Answer
a. 0.77A.
b. 8.2.
Work Step by Step
a. We are told the efficiency of the transformer, which relates the power in the secondary to the power in the primary.
$$P_S=75W=0.88P_P=0.88I_PV_P$$
Calculate the current in the primary.
$$I_P=\frac{P_S}{0.88V_P}=\frac{75W}{0.88(110V)}=0.77A$$
b. Start with equation 21-6.
$$\frac{N_P}{N_S}=\frac{V_P}{V_S}$$
We can relate the secondary voltage to the power and resistance using $P=V^2/R$
$$\frac{N_P}{N_S}=\frac{V_P}{V_S}=\frac{V_P}{\sqrt{P_SR_S}}$$
$$\frac{N_P}{N_S}=\frac{110V}{\sqrt{(75W)(2.4\Omega)}}=8.2$$
