DC circuit 3 loops 5 mins read ·
January 14, 2024
Analyze DC circuit with 3 loops.
simulator One of the hosted version of CircuitJS1 is available on https://www.falstad.com/circuit/ , where following code
$ 1 0.000005 0.7703437568215379 41 5 50 5e-11
v 80 176 80 80 0 0 40 6 0 0 0.5
r 80 80 240 80 0 1
r 240 80 240 176 0 1.5
r 240 176 80 176 0 2
r 240 176 400 176 0 2
v 400 176 400 80 0 0 40 12 0 0 0.5
r 80 176 80 288 0 16
v 400 288 80 288 0 0 40 16 0 0 0.5
r 240 80 400 80 0 1
w 400 288 400 176 0
will produce http://tinyurl.com/ytbrufmb with circuit diagram of
where ε 1 = 6 V \varepsilon_1 = 6 \ {\rm V} ε 1 = 6 V ε 2 = 12 V \varepsilon_2 = 12 \ {\rm V} ε 2 = 12 V ε 3 = 16 V \varepsilon_3 = 16 \ {\rm V} ε 3 = 16 V
elements Let us name the elements as follow
a--R1--b--R2--c
↑ | | | ↑
I1 E1 R3 E2 I2
| | |
d--R4--e--R5--f
| |
R6 | I6
| | ↓
g------E3-----h
according to the position of each element in previous figure.
Element ε 1 \varepsilon_1 ε 1 ε 2 \varepsilon_2 ε 2 ε 3 \varepsilon_3 ε 3 R 1 R_1 R 1 R 2 R_2 R 2 R 3 R_3 R 3 R 4 R_4 R 4 R 5 R_5 R 5 R 6 R_6 R 6 Value 6 12 16 1 1 1.5 2 2 12
equations from kcl Current subscript is chosen to be the same as the resistor it is passing through.
Point b b b I 1 + I 2 = I 3 . (1)
\tag{1} I_1 + I_2 = I_3.
I 1 + I 2 = I 3 . ( 1 ) Point e e e I 3 = I 4 + I 5 . (2) \tag{2}
I_3 = I_4 + I_5.
I 3 = I 4 + I 5 . ( 2 ) Point d d d I 6 + I 4 = I 1 . (3) \tag{3}
I_6 + I_4 = I_1.
I 6 + I 4 = I 1 . ( 3 ) Point f f f I 5 = I 2 + I 6 . (4) \tag{4}
I_5 = I_2 + I_6.
I 5 = I 2 + I 6 . ( 4 ) equation from kvl Loop begins from negative terminal of a battery.
Loop d − a − b − e − a d-a-b-e-a d − a − b − e − a ε 1 − I 1 R 1 − I 3 R 3 − I 4 R 4 = 0. (5) \tag{5}
\varepsilon_1 - I_1 R_1 - I_3 R_3 - I_4 R_4 = 0.
ε 1 − I 1 R 1 − I 3 R 3 − I 4 R 4 = 0. ( 5 ) Loop f − c − b − e − f f-c-b-e-f f − c − b − e − f ε 2 − I 2 R 2 − I 3 R 3 − I 5 R 5 = 0. (6) \tag{6}
\varepsilon_2 - I_2 R_2 - I_3 R_3 - I_5 R_5 = 0.
ε 2 − I 2 R 2 − I 3 R 3 − I 5 R 5 = 0. ( 6 ) Loop h − g − d − e − f − h h-g-d-e-f-h h − g − d − e − f − h ε 3 − I 6 R 6 + I 4 R 4 − I 5 R 5 = 0. (7) \tag{7}
\varepsilon_3 - I_6 R_6 + I_4 R_4 - I_5 R_5 = 0.
