A transformer with 300 turns on its primary coil and 30 turns on its secondary coil is designed to deliver an average power of 600 W with a peak voltage of 24 V.What is the peak current in the primary coil?
A)2.5 A
B)5.0 A
C)7.5 A
D)10.0 A
E)15.0 A

Question

Quizplus · Accepted Answer

We can use the formula for power in a transformer:
P = (Vp*Ip)/Np = (Vs*Is)/Ns
where P = power, Vp = peak voltage in primary, Ip = peak current in primary, Np = number of turns in primary, Vs = peak voltage in secondary, Is = peak current in secondary, and Ns = number of turns in secondary.

Plugging in the given values, we get:
600 = (24*Ip)/300 = (24*Is)/30
Simplifying, we get:
Ip = 5*Is

We need to find Ip, so we can substitute this into either equation to solve for Is. Let's use the second equation:
24 = 30*Is/600
Simplifying, we get:
Is = 1.2 A

Finally, we can substitute back to find Ip:
Ip = 5*Is = 5*1.2 = 6 A

Therefore, the peak current in the primary coil is 6 A, which corresponds to choice B.

A Transformer with 300 Turns on Its Primary Coil and 30