Question

An ideal step-down transformer has a primary coil of 320 turns and a secondary coil of 15 turns. Its primary coil is plugged into an outlet with 110 V (AC). If the current in the secondary coil is 20.69 A, what current does the primary coil draw?

Answer 1

Using transformer equations, it's calculated that the primary coil draws approximately 0.9691 A when an ideal step-down transformer with 320 turns on the primary and 15 turns on the secondary is supplied with 110 V and there is a current of 20.69 A in the secondary coil.

Step-by-step explanation:

The student asked how to calculate the current that the primary coil draws in an ideal step-down transformer. To solve this, we can use the transformer equations:

The voltage across the primary coil (Vp) divided by the voltage across the secondary coil (Vs) equals the number of turns in the primary coil (Np) divided by the number of turns in the secondary coil (Ns): Vp/Vs = Np/Ns.

The current in the secondary coil (Is) times the number of turns in the secondary coil (Ns) equals the current in the primary coil (Ip) times the number of turns in the primary coil (Np): Is × Ns = Ip × Np.

Given the transformer has 320 turns in the primary and 15 turns in the secondary, and the voltage supplied to the primary is 110 V, we can calculate the voltage in the secondary (Vs) using the first equation. With the provided secondary current of 20.69 A, we can then find the primary current (Ip) using the second equation.

For example, using the first equation:
Vs = (Ns/Np) × Vp = (15/320) × 110 V = 5.15625 V. Then, using the second equation:
Ip = (Is × Ns) / Np = (20.69 A × 15) / 320 = 0.9691 A.

Therefore, the primary coil draws a current of approximately 0.9691 A.