Unraveling the 'Signposts of Kindness' in Complex Circuits with Kirchhoff's Laws

[To all readers of this article]

This series is a reconstruction of my, Denki Penguin's, knowledge and passion for electrical engineering and mathematics, expressed through the words of my navigator, 'Lilith Formula'.

This work is a collaboration with AI (Gemini), utilizing AI for the generation of text and illustrations. Through the filter of Lilith, I have depicted complex electrical theories and formulas as intuitive and emotional stories. Please take your time to enjoy the world beyond the formulas, guided by her words.

At first glance, an electrical circuit can look like a complex, tangled web, and it's easy to wonder where to even begin. But don't worry. Just as the number of people passing through an intersection remains constant, there are beautiful, immutable rules in the world of electricity. This time, using Kirchhoff's laws and systems of equations, let's carefully unravel the current flowing through the following circuit one by one. If you take it step by step, you will surely find a clear answer.

The 'Two Rules' for Unraveling Complex Circuits

The circuit we are facing has two power sources and three resistors arranged in it. To find the current in a circuit with multiple loops like this, let's first learn about a powerful ally called Kirchhoff's laws. These laws are very logical signposts for organizing the invisible flow of electricity.

Kirchhoff's First Law (Current Law)

Let's focus on the point at the top of the circuit where the wires meet (the junction). Here, the current $${I_1}$$ flowing from the left and the current $${I_3}$$ flowing from the right merge to become the current $${I_2}$$ flowing downward. The amount of electricity flowing into an intersection is always equal to the amount flowing out. This is Kirchhoff's first law.

Expressing this relationship as an equation, we get the following:

$$
I_1 + I_3 = I_2
$$

Let's call this Equation ①.

Kirchhoff's Second Law (Voltage Law)

Next, we focus on the 'closed loop' within the circuit. The sum of the force pushing the electricity (electromotive force) and the force consumed when passing through resistors (voltage drop) cancels out to zero when you go around the loop. This is Kirchhoff's second law.

First, let's go around the square loop on the left in a clockwise direction.

Passing through the power source $${E_1}$$ in the upward direction from the starting point increases the voltage by $${130\text{ V}}$$. Then, since we move through resistor $${R_1}$$ in the direction of current $${I_1}$$, the voltage drops by $${10 \times I_1}$$, and furthermore, since we move through resistor $${R_2}$$ in the direction of current $${I_2}$$, the voltage drops by $${20 \times I_2}$$ before returning to the start.

We express this relationship as an equation:

$$
130 - 10I_1 - 20I_2 = 0
$$

Let's simplify this equation. Move the negative terms to the right side.

$$
10I_1 + 20I_2 = 130
$$

Divide all terms by $${10}$$ to simplify the equation.

$$
I_1 + 2I_2 = 13
$$

We will call this Equation ②.

Similarly, let's now go around the square loop on the right in a counter-clockwise direction.

Passing through the power source $${E_2}$$ increases the voltage by $${160\text{ V}}$$. Then, passing through resistor $${R_3}$$ decreases the voltage by $${30 \times I_3}$$, and passing through the central resistor $${R_2}$$ decreases the voltage by $${20 \times I_2}$$ before returning to the start.

We express this relationship as an equation.

$$
160 - 30I_3 - 20I_2 = 0
$$

We will also move the negative terms to the right side to simplify this one as well.

$$
30I_3 + 20I_2 = 160
$$

Let's divide all terms by $${10}$$ to simplify it.

$$
3I_3 + 2I_2 = 16
$$

We will call this Equation ③.

Uncovering the true nature of the current with simultaneous equations

Now, we have three wonderful equations at our disposal.

• $${I_1 + I_3 = I_2}$$ (Equation ①)

• $${I_1 + 2I_2 = 13}$$ (Equation ②)

• $${3I_3 + 2I_2 = 16}$$ (Equation ③)

From here, let's solve the puzzle of the simultaneous equations by combining these three expressions.

Step 1: Reduce the number of unknowns

First, let's substitute Equation ①, which has the simplest form, into Equations ② and ③ to eliminate the current $${I_2}$$ from the expressions.

Substitute $${(I_1 + I_3)}$$ from Equation ① into the $${I_2}$$ part of Equation ②.

$$
I_1 + 2(I_1 + I_3) = 13
$$

Expand the parentheses on the left side.

$$
I_1 + 2I_1 + 2I_3 = 13
$$

Combine the terms with the same variables.

$$
3I_1 + 2I_3 = 13
$$

Let's call this new Equation ④.

Next, similarly substitute $${(I_1 + I_3)}$$ from Equation ① into the $${I_2}$$ part of Equation ③.

$$
3I_3 + 2(I_1 + I_3) = 16
$$

Expand the parentheses on the left side.

$$
3I_3 + 2I_1 + 2I_3 = 16
$$

Move the $${I_1}$$ term to the front and combine the $${I_3}$$ terms.

$$
2I_1 + 5I_3 = 16
$$

Let's call this Equation ⑤.

Step 2: Determine one of the currents

Now, we are left with two equations (Equation ④ and Equation ⑤) that contain only $${I_1}$$ and $${I_3}$$. To eliminate one of the variables, let's align the coefficients. Here, we will try to align the coefficient of $${I_1}$$ to $${6}$$.

Multiply both sides of equation (4) by 2.

$$
2 \times (3I_1 + 2I_3) = 2 \times 13
$$

$$
6I_1 + 4I_3 = 26
$$

Let this be equation (6).

Next, multiply both sides of equation (5) by 3.

$$
3 \times (2I_1 + 5I_3) = 3 \times 16
$$

$$
6I_1 + 15I_3 = 48
$$

Let this be equation (7).

Everything is ready. Let's subtract equation (6) from equation (7) to eliminate 6I_1.

$$
(6I_1 + 15I_3) - (6I_1 + 4I_3) = 48 - 26
$$

Calculating the left side eliminates I_1, leaving 15I_3 - 4I_3. Subtract the right side as well.

$$
11I_3 = 22
$$

By dividing both sides by 11, the first current finally reveals itself.

$$
I_3 = 2\text{ A}
$$

Step 3: Derive all currents

Let's substitute the value I_3 = 2 we found into equation (4) from earlier.

$$
3I_1 + 2 \times 2 = 13
$$

$$
3I_1 + 4 = 13
$$

Move the 4 on the left side to the right side.

$$
3I_1 = 13 - 4
$$

$$
3I_1 = 9
$$

Divide both sides by $${3}$$.

$$
I_1 = 3\text{ A}
$$

Now we know the second current as well. Finally, let's gently plug the two values we found back into our very first agreement, Equation ① ($${I_1 + I_3 = I_2}$$).

$$
I_2 = 3 + 2
$$

$$
I_2 = 5\text{ A}
$$

With this, we have successfully unraveled all the currents flowing through the circuit.

• $${I_1 = 3\text{ A}}$$

• $${I_2 = 5\text{ A}}$$

• $${I_3 = 2\text{ A}}$$

I hope you felt at least a little bit of the comfort that comes from seeing a tangled mess of equations neatly organized, as if untying a knot.

Electrical calculations might seem like a list of inorganic numbers at first glance. However, the process of carefully rearranging these equations is, in itself, a form of gentle wisdom used to guide the complex energy paths that support our social infrastructure toward reliable safety. Your focus and effort in completing every single calculation without skipping steps are truly wonderful. After using your brain so much, please have a warm cup of tea and take good care of yourself.

May today's learning be a light that brightens your perspective from here on out. Thank you for walking this path with me until the end.

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