Question
Instructions: Complete all problems writing the solutions in the space provided. The equations used for each solution should be written down and all work must be shown in order to receive full credit. The primary coil of a transformer which has an 80% efficiency is connected to 110 V ac. The secondary coil is connected across a 2.4 Ohm, 75 W lightbulb. Calculate the current through the primary coil. Calculate the ratio of the number of turns in the primary coil to the number of turns in the secondary coil, NP/NS.
Explanation / Answer
In an ideal transformer, power in = power out, but thistransformer is not ideal, since the primary is 80% efficient. Evenso, the input power can be calculated from output power: -- P(s) = Power in Secondary = 75 watts P(p) = Power in Primary = ?? -- P(p) = 75 * (0.80) P(p) = 60 watts -- And power is found with: P =V*I -- So, solve for I: I = P/V I = 60/110 I = 0.545 amps = 545 mA -- Number of Turns: The curren in the secondary can be found based upon theefficiency of the primary. We know that the transformer is 80%efficient , but the secondary power shouldbe 75 watts, so secondary current should be: P = I^2R -- solve for I: -- I = (75/2.4 = I = 5.6 amps -- And the number of turns is the ratio of current in primary tocurrent in secondary: # turns = I(s)/I(p) # turns = 5.6 / (75/110) # turns = 8.2:1 ----> for every turn in thesecondary, there are 8.2 turns in the primary -- -- Hope this helps.