A single loop has an ideal source (positive terminal towards the top-left corner)

PHYSICS · Multi-part · Fermi University Physics · Electronics, Instrumentation & Communication · Circuit foundations and passive networks · Circuit foundations and passive networks

Electronics for Measurement, Ch The resistive network: two laws, one equivalent and the loading error

Problem

A single loop has an ideal 12.0 V12.0\,\mathrm{V} source (positive terminal towards the top-left corner), a 4.0 Ω4.0\,\Omega resistor and an 8.0 Ω8.0\,\Omega resistor, all in series, with a current II defined clockwise leaving the source's positive terminal, passing through the 4.0 Ω4.0\,\Omega resistor, then the 8.0 Ω8.0\,\Omega resistor, and returning to the source's negative terminal. Someone writes Kirchhoff's voltage law going clockwise as −I(4.0 Ω)+I(8.0 Ω)+12.0 V=0-I(4.0\,\Omega) + I(8.0\,\Omega) + 12.0\,\mathrm{V} = 0 and concludes I=−3.0 AI = -3.0\,\mathrm{A}. ((a)) Find the sign error in the equation. ((b)) Write the correct loop equation and find the correct value of II.

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