Understanding Kirchhoff’s Current Law with Calculations

In order to quantify current behavior, circuit analysis uses Kirchhoffs Current Law (KCL). The connecting point or junction, in an electrical or electronic circuit determines the relationship between currents according to this rule.

At any circuit junction: Σ I_OUT = Σ I_IN

The algebraic total of all currents entering the junction is represented by Σ I_IN.

Σ I_OUT is the algebraic total of all the currents leaving the junction.

To put it another way, KCL says that the total current flowing into and out of a junction must match precisely. The foundation of this idea is the idea of conservation of charge, which states that electrical charge cannot be generated or destroyed within of a circuit.

To demonstrate KCL in operation, we will examine a single junction. This will strengthen our comprehension of how KCL aids in the analysis of current flow in circuits.

KCL Single Junction Analysis

Think of a straightforward junction that has one outgoing current I_T, and two incoming currents, I₁ and I₂.

I_T = I₁ + I₂: The algebraic sum of the currents entering the junction (I₁ and I₂) equals the total current departing (I_T).

Conservation of Charge:

This formula illustrates the idea of charge conservation. Since current cannot be produced or destroyed.. the total current entering and exiting the system must be equal.

For instance:

It goes like this: I₁ = 3 A, I₂ = 2 A

I_T = I₁ + I₂ = 3 A + 2 A = 5 A is the solution.

KCL Is Relevant to All Junctions:

KCL is valid for any number of nodes or junctions. Every current flowing into and out of a junction adds up to zero.

Positive (+): The junction’s currents coming in.
Negative (-): The outgoing current from the junction.

The equivalent in mathematics:

The mathematical total of currents entering or exiting a junction is always zero regardless of direction:

Σ I_OUT = Σ I_IN = 0.

This is the foundation of Kirchhoffs Current Law (KCL) which is a key idea in the study of current flow in circuits.

KCL and Parallel Resistors: Analyzing Current Flow

Let’s examine how KCL works with resistors in parallel circuits, regardless of whether or not the resistor values are equal. Take a look at the circuit schematic above.

There are two separate junctions in this example of a basic parallel resistor:

Junction 1: Located at node B.
Junction 2: situated at node E.

To examine the currents at both junctions, we may apply KCL.

According to KCL the total current flowing in and out of a junction must be equal.

Splitting of Current at Junction B:

When the 12V supplys total current (I_T) reaches point A, it enters junction B.

The current separates into two branches at junction B. One branch travels down via resistor R1 and toward node E.

The remaining current ultimately reaches node E after passing via resistor R2 and node C.

There are two separate junctions in this example of a basic parallel resistor:

Junction 1: Located at node B.
Junction 2: situated at node E.

To examine the currents at both junctions, we may apply KCL.

According to KCL, the total current flowing in and out of a junction must be equal.

Splitting of Current at Junction B:

When the 12V supply’s total current (I_T) reaches point A, it enters junction B.

The current separates into two branches at junction B. One branch travels down via resistor R₁ and toward node E.

The remaining current ultimately reaches node E after passing via resistor R₂ and node C.

Analyzing the current branch B to E through resistor R₁ .

I_B-E = I₁ = V / R₁ = 12 / 4 = 3A

Analyzing the current branch C to D through resistor R₂ .

I_C-D = I₂ = V / R₂ = 12 / 6 = 2A

In our circuit, a single current, IT, enters the junction at node B. I1 and I2, the two currents, exit the intersection.

We calculated the branch currents out of the junction, which are I1 = 3 A and I₂ = 2 A.

By using KCL, the total current coming into the junction (Σ_IN) and the total current going out of the junction (Σ_OUT) must be equal. In this scenario, the total incoming current, I_T, or Σ_IN, should equal the sum of I₁ and I₂.

Verifying with the calculated values:

• Σ_IN (I_T) = I₁ + I₂
• 5 A (expected value) = 3 A + 2 A (calculated values)

This demonstrates that, at junction B, KCL is true since the entire incoming current equals the sum of the exiting currents.

Nodes B and E are the two separate junctions in our circuit. At this second junction, we can confirm KCL since the two branch currents (I1 and I2) coming from node B recombine at node E.

