Answer
a. $108:1$.
b. $P_{av}= 110.5W$.
c. $I_1 =0.921A$.
Work Step by Step
a. $\frac{N_2}{N_1}=\frac{V_2}{V_1}=\frac{13000V}{120V}=108:1$
b. $P_{av}=I_2V_2=(8.50\times10^{-3}A)(13000V)=110.5W$
c. $I_1 =\frac{ P_{av}}{V_1}=\frac{110.5W }{120V}=0.921A$
