From the loop equation $V_1+V_2+V_3+V_4+V_5+V_6+V_7-V_s=0$, the source voltage is equal to
A$V_s=V_1+V_2+V_3+V_4+V_5+V_6+V_7$
B$V_s=V_1+V_2-V_3$
C$V_s=0$
D$V_s=V_7-V_1-V_2$
Answer & Solution
Correct answer: A. $V_s=V_1+V_2+V_3+V_4+V_5+V_6+V_7$
Rearranging the KVL equation gives $V_s=V_1+V_2+V_3+V_4+V_5+V_6+V_7$. So the source voltage equals the total of all the drops around the loop.
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