In one line
Nodal analysis picks one node as reference (0 V), writes KCL at every other node in terms of unknown node voltages, and solves for those voltages.
Explained simply
Pick the lowest floor of a building as "ground level". Then the height of every other floor is measured from it. Water (current) flows from a higher floor to a lower one through each pipe (resistor), and the flow depends on the height difference. Once you know every floor's height, you know every flow.
Technically, with \( n \) nodes, choose one reference and write \( n-1 \) KCL equations. The current from node \( a \) to node \( b \) through \( R \) is \( (V_a - V_b)/R \). Solving gives all node voltages; branch currents follow from Ohm's law.
How it works
- Identify principal nodes (where three or more branches meet). Choose the reference node, usually the one with the most branches or the bottom line.
- Label the unknown node voltages \( V_1, V_2, \dots \).
- At each node, write the sum of currents leaving through every branch = 0 (a current source pushing current into the node appears on the other side as a known value).
- For a branch with a voltage source \( E \) and resistance \( R \), the leaving current is \( (V_1 - E)/R \) if the + end of \( E \) faces away from the node.
- Solve the equations; then find branch currents.
Nodal analysis is usually quicker when a circuit has fewer nodes than meshes, or when current sources are present.
Key formula
At node 1: \( \sum \dfrac{V_1 - V_k}{R_{1k}} = I_{\text{source into node}} \)
\( V_1 \) = voltage of the node (V) with respect to reference, \( V_k \) = voltage at the other end of each branch (V), \( R_{1k} \) = branch resistance (\( \Omega \)). In conductance form, \( G = 1/R \) in siemens (S).
Quick example
Same circuit as a typical mesh example: 10 V source in series with 2 \( \Omega \) feeds node \( V \); from \( V \) to ground are 4 \( \Omega \) and 2 \( \Omega \).
KCL at \( V \): \( \dfrac{V-10}{2} + \dfrac{V}{4} + \dfrac{V}{2} = 0 \). Multiply by 4: \( 2V - 20 + V + 2V = 0 \Rightarrow V = 4 \) V. Source current \( = (10-4)/2 = 3 \) A, current in 4 \( \Omega \) \( = 1 \) A. Only one equation was needed, compared with two in mesh analysis.
Common mistakes
- Forgetting to choose a reference node, or writing an equation at the reference node too.
- Sign errors with a voltage source in a branch: check which end is +.
- Writing currents entering and leaving with different conventions in the same equation.
How MU asks it
Usually a 7–10 mark numerical: "Using nodal analysis, find the voltage at node A and the current through the 3 \( \Omega \) resistor." Supernode is excluded in the MU syllabus. For full marks: mark the reference node clearly, label node voltages, write each KCL equation with all terms visible, solve, and convert to the asked current with direction.