Linearity & superposition

YOU WILL BE ABLE TO

State what makes a circuit linear.
Apply superposition: solve with one source at a time.
Kill sources correctly: voltage sources become shorts, current sources become opens.
Explain why power does NOT superpose.

Every circuit in this course is linear: made of resistors (v = iR is a straight line through the origin) and ideal sources. Linearity buys you a powerful and slightly magical property. In a circuit with several independent sources, any voltage or current equals the sum of the contributions each source would produce acting alone. That is the superposition principle, and it converts one hard problem into several easy ones.

"Acting alone" means every OTHER independent source is turned off — and turning off has a precise meaning. A dead voltage source maintains 0 V: that is a short circuit, a wire. A dead current source forces 0 A: that is an open circuit, a gap. Get these backwards and every answer that follows is wrong, so anchor them to the definitions, not to memory.

x = x^{(1)} + x^{(2)} + \cdots + x^{(N)}

superposition — x is any voltage or current; one term per independent source

The worked example: a voltage source and a current source both feed node v. Superposition solves them one at a time.

WORKED EXAMPLE

Superposition, start to finish

Find v in the figure: 12 V source, 2 mA source, R_1 = R_2 = 4 kΩ.
Voltage source alone (current source → open): R_1 and R_2 form a divider, v′ = 12 × 4/(4+4) = 6 V.
Current source alone (voltage source → short): 2 mA flows into R_1 ∥ R_2 = 2 kΩ, so v″ = 2 mA × 2 kΩ = 4 V.
Total: v = 6 + 4 = 10 V.
Verify by nodal analysis on the full circuit: v/4 + (v − 12)/4 = 2 → 2v − 12 = 8 → v = 10 V. ✓

v = 10 V

WORKED EXAMPLE

When superposition earns its keep

Superposition shines when sources are of mixed types, or when you only need one output and each single-source circuit collapses to a divider.
It is also the reasoning tool behind sentences like "the sensor's signal rides on a 1.65 V offset" — an AC contribution superposed on a DC one, which is exactly how ME-3's amplifiers are analysed.
When all sources are the same type and you need everything, nodal or mesh on the full circuit is usually less work — superposition solves N circuits instead of one.

A tool for insight and mixed-source problems, not a universal shortcut.

PRACTICE

To apply superposition you must turn off all but one source. What does each source type become, and why?

A 10 V source in series with R_1 = 1 kΩ feeds a node; R_2 = 1 kΩ runs from that node to ground; a 4 mA source also enters the node. Use superposition to find the node voltage.

In practice problem 2, compute the power in R_2 from each contribution separately, sum them, and compare with the true power. What do you conclude?

A 6 V source in series with 2 kΩ feeds a node; a 3 mA source enters the same node; R_2 = 2 kΩ runs to ground. Find the current through R_2 by superposition.