Answer
See answers.
Work Step by Step
a. The secondary voltage is 12 V, and the primary voltage is 120 V. This is a step-down transformer.
b. Assume perfect efficiency. The power in the primary and in the secondary are both the same, 45 W. Find the current in the secondary by using P=IV.
$$P_S=I_SV_S$$
$$I_S=\frac{P_S}{V_S}=\frac{45W}{12V}=3.8A$$
c. $$P_P=I_PV_P$$
$$I_P=\frac{P_P}{V_P}=\frac{45W}{120V}=0.38A$$
d. Find the resistance of the bulb, using Ohm’s law. The bulb’s voltage is 12 V because it is in the secondary circuit.
$$V_S=I_SR$$
$$R=\frac{V_S}{I_S}=\frac{12V}{3.75A}=3.2\Omega$$
