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In the loop shown, why does Kirchhoff's Voltage Law give an equation of the form $V_1+V_2+V_3+V_4+V_5+V_6+V_7-V_s=0$? ![](https://qallery.app/diagrams/v2_74feee0d9633/img-001.jpeg)

ABecause all resistor voltages are rises and the source is a drop
BBecause the chosen loop direction treats the element voltages as drops and the source as a rise
CBecause Kirchhoff's Voltage Law only applies to resistors
DBecause the source voltage must always be numerically smaller than the sum of the drops
Answer & Solution
Correct answer: B. Because the chosen loop direction treats the element voltages as drops and the source as a rise
In the shown sign convention, traversing the loop counts the listed element voltages as drops, while crossing the source corresponds to a voltage rise. Therefore the algebraic sum is written as total drops minus source rise equals zero: $V_1+V_2+V_3+V_4+V_5+V_6+V_7-V_s=0$.
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