a known $R_7$ and $I_7 = I_t$:

$$ P_7 = {I_7}^2 R_7 $$

In [14]:
I7 = It
P7 = R7 * I7**2
print(f'P7 = {round(P7,2)} W')
P7 = 306.27 W

Current $I_{45}$ is equal to current $I_4$. Voltage $V_{45} = V_{2345}$. Using Ohm's Law again:

$$ V = IR $$

$$ I = \frac{V}{R} $$

$$ I_{45} = \frac{V_{45}}{R_{45}} $$

In [15]:
V45 = V2345
I45 = V45/R45
print(f'I45 = {round(I45,3)} A')
I45 = 20.721 A

One more time using the power law:

$$ P = I^2 R $$

With a known $R_4$ and $I_4 = I_{45}$:

$$ P_4 = {I_4}^2 R_4 $$

In [16]:
I4 = I45
P4 = R4 * I4**2
print(f'P4 = {round(P4,4)} W')
P4 = 6.5261 W

Final Answer

Let's print out all of the final values to three significant figures including units:

In [17]:
print(f'V6 = {round(V6,3)} V')
print(f'V7 = {round(V7,2)} V')
print(f'I3 = {round(I3,0)} A')
print(f'I6 = {round(I6,0)} A')
print(f'P4 = {round(P4,2)} W')
print(f'P7 = {round(P7,0)} W')
V6 = 0.448 V
V7 = 1.51 V
I3 = 156.0 A
I6 = 203.0 A
P4 = 6.53 W
P7 = 306.0 W

Conclusion

SchemDraw is a great package for making circuit diagrams in Python. Python is also useful for doing calculations that involve lots of different values. Although none of the calculations in this problem were particularly difficult, keeping track of all the values as variables in Python can cut down on errors when there multiple calculations and many parameters to keep track of.