use Kirchhoff’s laws to solve simple circuit problems

Kirchhoff’s Laws ⚡️

In a circuit, electricity behaves a lot like water flowing through pipes.

Kirchhoff’s Laws give us the rules for how that “water” (current) and the “pressure” (voltage) move around a network of wires.

They are the key to solving any circuit problem, no matter how many batteries, resistors or switches you have.

1️⃣ Kirchhoff’s Current Law (KCL)

KCL says: At any junction (node) the total current flowing in equals the total current flowing out.

Think of a junction as a fork in a water pipe: the water that enters the fork must leave through the other pipes.

Mathematically:

\$\sum I{\text{in}} = \sum I{\text{out}}\$

or, if we assign a sign convention (inward +, outward –):

\$\sum I = 0\$

2️⃣ Kirchhoff’s Voltage Law (KVL)

KVL says: In any closed loop, the sum of all voltage rises equals the sum of all voltage drops.

Imagine walking around a loop of a river: the total uphill climb (voltage rise) you make must equal the total downhill drop (voltage drop) you experience.

Mathematically:

\$\sum V{\text{rise}} = \sum V{\text{drop}}\$

or, with a sign convention (rise +, drop –):

\$\sum V = 0\$

🔌 Example 1 – Simple Series Circuit

Problem: A 9 V battery powers two resistors in series: \$R1 = 3\,\Omega\$ and \$R2 = 6\,\Omega\$. Find the current and the voltage drop across each resistor.

  1. Apply KVL around the loop:

    \$V{\text{battery}} - V{R1} - V{R_2} = 0\$

  2. Use Ohm’s law \$V = IR\$ for each resistor.

    Let \$I\$ be the current (same through series).

    Then \$V{R1} = I R1\$, \$V{R2} = I R2\$.

  3. Substitute into KVL:

    \$9\,\text{V} - I(3\,\Omega) - I(6\,\Omega) = 0\$

    \$9 = I(9) \;\Rightarrow\; I = 1\,\text{A}\$

  4. Find voltage drops:

    \$V{R1} = 1\,\text{A} \times 3\,\Omega = 3\,\text{V}\$

    \$V{R2} = 1\,\text{A} \times 6\,\Omega = 6\,\text{V}\$

Result: Current = 1 A, \$V{R1}=3\$ V, \$V{R2}=6\$ V. The voltage drops add up to the battery voltage (3 V + 6 V = 9 V), confirming KVL.

🔌 Example 2 – Parallel Branches with a Junction

Problem: A 12 V battery supplies a circuit with a 4 Ω resistor (\$R1\$) in series with a junction that splits into two parallel branches: \$R2 = 6\,\Omega\$ and \$R_3 = 12\,\Omega\$. Find the current through each resistor.

ResistorValueCurrent
\$R_1\$4 Ω\$I_1\$
\$R_2\$6 Ω\$I_2\$
\$R_3\$12 Ω\$I_3\$

Solution Steps:

  1. Apply KCL at the junction:

    \$I1 = I2 + I_3\$

  2. Apply KVL around the outer loop (battery → \$R_1\$ → junction → back to battery):

    \$12\,\text{V} - I1(4\,\Omega) - V{\text{junction}} = 0\$

    The voltage at the junction relative to the battery terminal is \$V_{\text{junction}}\$.

  3. For each parallel branch, apply KVL from the junction to the battery:

    \$V{\text{junction}} = I2(6\,\Omega)\$

    \$V{\text{junction}} = I3(12\,\Omega)\$

  4. From the two equations in (3) we get:

    \$I2 = \frac{V{\text{junction}}}{6\,\Omega}\$

    \$I3 = \frac{V{\text{junction}}}{12\,\Omega}\$

  5. Substitute \$I2\$ and \$I3\$ into the KCL equation (1):

    \$I1 = \frac{V{\text{junction}}}{6} + \frac{V{\text{junction}}}{12} = \frac{V{\text{junction}}}{4}\$

  6. Now use the outer loop KVL (2):

    \$12 = I1(4) + V{\text{junction}} = \frac{V{\text{junction}}}{4}(4) + V{\text{junction}} = V{\text{junction}} + V{\text{junction}} = 2V_{\text{junction}}\$

    Hence \$V_{\text{junction}} = 6\,\text{V}\$.

  7. Finally, compute currents:

    \$I_1 = \frac{6}{4} = 1.5\,\text{A}\$

    \$I_2 = \frac{6}{6} = 1.0\,\text{A}\$

    \$I_3 = \frac{6}{12} = 0.5\,\text{A}\$

Check: \$I1 = I2 + I_3\$ → \$1.5 = 1.0 + 0.5\$ ✔️.

The voltage drops: \$V{R1}=I1(4)=6\$ V, \$V{R2}=I2(6)=6\$ V, \$V{R3}=I_3(12)=6\$ V. All add to 12 V, confirming KVL.

📚 Summary of Steps

  • Draw the circuit clearly, label all currents and voltages.

  • Choose a sign convention (inward +, outward – for currents; rise +, drop – for voltages).

  • Write KCL equations at each junction.

  • Write KVL equations for each independent loop.

  • Use Ohm’s law \$V = IR\$ to relate unknowns.

  • Solve the resulting system of linear equations (often with substitution or matrix methods).

  • Check your solution by verifying that all KCL and KVL equations are satisfied.

With practice, Kirchhoff’s Laws become a powerful toolkit that lets you tackle any circuit – from simple LED lights to complex power grids. Happy circuit solving! 🚀