KCL at Junction E:

• Σ_IN (I_T) = Σ_OUT
• Where:
• Σ_IN = Sum of currents entering junction E (represented by the total incoming current I_T)
• Σ_OUT = Sum of currents leaving junction E (I₁ and I₂)

Checking the Equation:

We already knew that I₂ = 2 A and I₁ = 3 A. Using these parameters:

Σ_IN (I_T) = I₁ + I₂ 

5 A (expected value) = 3 A + 2 A (calculated values)

This demonstrates that KCL is also true at junction E. The sum of the outgoing currents (I₁ and I₂) equals the total incoming current (I_T).

Moreover, the starting current (I_T) leaving point A is identical to the value of Σ_IN (I_T) at junction E (5 A). This consistency supports the validity of KCL across the circuit and demonstrates the conservation of charge concept.

KCL application to more complex circuits

For examining current flow in complex circuits, Kirchhoff’s Current Law (KCL) is an effective tool. Remember that KCL stipulates that the total currents exiting a junction (node) must equal the algebraic sum of all currents entering that junction. To put it another way, a node’s net current is zero.

Now let’s look at the circuit below (see picture). To find the currents passing through each branch, we may utilize KCL.

Solving a Kirchhoff’s Current Law Problem

This circuit depicts how current flows across various branches and connections. Let us break it down:

Nodes: There are four separate junctions (connecting points) named A, C, E, and F.

Current Path: At node A, the voltage source’s total current, I_T (110V), separates.

A portion of the I_T flows via resistor R₁.

The remaining current flows through resistor R₂.

Current Recombination: The currents from R₁ and R₂ recombine at node C.

Second Splitting: From node C, the current divides again:

Some current passes via the resistor R3.

The remaining current travels through parallel resistors R₄ and R₅.

Final Recombination: The currents from R₃, R₄, and R₅ all meet at node F.

Before we can calculate the individual currents via each resistor, we need to figure out the overall current of the circuit (I_T). We may do this by applying Ohm’s Law (I = V/R), which states:

I = Current V = Voltage (110V in this example).
R = Total circuit resistance (which we have to determine)

The next step is to calculate the circuit’s total resistance by considering the resistors linked in series and parallel inside the circuit. After we have the total resistance, we can use Ohm’s Law to calculate the total current (I_T).

Circuit Resistance R_AC

• 1 / R_(AC) = 1 / R1 + 1 / R2 = 1 / 2.4 + 1 / 1.7
• 1 / R_(AC) = 1 Ω
• ∴ R_(AC) = 1 Ω

Therefore, we find that the equivalent circuit resistance between nodes A and C is 1 Ohm.

Circuit Resistance R_CF

• 1 / R_(CF) = 1 / R3 + 1 / R4 + 1 / R5
• = 1 / 60 + 1 / 20 + 1 / 30
• = 0.1
• ∴ R_(CF) = 10 Ω

Therefore, we can see that the equivalent circuit resistance between nodes C and F is 10 Ohms, which means that the total circuit current, I_T can be calculated as:

• R_T = R_(AC) + R_(CF)
• = 1 + 10
• = 11 Ω
• I_T = V / R_T
• = 110 / 11
• = 10 Amps

From the above calculations we are able to finalize the equivalent circuit as shown below:

Kirchhoff’s Current Law Equivalent Circuit

From the above calculations we get, V = 110V, R_AC = 1Ω, R_CF = 10Ω and I_T = 10A.

Now that we have the supply current and corresponding parallel resistances, we can use Kirchhoff’s junction rule to calculate and check the individual branch currents.

• V_AC = I_T * R_AC
• = 10 * 1
• 10 Volts
• V_CF = I_T * R_CF
• = 10 * 10
• = 100 Volts
• I₁ = V_AC / R₁
• = 10 / 2.4
• = 4.16 Amps
• I₂ = V_AC / R₂
• = 10 / 1.7
• = 5.88 Amps
• I₃ = V_CF / R3
• = 100 / 60
• = 1.66 Amps
• I₄ = V_CF / R4
• = 100 / 20
• = 5 Amps
• I₅ = V_CF / R5
• = 100 / 30
• = 3.33 Amps

Thus, I₁ = 4.16 A, I₂ = 5.88 A, I₃ = 1.66 A, I₄ = 5 A, and I₅ = 3.33 A.

Kirchhoff’s current law may be verified around the circuit by calculating the currents entering and exiting the junction as follows, using node C as our reference point.

