In this set of notes, we are going to solve a complex circuit diagram problem.
The visualization (circuit schematics) are constructed using Python and a package called the SchemDraw. SchemDraw is a specialized Python package for drawing circuit diagrams. For SchemDraw documentation see: https://cdelker.bitbucket.io/SchemDraw/SchemDraw.html
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')
First we'll find the total resistance of the circuit Rt by combining the individual resistances.
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
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 reistors $R_4$ and $R_5$ are in simple series:
$$ R_{45} = R_4 + R_5 $$$$ R_{45} = 0.0152 \ \Omega + 11.9 \ \Omega $$$$ R_{45} = 0.0271 \ \Omega $$Since resistors $R_6$ and $R_7$ are in simple series:
$$ R_{67} = R_6 + R_7 $$$$ R_{67} = 0.00220 \ \Omega + 0.00740 \ \Omega $$$$ R_{67} = 0.00960 \ \Omega $$
In [4]:
R45 = R4 + R5
R67 = R6 + R7
R45, R67
Out[4]:
Let's redraw the circuit diagram to show our 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.pdf')
d.save('5_resistors_3_loops.png')
Let's redraw the circuit diagram to show our combined resistors.
| Vt = | 5.20 V |
|---|---|
| R1 = | 13.2 mΩ |
| R2 = | 21.0 mΩ |
| R3 = | 3.60 mΩ |
| R45 = | 27.1 mΩ |
| R67 = | 9.60 mΩ |
Next we'll 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:
$$ \frac{1}{R_{2345}} = \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_{45}} $$$$ R_{2345} = \frac{1}{\frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_{45}}} $$$$ R_{2345} = \frac{1}{\frac{1}{0.0210 \ \Omega} + \frac{1}{0.00360 \ \Omega} + \frac{1}{0.0271 \ \Omega}} $$$$ R_{2345} = 0.00276016 \ \Omega $$
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)
R2345
Out[6]:
OK, now let's simplify the circuit diagram even more. This time combining $R_2$, $R_3$ 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.pdf')
d.save('3_resistors_1_loop.png')
Let's redraw the circuit diagram to show our combined resistors.
| Vt = | 5.20 V |
|---|---|
| R1 = | 13.2 mΩ |
| R2345 = | 2.76016 mΩ |
| R67 = | 9.60 mΩ |
Finally, we'll 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 $R_{1234567}$
$$ R_t = R_{1234567} $$$$ R_{t} = R_1 + R_{2345} + R_{67} $$$$ R_{t} = 0.0132 \ \Omega + 0.00276016 \ \Omega + 0.00960 \ \Omega $$$$ R_{t} = 0.0255601 \ \Omega $$
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
Rt
Out[8]:
In [9]:
d = schem.Drawing(unit=2.5)
R67 = d.add(e.LINE, d='right')
Rt = d.add(e.RES, d='up' , botlabel='$R_{t}$')
R1 = d.add(e.LINE, d='left', tox=R67.start)
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('1_resistor_1_loop.pdf')
d.save('1_resistor_1_loop.png')
Let's redraw the circuit diagram to show our combined resistors.
| Vt = | 5.20 V |
|---|---|
| Rt = | 25.56016 mΩ |
In [10]:
It = Vt/Rt
It
The total current of the circuit, $I_t$ is the same as the current running through resistor $R_6$ and resistor $R_7$.
$$ I_t = I_6 = I_7 $$$$ I_t = 203.44 \ A $$$$ I_6 = 203.44 \ A $$$$ I_7 = 203.44 \ A $$We can apply Ohm's law to find $V_6$ now that we have $I_6$ and $I_7$.
