In the loop shown in the figure, which expression matches Kirchhoff's Voltage Law as printed in the notes? 
A$V_1+V_2+V_3+V_4+V_5+V_6+V_7-V_s=0$
B$V_s+V_1+V_2+V_3+V_4+V_5+V_6+V_7=0$
C$V_1+V_2+V_3-V_4-V_5-V_6-V_7=0$
D$V_s=0$
Answer & Solution
Correct answer: A. $V_1+V_2+V_3+V_4+V_5+V_6+V_7-V_s=0$
The figure and printed relation indicate that the source rise balances the sum of the potential drops across the elements. Therefore the loop equation is $V_1+V_2+V_3+V_4+V_5+V_6+V_7-V_s=0$.
Related questions
For a battery of emf ε and internal resistance r driving current I through external resistA galvanometer of resistance G is converted into a voltmeter reading full-scale voltage V A galvanometer of resistance G shows full-scale deflection at current I_g. To convert it iThe relation between current density j, drift velocity v_d, number density n and charge e A wire has a resistance R. It is stretched uniformly so that its length becomes 2L. The neKirchhoff current law at a junction is a statement ofTwo cells of emf ε and internal resistance r each are connected in parallel. The equivalenA potentiometer of length L is used to compare emfs. If the balance lengths for cells of e