Question

An alternating pd of rms values 240v is applied to the terminals of an ideal transformer having 600 turns in the primary and 45 in the secondary. Across the secondary is connected a load of resistance 15ohms. Determine (a) the rms pd over the secondary (b) the rms current through the load (c) the power of the secondary circuit (d) the rms current, taken by the primary

Answer 1

(a) The voltage on the secondary will be

$$V_2=V_1\frac{N_2}{N_1}=240\frac{45}{600}=18\text{ V}.$$

(b) The current can be found with Ohm's law:

$$I_2=\frac{V_2}{R_L}=\frac{18}{15}=1.2\text{ A}.$$

(c) The power:

$$P_2=I_2V_2=1.2\cdot18=21.6\text{ W}.$$

(d) Since the transformer converts the same power to different voltage/current, the power of the primary is 21.6 as well while the current is

$$I_1=\frac{P}{V_1}=\frac{P_2}{V_1}=0.09\text{ A}.$$