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If in the shown loop $V_1=1\,\text{V}$, $V_2=2\,\text{V}$, $V_3=3\,\text{V}$, $V_4=4\,\text{V}$, $V_5=5\,\text{V}$, $V_6=6\,\text{V}$, and $V_7=7\,\text{V}$, then the source voltage $V_s$ is 
A$7\,\text{V}$
B$14\,\text{V}$
C$21\,\text{V}$
D$28\,\text{V}$
Answer & Solution
Correct answer: D. $28\,\text{V}$
Using the printed KVL relation, $V_s=V_1+V_2+V_3+V_4+V_5+V_6+V_7$. Therefore $V_s=1+2+3+4+5+6+7=28\,\text{V}$.