📘 Kirchhoff’s Laws – Complete Concept & Detailed Worked Examples
🔹 1. Introduction
Kirchhoff’s Laws are the foundation of Electrical Network Analysis. Almost every advanced theorem (Superposition, Thevenin, Norton, Mesh, Node Analysis) is built on KCL and KVL. In GATE, IES, and PSU exams, at least 2–4 questions directly or indirectly use Kirchhoff’s laws.
🔹 2. Kirchhoff’s Current Law (KCL)
KCL states that the algebraic sum of currents at a node is zero. Sum of Incoming Currents = Sum of Outgoing Currents.
🔸 Physical Meaning
This law is based on conservation of charge. Charge cannot accumulate at a node in steady state.
🔸 Worked Example 1 (Basic)
At a node, 5A and 3A enter. One current 4A leaves. Find remaining current.
Incoming = 5 + 3 = 8A Outgoing = 4 + I 8 = 4 + I I = 4A
🔸 Worked Example 2 (Node Voltage Method)
Node V connected to 20V through 5Ω, and to ground through 10Ω. Find node voltage.
Step 1: Apply KCL
(V − 20)/5 + V/10 = 0
Step 2: Multiply by 10
2(V − 20) + V = 0 2V − 40 + V = 0 3V = 40 V = 13.33V
🔸 Worked Example 3 (Two Nodes)
Node V1 connected to 10V via 2Ω and to V2 via 4Ω. V2 connected to ground via 6Ω. Find V2.
KCL at V1:
(V1 − 10)/2 + (V1 − V2)/4 = 0
KCL at V2:
(V2 − V1)/4 + V2/6 = 0
Solve simultaneous equations → V2 = 4V
🔹 3. Kirchhoff’s Voltage Law (KVL)
KVL states that algebraic sum of voltages in a closed loop is zero.
🔸 Physical Meaning
Based on conservation of energy. Energy supplied = Energy consumed.
🔸 Worked Example 4 (Single Loop)
20V source, 4Ω and 6Ω in series. Find current.
20 − 4I − 6I = 0 20 − 10I = 0 I = 2A
🔸 Worked Example 5 (Two Loop Mesh)
Two loops share 5Ω resistor. Left loop has 10V source.
(2+5)I1 − 5I2 = 10 −5I1 + 5I2 = 0 Solve → I1 = 1A
🔹 4. Common Mistakes in GATE
- Wrong current direction assumption
- Sign errors in KVL
- Not multiplying equation properly
- Ignoring dependent sources
🎯 Final Summary
KCL → Used for Node Analysis KVL → Used for Mesh Analysis Every advanced theorem is built on these two laws.
Strong KCL & KVL = Strong Network Theory Foundation
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