Answer
a) $ I_1 + I_2 = I_3$
b) $v_2I_3+v_3I_2+v_4I_1+v_5I_2=v_1I_3$
c) $(-v_1+v_4+v_2)I_1 + (v_5+v_3+v_2-v_1)I_2=0$
d) Equating the coefficient in from of $I_1$ and the coefficient in parenthesis in front of $I_2$ in part c gives Kirchhoff's Voltage Law. Thus, we see that Kirchhoff's Voltage Law agrees with conservation of energy.
Work Step by Step
a) We use Kirchhoff's junction rule to find: $ I_1 + I_2 = I_3$.
b) We know that the sum of all of the power in the circuit is 0, for energy must be conserved. Thus, we find:
$P_1+P_2 + P_3+P_4+P_5=0$
We plug in the known values for power and simplify to find:
$v_2I_3+v_3I_2+v_4I_1+v_5I_2-v_1I_3 = 0 $
$v_2I_3+v_3I_2+v_4I_1+v_5I_2=v_1I_3$
c) We know that $I_1+I_2=I_3$ from part a. Substituting this into part b gives:
$(-v_1+v_4+v_2)I_1 + (v_5+v_3+v_2-v_1)I_2=0$
d) Equating the coefficient in from of $I_1$ and the coefficient in parenthesis in front of $I_2$ in part c gives Kirchhoff's Voltage Law. After all:
$-v_1+v_4+v_2 = v_5+v_3+v_2-v_1 \\ v_4 = v_5 + v_3$