ε 3 − I 6 R 6 + I 4 R 4 − I 5 R 5 = 0. ( 7 ) equations reduction Since there are only three unknowns use Equations (1)-(4) in Equations (5)-(6) so that it contains only I 1 I_1 I 1 I 2 I_2 I 2 I 6 I_6 I 6
(1) & (3) → (5)
ε 1 − I 1 R 1 − ( I 1 + I 2 ) R 3 − ( I 1 − I 6 ) R 4 = 0 I 1 R 1 + ( I 1 + I 2 ) R 3 + ( I 1 − I 6 ) R 4 = ε 1 ( R 1 + R 3 + R 4 ) I 1 + R 3 I 2 − R 4 I 6 = ε 1 . (8) \tag{8}
\begin{array}{c}
\varepsilon_1 - I_1 R_1 - (I_1 + I_2) R_3 - (I_1 - I_6) R_4 = 0 \newline
I_1 R_1 + (I_1 + I_2) R_3 + (I_1 - I_6) R_4 = \varepsilon_1 \newline
(R_1 + R_3 + R_4) I_1 + R_3 I_2 - R_4 I_6 = \varepsilon_1.
\end{array}
ε 1 − I 1 R 1 − ( I 1 + I 2 ) R 3 − ( I 1 − I 6 ) R 4 = 0 I 1 R 1 + ( I 1 + I 2 ) R 3 + ( I 1 − I 6 ) R 4 = ε 1 ( R 1 + R 3 + R 4 ) I 1 + R 3 I 2 − R 4 I 6 = ε 1 . ( 8 ) (1) & (4) → (6)
ε 2 − I 2 R 2 − ( I 1 + I 2 ) R 3 − ( I 2 + I 6 ) R 5 = 0 I 2 R 2 + ( I 1 + I 2 ) R 3 + ( I 2 + I 6 ) R 5 = ε 2 R 3 I 1 + ( R 2 + R 3 + R 5 ) I 2 + R 5 I 6 = ε 2 . (9) \tag{9}
\begin{array}{c}
\varepsilon_2 - I_2 R_2 - (I_1 + I_2) R_3 - (I_2 + I_6) R_5 = 0 \newline
I_2 R_2 + (I_1 + I_2) R_3 + (I_2 + I_6) R_5 = \varepsilon_2 \newline
R_3 I_1 + (R_2 + R_3 + R_5) I_2 + R_5 I_6 = \varepsilon_2.
\end{array}
ε 2 − I 2 R 2 − ( I 1 + I 2 ) R 3 − ( I 2 + I 6 ) R 5 = 0 I 2 R 2 + ( I 1 + I 2 ) R 3 + ( I 2 + I 6 ) R 5 = ε 2 R 3 I 1 + ( R 2 + R 3 + R 5 ) I 2 + R 5 I 6 = ε 2 . ( 9 ) (3) & (4) → (7)
ε 3 − I 6 R 6 + ( I 1 − I 6 ) R 4 − ( I 2 + I 6 ) R 5 = 0 I 6 R 6 − ( I 1 − I 6 ) R 4 + ( I 2 + I 6 ) R 5 = ε 3 − R 4 I 1 + R 5 I 2 + ( R 4 + R 5 + R 6 ) I 6 = ε 3 . (10) \tag{10}
\begin{array}{c}
\varepsilon_3 - I_6 R_6 + (I_1 - I_6) R_4 - (I_2 + I_6) R_5 = 0 \newline
I_6 R_6 - (I_1 - I_6) R_4 + (I_2 + I_6) R_5 = \varepsilon_3 \newline
-R_4 I_1 + R_5 I_2 + (R_4 + R_5 + R_6) I_6 = \varepsilon_3.
\end{array}
ε 3 − I 6 R 6 + ( I 1 − I 6 ) R 4 − ( I 2 + I 6 ) R 5 = 0 I 6 R 6 − ( I 1 − I 6 ) R 4 + ( I 2 + I 6 ) R 5 = ε 3 − R 4 I 1 + R 5 I 2 + ( R 4 + R 5 + R 6 ) I 6 = ε 3 . ( 10 ) new variables To simplity a new and more compact variable is defined
R a b c = R a + R b + R c . (11) \tag{11}
R_{abc} = R_a + R_b + R_c.
