If the voltage drops across the passive elements in the loop are $V_1=1\,\text{V}$, $V_2=2\,\text{V}$, $V_3=1\,\text{V}$, $V_4=3\,\text{V}$, $V_5=2\,\text{V}$, $V_6=1\,\text{V}$, and $V_7=4\,\text{V}$, then the source voltage $V_s$ is
A$10\,\text{V}$
B$12\,\text{V}$
C$14\,\text{V}$
D$16\,\text{V}$
Answer & Solution
Correct answer: C. $14\,\text{V}$
From KVL, $V_1+V_2+V_3+V_4+V_5+V_6+V_7-V_s=0$, so $V_s$ equals the sum of all listed drops. Adding them gives $1+2+1+3+2+1+4=14\,\text{V}$.
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