At Node C, Σ I_IN = Σ I_OUT

I_T = I₁ + I₂ = I₃ + I₄ + I₅

∴ 10 = (4.16 + 5.88) = (1.66 + 5 + 3.33)

Since the currents entering the junction are positive and the currents exiting the junction are negative, we can also perform a double check to verify whether Kirchhoff’s Current Law is valid. The algebraic total is I₁ + I₂ – I₃ – I₄ – I₅ = 0, which is 4.16 + 5.88 – 1.66 – 5 – 3.33 = 0.

This means that Kirchhoff’s current law (KCL), which stipulates that the algebraic sum of the currents at a circuit network junction point always equals zero, is verified to be true in this particular case through analysis.

Solving another Kirchhoff’s Current Law Problem

Using just Kirchhoff’s Current Law, determine the currents circulating across the circuit below.

I_T is the total current (powered by the 9V supply voltage) passing through the circuit. Since I₁ equals I_T at point A, resistor R₁ will have an I₁*R voltage drop across it.

Due to the circuit’s two branches, three nodes (B, C, and D), and two separate loops, the I*R voltage drops along the two loops will be as follows:

Loop ABC ⇒ 9 = 4I₁ + 6I₂
Loop ABD ⇒ 9 = 4I₁ + 12I₃

We may thus substitute current I₁ for (I₂ + I₃) in each of the following loop equations to further simplify because Kirchhoff’s current rule indicates that at node B, I₁ = I₂ + I₃.

Kirchhoff’s Loop Equations

• Loop (ABC)—————————————Loop (ABD)
• 9 = 4I1 + 6I2————————————–9 = 4I1 + 12I3
• 9 = 4(I₂ + I3) + 6I₂——————————-9 = 4(I₂ + I₃) + 12I₃
• 9 = 4I₂ + 4I₃ + 6I₂——————————-9 = 4I₂ + 4I₃ + 12I₃
• 9 = 10I₂ + 4I₃————————————9 = 4I₂ + 16I₃

The above calculations provide us with two simultaneous formulas related to the circuit currents, as shown below:

• Eq. No 1 :    9 = 10I₂ + 4I₃
• Eq. No 2 :    9 = 4I₂ + 16I₃

We could reduce the two equations to obtain the values of I2 and I3 by multiplying the first equation (Loop ABC) by 4 and deducting Loop ABD from Loop ABC.

• Eq. No 1 :    9 = 10I₂ + 4I₃ ( x4 )    ⇒   36 = 40I₂ + 16I₃
• Eq. No 2 :    9 = 4I₂ + 16I₃ ( x1)    ⇒   9 = 4I₂ + 16I₃
• Eq. No 1 – Eq. No 2   ⇒   27 = 36I₂ + 0

When we replace I₂ in terms of I₃ we are able to get the value of I₂ as 1.0 Amp.

Nest, by multiplying the first equation (Loop ABC) by 4 and the second equation (Loop ABD) by 10, we can now use the same process to obtain the value of I3. As before, we could reduce both equations to obtain the values of I2 and I3 by deducting Loop ABC from Loop ABD.

• Eq. No 1 :    9 = 10I₂ + 4I₃ ( x4 )    ⇒   36 = 40I₂ + 16I₃
• Eq. No 2 :    9 = 4I₂ + 16I₃ ( x10 )    ⇒   90 = 40I₂ + 160I₃
• Eq. No 2 – Eq. No 1   ⇒   54 = 0 + 144I₃

Hence, when we substitute I₃ for I₂, we obtain the value of I₃ as 0.5 Amps.

Given that I₁ = I₂ + I₃ according to Kirchhoff’s junction rule

The formula for the supply current passing via resistor R₁ is 1.0 + 0.5 = 1.5 Amps.

Because I₁ = I_T = 1.5 Amps, I₂ = 1.0 Amps, and I₃ = 0.5 Amps, we can calculate the I*R voltage drops across the devices and at the different locations (nodes) across the circuit using the above data.

If we could have applied Ohm’s Law alone, we could have solved the second example circuit quickly and easily. However, Kirchhoff’s Current Law allows us to solve more complicated circuits when Ohm’s Law is not sufficient.

References: Kirchhoff’s circuit laws

Current and voltage circuit laws