R ab c = R a + R b + R c . ( 11 )
Using this equation we can have from (8), (9), (10)
R 134 I 1 + R 3 I 2 − R 4 I 6 = ε 1 , R 3 I 1 + R 235 I 2 + R 5 I 6 = ε 2 , − R 4 I 1 + R 5 I 2 + R 456 I 6 = ε 3 , (12) \tag{12}
\begin{array}{c}
R_{134} I_1 + R_3 I_2 - R_4 I_6 = \varepsilon_1,\newline
R_3 I_1 + R_{235} I_2 + R_5 I_6 = \varepsilon_2,\newline
-R_4 I_1 + R_5 I_2 + R_{456} I_6 = \varepsilon_3,
\end{array}
R 134 I 1 + R 3 I 2 − R 4 I 6 = ε 1 , R 3 I 1 + R 235 I 2 + R 5 I 6 = ε 2 , − R 4 I 1 + R 5 I 2 + R 456 I 6 = ε 3 , ( 12 )
which later can be presented in matrix form as follow
[ R 134 R 3 − R 4 R 3 R 235 R 5 − R 4 R 5 R 456 ] [ I 1 I 2 I 6 ] = [ ε 1 ε 2 ε 6 ] . (13) \tag{13}
\left[
\begin{array}{ccc}
R_{134} & R_3 & -R_4 \newline
R_3 & R_{235} & R_5 \newline
-R_4 & R_5 & R_{456}
\end{array}
\right]
\left[
\begin{array}{c}
I_1 \newline
I_2 \newline
I_6
\end{array}
\right] =
\left[
\begin{array}{c}
\varepsilon_1 \newline
\varepsilon_2 \newline
\varepsilon_6
\end{array}
\right].
R 134 R 3 − R 4 R 3 R 235 R 5 − R 4 R 5 R 456 I 1 I 2 I 6 = ε 1 ε 2 ε 6 . ( 13 )
The last equation can be solved using invers matrix or iterative method for SLE.
matrices From previous table, Equation (13) can be written with its elements value
[ 4.5 1.5 − 2 1.5 4.5 2 − 2 2 16 ] [ I 1 I 2 I 6 ] = [ 6 12 16 ] . (14) \tag{14}
\left[
\begin{array}{ccc}
4.5 & 1.5 & -2 \newline
1.5 & 4.5 & 2 \newline
-2 & 2 & 16
\end{array}
\right]
\left[
\begin{array}{c}
I_1 \newline
I_2 \newline
I_6
\end{array}
\right] =
\left[
\begin{array}{c}
6 \newline
12 \newline
16
\end{array}
\right].
4.5 1.5 − 2 1.5 4.5 2 − 2 2 16 I 1 I 2 I 6 = 6 12 16 . ( 14 )
It would be clearer if every matrix in above equation is uniquely named.
numpy There is solve() function in NumPy to solve SLE.
import numpy as np
R = np. array(
[
[ 4.5 , 1.5 , - 2 ],
[ 1.5 , 4.5 , 2 ],
[ - 2 , 2 , 16 ]
]
)
E = np. array(
[
6 ,
12 ,
16
]
)
I = np. linalg. solve(R,E)
print(I)
that is available on https://onecompiler.com/python/3zze7accv and
with the result is as follow
[1.1 1.9 0.9]
which shows that I 1 = 1.1 A I_1 = 1.1 \ {\rm A} I 1 = 1.1 A I 2 = 1.9 A I_2 = 1.9 \ {\rm A} I 2 = 1.9 A I 6 = 0.9 A I_6 = 0.9 \ {\rm A} I 6 = 0.9 A
challenges Solve Equations (12) using substitution or elimination method to find I 1 I_1 I 1 I 2 I_2 I 2 I 3 I_3 I 3 Execute the code on https://onecompiler.com/python/3zze7accv and explain what the matrices R \mathbf{R} R E \mathbf{E} E I \mathbf{I} I Relate the R \mathbf{R} R E \mathbf{E} E I \mathbf{I} I Compare the result from given code with the simulator on http://tinyurl.com/ytbrufmb . Calculate the values of I 1 I_1 I 1 I 2 I_2 I 2 I 3 I_3 I 3 ε 1 = 12 V \varepsilon_1 = 12 \ {\rm V} ε 1 = 12 V ε 2 = 24 V \varepsilon_2 = 24 \ {\rm V} ε 2 = 24 V ε 3 = 32 V \varepsilon_3 = 32 \ {\rm V} ε 3 = 32 V refs