Solving For Total Current In Complex Circuits

Problem: Find the total current in the circuit shown above.  Assume that R1 = R3 = R4 = R5 = R6 = 2 ohms and V1 = 8 volts.

Normally, one would resolve this problem using a circuit emulator like LTSPICE.   However, in this post, the equation will be derived via defining three loops around the circuit.  The DC power supply V1 will be a part of every loop.

So lets define the three loops.

The current through the first loop is arbitrarily labeled i1.

The first loop includes the power supply labeled V1 and the two resistors labeled R6 and R4.

The current through the second loop is labeled i2.

The second loop includes the power supply labeled V1 and the two resistors labeled R3 and R5.

The current through the third loop is labeled i3.

The third loop includes the power supply labeled V1 and the three resistors labeled R3, R1 and R4.

For this problem, we'll assume conventional current flow from the plus terminal of V1 through the resistors to the negative terminal of V1.

First lets develop the equation for the first loop;

The total current through R6 is i1.  The total current through R4 is the sum of i1 and i3. Hence the loop equation is
V1 = i1*R6 + (i1 + i3)R4
V1 = i1*R6 + i1*R4 + i3*R4
V1 = i1(R6 + R4) + i3*R4

The second loop includes the power supply labeled V1 and the two resistors labeled R3 and R5.
V1 = (i2 + i3)R3 + i2*R5
V1 = i2*R3 + i3*R3 + i2*R5
V1 = i2(R3 + R5) + i3*R3

The third loop includes the power supply labeled V1 and the three resistors labeled R3, R1 and R4.
V1 = (i2 + i3)R3 + i3*R1 + (i1 + i3)R4
V1 = i2*R3 + i3*R3 + i3*R1 + i1*R4 + i3* R4
V1 = i1*r4 + i2*R3 + i3(R1 + R3 + R4)

Since all resistors are equal, we could use the label R to represent any resistor
For the first loop
V1 = i1(R6 + R4) + i3*R4
V1 = i1* 2R + i3*R
For the second loop
V1 = i2(R3 + R5) + i3*R3
V1 = i2*2R + i3*R
And for the third loop
V1 = i1*r4 + i2*R3 + i3(R1 + R3 + R4)
V1 = i1*R + i2*R + i3*3R

So our three loop equations are
For the first loop
V1 = i1*2R + i3*R
For the second loop
V1 = i2*2R + i3*R
For the third loop
V1 = i1*R + i2*R + i3*3R

The loop current i2 does not flow in the first loop. Therefore we can write the first loop equation as follows
V1 = i1*2R + i2*0 + i3*R
Likewise we can write the second loop equation as follows
V1 = i1*0 +  i2*2R + i3*R

Now our three loop equations are
loop one
V1 = i1*2R + i2*0 + i3*R
loop two
V1 = i1*0 +  i2*2R + i3*R
loop three
V1 = i1*R + i2*R + i3*3R

Now we can use determinants to solve for the loop currents and then add the three loop currents to find the total current.

Using determinants to determine i1

First, we'll solve for the denominator

= 2R*2R*3R + 0*R*R + 0*R*R - R*2R*R - 0*0*3R - 2R*R*R
= 4*4*4 + 0 + 0 - 2*4*2 - 0 - 4*2*2
= 64 - 16 - 16
= 64 - 32
= 32

Next, we'll solve for the numerator

= V1*2R*2R + 0*R*V1 + R*V1*R - R*2R*V1 - 0*R*3R - V1*R*R
=  8*4*4 + 0 + 2*8*2 - 2*4*8 - 0 - 8*2*2
= 128 + 32 - 64 - 32
= 160 - 32
= 128

i1 = 128/32
i1 = 4 amps

= 2R*V1*3R + V1*R*R + R*0*V1 - R*V1*R - V1*0*3R - 2R*V1*R

= 4*8*6 + 8*2*2 + 0 - 2*8*2 - 0 - 4*8*2

= 4*8*6 - 4*8*2

= 192 - 64

= 128

128/32 = 4 amps

i2 = 4 amps

Solving for the numerator

= 2R*2R*V1 + 0*V1*R + V1*0*R - V1*2R*R - 0*0*R - 2R*V1*R
= 4*4*8 + 0 + 0 - 8*4*2 - 0 - 4*8*2
= 128 - 64 - 64
= 0
 i3 = 0 amps

The total current
it = i1 + i2 + i3 = 4+4+0 = 8 amps

Here is a link to an excellent article on simulateous equations and third order determinants.

http://math-equation.net/syst2.html