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: $R_1 = 3\,\Omega$ and $R_2 = 6\,\Omega$. Find the current and the voltage drop across each resistor.

1. Apply KVL around the loop: $$V_{\text{battery}} - V_{R_1} - V_{R_2} = 0$$
2. Use Ohm’s law $V = IR$ for each resistor. Let $I$ be the current (same through series). Then $V_{R_1} = I R_1$, $V_{R_2} = I R_2$.
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: $$V_{R_1} = 1\,\text{A} \times 3\,\Omega = 3\,\text{V}$$ $$V_{R_2} = 1\,\text{A} \times 6\,\Omega = 6\,\text{V}$$

Result: Current = 1 A, $V_{R_1}=3$ V, $V_{R_2}=6$ 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 ($R_1$) in series with a junction that splits into two parallel branches: $R_2 = 6\,\Omega$ and $R_3 = 12\,\Omega$. Find the current through each resistor.

Solution Steps:

1. Apply KCL at the junction: $$I_1 = I_2 + I_3$$
2. Apply KVL around the outer loop (battery → $R_1$ → junction → back to battery): $$12\,\text{V} - I_1(4\,\Omega) - V_{\text{junction}} = 0$$ 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: $$V_{\text{junction}} = I_2(6\,\Omega)$$ $$V_{\text{junction}} = I_3(12\,\Omega)$$
4. From the two equations in (3) we get: $$I_2 = \frac{V_{\text{junction}}}{6\,\Omega}$$ $$I_3 = \frac{V_{\text{junction}}}{12\,\Omega}$$
5. Substitute $I_2$ and $I_3$ into the KCL equation (1): $$I_1 = \frac{V_{\text{junction}}}{6} + \frac{V_{\text{junction}}}{12} = \frac{V_{\text{junction}}}{4}$$
6. Now use the outer loop KVL (2): $$12 = I_1(4) + V_{\text{junction}} = \frac{V_{\text{junction}}}{4}(4) + V_{\text{junction}} = V_{\text{junction}} + V_{\text{junction}} = 2V_{\text{junction}}$$ 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: $I_1 = I_2 + I_3$ → $1.5 = 1.0 + 0.5$ ✔️. The voltage drops: $V_{R_1}=I_1(4)=6$ V, $V_{R_2}=I_2(6)=6$ V, $V_{R_3}=I_3(12)=6$ 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! 🚀