One source at a time

Superposition trades one hard problem for several easy ones.

• In a linear circuit the response is the algebraic sum of the responses to each independent source acting alone.
• To silence a source, set its value to zero: a voltage source becomes a short circuit (a plain wire), a current source becomes an open circuit (a gap).
• Dependent sources are never turned off. They are part of the circuit, not an input to it.
• Solve the circuit once per independent source, then add the contributions with their signs.

When it applies

The conditions are the same linearity conditions from the previous lesson.

• Every element must be linear: resistors, capacitors and inductors qualify, diodes and transistors do not.
• Elements must be bilateral, so the magnitude of the current does not depend on the polarity of the source.
• Superposition finds currents, voltage drops and node voltages. It cannot find power.

The power trap

This is the mistake that costs marks in every exam.

• Power is p = i²R, which is quadratic, so (i1 + i2)² ≠ i1² + i2².
• Find the total current first by superposition, then compute the power once from that total.
• Adding the powers from each pass gives a number that is simply wrong, and the cross term 2·i1·i2·R is exactly what you dropped.