$$ V_6 = I_6 R_6 $$$$ V_6 = (203.44 \ A) (0.00220 \ \Omega) $$$$ V_6 = 0.447571 \ V $$$$ V_7 = I_7 R_7 $$$$ V_7 = (203.44 \ A) (0.00740 \ \Omega) $$$$ V_7 = 1.50547 \ V $$| Parameter | Value | Engineering Notation |
|---|---|---|
| V6 | 0.447571 V | 44.8 mV |
| V7 | 1.50547 V | 1.51 V |
| I3 | ? | |
| I6 | ? | |
| P4 | ? | |
| P7 | ? |
In [ ]:
I6 = It
I7 = It
V6 = I6 * R6
V7 = I7 * R7
V6, V7
The total current of the circuit, $I_t$ is the same as the current running through resistor $R_{2345}$.
$$ I_t = I_{2345} $$$$ I_t = 203.44 \ A $$$$ I_{2345} = 203.44 \ A $$We can apply Ohm's law to find $V_{2345}$ now that we have $I_{2345}$.
$$ V_{2345} = I_{2345} R_{2345} $$$$ V_{2345} = (203.44 \ A) (0.00276016 \ \Omega) $$$$ V_{2345} = 0.561532 \ V $$
In [ ]:
I2345 = It
V2345 = I2345 * R2345
V2345
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:
$$ V = IR $$$$ I = \frac{V}{R} $$$$ I_3 = \frac{V_3}{R_3} $$$$ I_3 = 155.981 \ A $$Current $I_6$ is the same as current $I_t$:
$$ I_t = I_6 $$$$ I_t = 203.44 \ A$$$$ I_6 = 203.44 \ A $$| Parameter | Value | Engineering Notation |
|---|---|---|
| V6 | 0.447571 V | 44.8 mV |
| V7 | 1.50547 V | 1.51 V |
| I3 | 155.981 A | 156 A |
| I6 | 203.44 A | 203 A |
| P4 | ? | |
| P7 | ? |
In [ ]:
V3 = V2345
I3 = V3 / R3
I6 = It
I3, I6
Power is equal to Voltage times Current:
$$ P = VI $$According to Ohm's law, $V = IR$. If we substitue in $V$ as $IR$ into the Power equation we get:
$$ P = (IR)(I) $$$$ P = I^2 R $$With a known $R_7$ and $I_7 = I_t = 203.44 \ A$:
$$ P_7 = {I_7}^2 R_7 $$ $$ P_7 = ({203.44}^2)(0.00740 \ \Omega) $$$$ P_7 = 306.27 W $$
In [ ]:
I7 = It
P7 = R7 * I7**2
P7
Current $I_{45}$ is equal to current $I_4$. Voltage $V_{45} = V_{2345}$. Using Ohm's Law again:
$$ I_{45} = \frac{V_{45}}{R_{45}} $$$$ I_{45} = \frac{0.561532 V}{0.0271 \ \Omega} $$$$ I_{45} = 20.721 \ A $$$$ I_{4} = 20.721 \ A $$
In [ ]:
V45 = V2345
I45 = V45/R45
I45
One more time using the power law one more time:
$$ P = I^2 R $$With a known $R_4$ and $I_4 = I_{45}$:
$$ P_4 = {I_4}^2 R_4 $$$$ P_4 = ({20.721 \ A})^2 (0.0152 \ \Omega) $$$$ P_4 = 6.52611 \ W $$
In [ ]:
I4 = I45
P4 = R4 * I4**2
P4
| Parameter | Value | Engineering Notation |
|---|---|---|
| V6 | 0.447571 V | 44.8 mV |
| V7 | 1.50547 V | 1.51 V |
| I3 | 155.981 A | 156 A |
| I6 | 203.44 A | 203 A |
| P4 | 6.52611 W | 6.53 W |
| P7 | 306.27 W | 306 W |
Let's print out all of our final values including units:
In [ ]:
print(f'V6 = {round(V6,3)} V')
print(f'V7 = {round(V7,3)} V')
print(f'I3 = {round(I3,3)} A')
print(f'I6 = {round(I6,3)} A')
print(f'P4 = {round(P4,3)} W')
print(f'P7 = {round(P7,3)} W')