A near-zero-resistance path across the source terminals. \(R_\text{EQ}\) collapses, so \(I = V/R\) grows until limited only by the wire's own small resistance — enough to melt insulation and start fires.
Both interrupt excessive current. A fuse is a thin wire that melts and must be replaced. A breaker is a spring-loaded switch that can be reset. Breakers cost more up front and less over a service life.
It must be higher than the expected current (or it trips during normal operation) and lower than the wiring's failure point (or the wire fails before the protection does).
\([\,1.1\,I_s,\ 1.5\,I_s\,]\) — between 10% and 50% above the nominal source current. Then choose an available part from the provided table.
That it has not opened — model it as a short circuit (a plain wire) while you analyze the rest of the circuit.
A larger-diameter conductor carries more current before failing, and copper carries more than aluminum of the same dimensions. In practice you consult an AWG chart; in this course you use the rating table instead.
The window is \(\left[\,1.1(2.00),\ 1.5(2.00)\,\right] = \left[\,2.2,\ 3.0\,\right]\ \text{A}\). Both 2.5 A and 3 A land inside it, so take the smaller — the lowest rating in the window opens soonest in a fault. The 2 A part is below the window and would nuisance-trip at the nominal current; 3.5 A and 4 A are above it, so the wiring fails first.
Take the smallest available part above the window. It still protects the circuit; a part below the window would trip during normal operation, which is the worse failure.
\(R_\text{EQ}\) does not depend on \(V_s\), so it stays 512.5 Ω as you turn the source up. The fuse opens once the source current reaches its rating: \(I_s = V_s/R_\text{EQ} \geq 0.015\ \text{A}\), so \(V_s \geq (0.015)(512.5) = 7.69\ \text{V}\). Any source at or above about 7.69 V blows it. Same three steps as sizing a fuse, run backward — and that is the shape Quiz 1 asks.
In series with it, so all of the load current passes through the protective device.