INTRODUCTION TO ELECTRONICS: Kirchhoff’s Laws ( Part 5 )

We are now ready to complete the Part 3 exercise using Kirchhoff’s Node and Loop Rules:

Due to the presence of nodes at points C and E, differing current ( I ) values will be used to evaluate the voltage ( V ) drops that occur around each loop. There are three unique circuit pathways through which conventional current may travel:

1. The outer perimeter of the circuit consists of the large loop and the upper portion of the smaller loop.

2. The large-loop branch along with the lower portion of the smaller loop.

3. The smaller loop as an isolated unit.

We must first sketch a current that branches outward as it crosses Node C and recombines at Node E. The A, B, and D regions along the circuit mark places where voltage drops will have occurred due to current passing through a circuit element. These are visual clues only, and they may be ignored for convenience. Additionally, the units associated with voltage and resistance will be omitted for the sake of simplicity:

1. 40 – 10I₁ – 20I₁ – 20 – 10I₂ = 0

20 – 30I₁ – 10I₂ = 0

2. 40 – 10I₁ – 20I₁ – 30I₃ = 0

40 – 30I₁ – 30I₃ = 0

3. 20 – 30I₃ + 10I₂ = 0

*** Notice how the small loop was traversed in the counterclockwise direction. This was done to obtain a 20 V rise as the voltage source is crossed. ***

We now have three equations and three unknowns. If the number of unknown current values is reduced to two, we can solve one of the two remaining current values algebraically. The quantity of current passing into and out of Node C is as follows:

I₁ = I₂ + I₃ 

*** The I₂ and I₃ currents combine at Node E to become I₁ once again. This is a mere reversal of the current equation just obtained, and it may be omitted. ***

We must now eliminate a variable via addition or subtraction of two of the loop equations above. Be sure to use two independent equations during this step! Equations 1 and 2 are both inclusive of the 40 V branch element; thus, we must add one of these two equations to equation 3. Let’s first eliminate the I₃ term from equation 3:

20 – 30I₃ + 10I₂ = 0

I₃ = I₁ – I₂ 

20 – ( 30 )( I₁ – I₂ ) + 10I₂ = 0

20 – 30I₁ + 30I₂ + 10I₂ = 0

20 – 30I₁ + 40I₂ = 0

Since equation 1 contains I₁ and I₂ terms, we now add it to equation 3:

     20 – 30I₁ + 40I₂ = 0

+  ( 4 )( 20 – 30I₁ – 10I₂ ) = 0 

And,

    20 – 30I₁ + 40I₂ = 0

+  80 – 120I₁ – 40I₂ = 0 

100 – 150I₁ = 0

Solving for I₁, we have…

I₁ = ( – 100 / – 150 ) = 0.667 ≅ ⅔ A

Now that I₁ has been determined, should we substitute it into equation 1 or 2? Notice that substitution into equation 1 will eliminate the 20 V term and leave us with a false answer. We used equation 1 to derive I₁, so equation 2 is the equation in which we substitute I₁:

40 – 30I₁ – 30I₃ = 0

40 – ( 30 )( ⅔ A ) – 30I₃ = 0

40 – 20 – 30I₃ = 0

20 – 30I₃ = 0

– 30I₃ = – 20

I₃ = ( – 20 / – 30 ) = ≅ ⅔ A

Fortunately, both current values were positive, so no current reversals are necessary; however, there is one surprising conclusion:

I₁ = I₂ + I₃ 

⅔ A = I₂ + ⅔ A

I₂ = ⅔ A – ⅔ A = 0

NO CURRENT TRAVELS ACROSS THE UPPERMOST BRANCH!

As a consequence, the voltage drop going from Node C to E via the upper or lower branch of the smaller loop yields a – 20 V drop; thus, if we begin our travels by crossing the low-to-high energy barrier of the 40 V source, our travel around path A-B-C-E is as follows:

40 V – ( ⅔ A )( 10 Ω ) – ( ⅔ A )( 20 Ω ) – ( ⅔ A )( 30 Ω ) = ?

40 V – 6.7 V – 13 V – 20 V = ?

33 V – 13 V – 20 V = ?

20 V – 20 V = 0 V at point E, which is what we’d expect.